Step 1: Setup and Identification of the Integral

We are given the definite integral:

$I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} \, dx$

[By Fundamental Theorem of Calculus]

Step 2: Algebraic Simplification

To facilitate integration, we divide both the numerator and the denominator by $\cos^4 x$. This conversion is standard practice to transform trigonometric powers into functions involving $\tan x$ and $\sec^2 x$.

$I = \int_{0}^{\frac{\pi}{4}} \frac{\frac{\sin x \cos x}{\cos^4 x}}{\frac{\cos^4 x}{\cos^4 x} + \frac{\sin^4 x}{\cos^4 x}} \, dx$

$I = \int_{0}^{\frac{\pi}{4}} \frac{\tan x \sec^2 x}{1 + \tan^4 x} \, dx$

Step 3: Variable Substitution

Let $u = \tan^2 x$. [By Change of Variable Theorem]

Differentiating with respect to $x$:

$\frac{du}{dx} = 2 \tan x \cdot \sec^2 x \implies \frac{1}{2} du = \tan x \sec^2 x \, dx$

Determining new limits:

• When $x = 0$, $u = \tan^2(0) = 0$.
• When $x = \frac{\pi}{4}$, $u = \tan^2(\frac{\pi}{4}) = (1)^2 = 1$.

Step 4: Evaluating the Transformed Integral

Substitute $u$ and the differential $du$ into the integral:

$I = \frac{1}{2} \int_{0}^{1} \frac{du}{1 + u^2}$

[Using the standard integral identity $\int \frac{1}{1+x^2} dx = \tan^{-1} x + C$]

$I = \frac{1}{2} \left[ \tan^{-1} u \right]_{0}^{1}$

$I = \frac{1}{2} (\tan^{-1} 1 - \tan^{-1} 0)$

Step 5: Final Numerical Evaluation

Since $\tan^{-1}(1) = \frac{\pi}{4}$ and $\tan^{-1}(0) = 0$:

$I = \frac{1}{2} \left( \frac{\pi}{4} - 0 \right)$

$I = \frac{\pi}{8}$

Final Solution: \frac{\pi}{8}

Step 1: Algebraic Manipulation of the Integrand

To integrate $I = \int \frac{dx}{\cos(x+a)\cos(x+b)}$, we must express the numerator in a form that allows for the simplification of the trigonometric products in the denominator. We utilize the constant $\sin(b-a)$ as a scaling factor [By the property of linear independence of trigonometric functions].

$I = \frac{1}{\sin(b-a)} \int \frac{\sin(b-a)}{\cos(x+a)\cos(x+b)} dx$

Step 2: Trigonometric Substitution

We rewrite the numerator $(b-a)$ as $(x+b) - (x+a)$ to match the arguments present in the denominator:

$I = \frac{1}{\sin(b-a)} \int \frac{\sin((x+b) - (x+a))}{\cos(x+a)\cos(x+b)} dx$

Step 3: Expansion via Sine Addition Formula

Applying the trigonometric identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$ [Per the Compound Angle Formula], where $A = (x+b)$ and $B = (x+a)$:

$I = \frac{1}{\sin(b-a)} \int \frac{\sin(x+b)\cos(x+a) - \cos(x+b)\sin(x+a)}{\cos(x+a)\cos(x+b)} dx$

Step 4: Separating the Integral

Distribute the denominator across the terms in the numerator:

$I = \frac{1}{\sin(b-a)} \left[ \int \frac{\sin(x+b)\cos(x+a)}{\cos(x+a)\cos(x+b)} dx - \int \frac{\cos(x+b)\sin(x+a)}{\cos(x+a)\cos(x+b)} dx \right]$

$I = \frac{1}{\sin(b-a)} \left[ \int \frac{\sin(x+b)}{\cos(x+b)} dx - \int \frac{\sin(x+a)}{\cos(x+a)} dx \right]$

$I = \frac{1}{\sin(b-a)} \left[ \int \tan(x+b) dx - \int \tan(x+a) dx \right]$

Step 5: Final Integration

Applying the standard integral result $\int \tan(u) du = \ln|\sec(u)| + C$ or $-\ln|\cos(u)| + C$ [Per the Fundamental Theorem of Calculus]:

$I = \frac{1}{\sin(b-a)} \left[ \ln|\sec(x+b)| - \ln|\sec(x+a)| \right] + C$

Using the logarithmic property $\ln(m) - \ln(n) = \ln(\frac{m}{n})$:

$I = \frac{1}{\sin(b-a)} \ln \left| \frac{\sec(x+b)}{\sec(x+a)} \right| + C$

Substituting $\sec(\theta) = \frac{1}{\cos(\theta)}$:

$I = \frac{1}{\sin(b-a)} \ln \left| \frac{\cos(x+a)}{\cos(x+b)} \right| + C$

Final Solution: $\frac{1}{\sin(b-a)} \ln \left| \frac{\cos(x+a)}{\cos(x+b)} \right| + C$

Step 1: Setting up the Integral

We are tasked with evaluating the indefinite integral of the function:

$I = \int \frac{1}{\sqrt{x+a} + \sqrt{x+b}} \, dx$

Step 2: Rationalizing the Integrand

To simplify the integrand, multiply the numerator and the denominator by the conjugate of the denominator, which is $(\sqrt{x+a} - \sqrt{x+b})$. [Per the Algebraic Identity $(u+v)(u-v) = u^2 - v^2$]:

$I = \int \frac{1(\sqrt{x+a} - \sqrt{x+b})}{(\sqrt{x+a} + \sqrt{x+b})(\sqrt{x+a} - \sqrt{x+b})} \, dx$

$I = \int \frac{\sqrt{x+a} - \sqrt{x+b}}{(x+a) - (x+b)} \, dx$

$I = \int \frac{\sqrt{x+a} - \sqrt{x+b}}{a-b} \, dx$

Step 3: Distributing the Integral

Since $(a-b)$ is a constant with respect to $x$, it can be factored out of the integral:

$I = \frac{1}{a-b} \left[ \int (x+a)^{1/2} \, dx - \int (x+b)^{1/2} \, dx \right]$

Step 4: Applying Integration Power Rule

Using the power rule for integration $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$, where $n = 1/2$:

$I = \frac{1}{a-b} \left[ \frac{(x+a)^{3/2}}{3/2} - \frac{(x+b)^{3/2}}{3/2} \right] + C$

$I = \frac{1}{a-b} \cdot \frac{2}{3} \left[ (x+a)^{3/2} - (x+b)^{3/2} \right] + C$

Step 5: Geometric Representation of the Functions

The following SVG represents the curves $y_1 = \frac{1}{\sqrt{x+a}}$ and $y_2 = \frac{1}{\sqrt{x+b}}$.

Step 6: Final Algebraic Consolidation

The constant $C$ is the arbitrary constant of integration. The result is finalized by simplifying the scalar factor.

Final Solution: $\frac{2}{3(a-b)} \left[ (x+a)^{3/2} - (x+b)^{3/2} \right] + C$

Step 1: Partial Fraction Decomposition Setup

Given the rational function $f(x) = \frac{x^2+x+1}{(x+1)^2(x+2)}$, since the degree of the numerator (2) is less than the degree of the denominator (3), the function is a proper fraction. Per the theory of Partial Fractions, we decompose it as follows:

$\frac{x^2+x+1}{(x+1)^2(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2}$ [By Partial Fraction Decomposition Theorem]

Step 2: Determining Constants A, B, and C

Equating the numerators by multiplying both sides by $(x+1)^2(x+2)$:

$x^2+x+1 = A(x+1)(x+2) + B(x+2) + C(x+1)^2$

• To find $B$, set $x = -1$: $(-1)^2 + (-1) + 1 = B(-1+2) \implies 1 = B(1) \implies \mathbf{B = 1}$
• To find $C$, set $x = -2$: $(-2)^2 + (-2) + 1 = C(-2+1)^2 \implies 4 - 2 + 1 = C(1) \implies \mathbf{C = 3}$
• To find $A$, equate the coefficients of $x^2$: $1 = A + C \implies 1 = A + 3 \implies \mathbf{A = -2}$

Step 3: Integration of Partial Fractions

The integral becomes:

$\int \frac{x^2+x+1}{(x+1)^2(x+2)} dx = \int \left( \frac{-2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2} \right) dx$

Applying the linearity property of integration [$\int (af+bg) dx = a\int f dx + b\int g dx$]:

$I = -2 \int \frac{1}{x+1} dx + \int (x+1)^{-2} dx + 3 \int \frac{1}{x+2} dx$

Step 4: Final Evaluation

Executing the standard integration rules [$\int \frac{1}{x} dx = \ln|x| + C$ and $\int x^n dx = \frac{x^{n+1}}{n+1}$]:

$I = -2 \ln|x+1| + \frac{(x+1)^{-1}}{-1} + 3 \ln|x+2| + C$

$I = 3 \ln|x+2| - 2 \ln|x+1| - \frac{1}{x+1} + C$

Final Solution: $3 \ln|x+2| - 2 \ln|x+1| - \frac{1}{x+1} + C$

Step 1: Algebraic Transformation of the Integrand

Given the integral: $I = \int \frac{dx}{e^x + e^{-x}}$

To simplify the expression, we multiply both the numerator and the denominator by $e^x$:

$I = \int \frac{e^x}{e^x(e^x + e^{-x})} dx$

[By the property of algebraic manipulation: $\frac{a}{b} = \frac{ac}{bc}$, provided $c \neq 0$]

$I = \int \frac{e^x}{e^{2x} + e^x \cdot e^{-x}} dx$

$I = \int \frac{e^x}{e^{2x} + 1} dx$

[Using the law of exponents: $e^x \cdot e^{-x} = e^{x-x} = e^0 = 1$]

Step 2: Substitution Method

Let $u = e^x$. Then, differentiating with respect to $x$ yields:

$\frac{du}{dx} = e^x \implies du = e^x dx$

[By the Chain Rule of differentiation]

Step 3: Transforming the Integral

Substituting $u$ and $du$ into the integral expression:

$I = \int \frac{du}{u^2 + 1}$

Step 4: Integration using Standard Formulas

The integral of the form $\int \frac{1}{u^2 + a^2} du$ is given by the standard result $\frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$. Here, $a = 1$.

[Per the standard integral table for inverse trigonometric functions]

$I = \tan^{-1}(u) + C$

Step 5: Re-substitution and Final Result

Replace $u$ with $e^x$ to return to the original variable:

$I = \tan^{-1}(e^x) + C$

Visual Representation of the Function Path

Final Solution: \tan^{-1}(e^x) + C

Step 1: Substitution for Simplification

Given the integral: $I = \int \tan^{-1} \sqrt{\frac{1-x}{1+x}} \, dx$

To eliminate the radical, we employ a trigonometric substitution. Let $x = \cos \theta$. Then $dx = -\sin \theta \, d\theta$. [By the substitution method of integration].

The term under the radical becomes:

$\sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \sqrt{\frac{2 \sin^2 (\theta/2)}{2 \cos^2 (\theta/2)}} = \sqrt{\tan^2 (\theta/2)} = \tan (\theta/2)$

[Using trigonometric identities: $1-\cos \theta = 2 \sin^2 (\theta/2)$ and $1+\cos \theta = 2 \cos^2 (\theta/2)$].

Step 2: Substituting into the Integral

Substituting these into the original expression:

$I = \int \tan^{-1} (\tan (\theta/2)) \cdot (-\sin \theta) \, d\theta$ $I = -\int \frac{\theta}{2} \cdot \sin \theta \, d\theta$ $I = -\frac{1}{2} \int \theta \sin \theta \, d\theta$

Step 3: Integration by Parts

Apply the Integration by Parts formula: $\int u \, dv = uv - \int v \, du$. [Per the product rule for integration].

Let $u = \theta$ and $dv = \sin \theta \, d\theta$. Then $du = d\theta$ and $v = -\cos \theta$.

$I = -\frac{1}{2} \left[ \theta (-\cos \theta) - \int (-\cos \theta) \, d\theta \right]$ $I = -\frac{1}{2} \left[ -\theta \cos \theta + \sin \theta \right] + C$ $I = \frac{1}{2} \theta \cos \theta - \frac{1}{2} \sin \theta + C$

Step 4: Back-substitution

Recall $x = \cos \theta$, therefore $\theta = \cos^{-1} x$ and $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2}$.

$I = \frac{1}{2} \cos^{-1} x \cdot x - \frac{1}{2} \sqrt{1-x^2} + C$ $I = \frac{1}{2} x \cos^{-1} x - \frac{1}{2} \sqrt{1-x^2} + C$

Step 5: Geometric Representation of the Substitution

Final Solution: $\frac{1}{2} \left( x \cos^{-1} x - \sqrt{1-x^2} \right) + C$

Step 1: Establishing the Initial Integral

We are required to evaluate the indefinite integral:

$I = \int \frac{1}{x\sqrt{ax - x^2}} \, dx$

Following the hint provided, we perform the substitution $x = \frac{a}{t}$.

Step 2: Differentiation and Substitution

Differentiating $x = \frac{a}{t}$ with respect to $t$:

$\frac{dx}{dt} = -\frac{a}{t^2} \implies dx = -\frac{a}{t^2} \, dt$

Now, substitute $x$ and $dx$ into the integral:

$I = \int \frac{1}{(\frac{a}{t})\sqrt{a(\frac{a}{t}) - (\frac{a}{t})^2}} \cdot \left( -\frac{a}{t^2} \right) \, dt$

Step 3: Simplifying the Algebraic Expression

Simplify the term inside the square root:

$I = \int \frac{1}{\frac{a}{t}\sqrt{\frac{a^2}{t} - \frac{a^2}{t^2}}} \cdot \left( -\frac{a}{t^2} \right) \, dt$

$I = \int \frac{1}{\frac{a}{t} \cdot \frac{a}{t} \sqrt{t - 1}} \cdot \left( -\frac{a}{t^2} \right) \, dt$ [By extracting $a/t$ from the root]

$I = \int \frac{1}{\frac{a^2}{t^2} \sqrt{t-1}} \cdot \left( -\frac{a}{t^2} \right) \, dt$

Canceling $\frac{1}{t^2}$ and simplifying the constants:

$I = -\int \frac{a}{a^2 \sqrt{t-1}} \, dt = -\frac{1}{a} \int (t-1)^{-1/2} \, dt$

Step 4: Integration

Apply the power rule for integration $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$:

$I = -\frac{1}{a} \cdot \frac{(t-1)^{1/2}}{1/2} + C$

$I = -\frac{2}{a} \sqrt{t-1} + C$

Step 5: Back-Substitution

Since $x = \frac{a}{t}$, we have $t = \frac{a}{x}$. Substitute this back into the expression:

$I = -\frac{2}{a} \sqrt{\frac{a}{x} - 1} + C$

$I = -\frac{2}{a} \sqrt{\frac{a-x}{x}} + C$

Final Solution: $-\frac{2}{a} \sqrt{\frac{a-x}{x}} + C$

Step 1: Algebraic Simplification of the Integrand

Given the integrand $I = \int \frac{\sin^8 x - \cos^8 x}{1 - 2\sin^2 x \cos^2 x} \, dx$, we first simplify the numerator using the difference of squares identity $a^2 - b^2 = (a-b)(a+b)$.

Let $a = \sin^4 x$ and $b = \cos^4 x$:

$\sin^8 x - \cos^8 x = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)$

Further factorizing $(\sin^4 x - \cos^4 x)$:

$(\sin^4 x - \cos^4 x) = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)$

[By the Pythagorean identity: $\sin^2 x + \cos^2 x = 1$]

Therefore, $(\sin^4 x - \cos^4 x) = (\sin^2 x - \cos^2 x)(1) = -(\cos^2 x - \sin^2 x) = -\cos(2x)$.

Step 2: Processing the Remaining Terms

For the $(\sin^4 x + \cos^4 x)$ term, we complete the square:

$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$.

[By the identity $(a+b)^2 = a^2 + b^2 + 2ab$]

Step 3: Substituting back into the Integral

Substitute these expressions into the original integral:

$I = \int \frac{(\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)}{1 - 2\sin^2 x \cos^2 x} \, dx$

$I = \int \frac{-(\cos^2 x - \sin^2 x)(1 - 2\sin^2 x \cos^2 x)}{1 - 2\sin^2 x \cos^2 x} \, dx$

Canceling the common term $(1 - 2\sin^2 x \cos^2 x)$, we obtain:

$I = \int -(\cos^2 x - \sin^2 x) \, dx$

[By the Double Angle Identity: $\cos(2x) = \cos^2 x - \sin^2 x$]

Step 4: Final Integration

$I = \int -\cos(2x) \, dx$

Integrating with respect to $x$ using the rule $\int \cos(ax) \, dx = \frac{\sin(ax)}{a} + C$:

$I = -\frac{\sin(2x)}{2} + C$

Final Solution: -\frac{1}{2}\sin(2x) + C

Step 1: Algebraic Manipulation of the Integrand

To evaluate the integral $I = \int \frac{1}{x^2(x^4+1)^{\frac{3}{4}}} dx$, we manipulate the term inside the parenthesis to extract $x^4$.

$I = \int \frac{1}{x^2 \left[ x^4 \left( 1 + \frac{1}{x^4} \right) \right]^{\frac{3}{4}}} dx$

[By algebraic factorization property $ab^n = a^n b^n$]

$I = \int \frac{1}{x^2 \cdot (x^4)^{\frac{3}{4}} \cdot (1 + x^{-4})^{\frac{3}{4}}} dx$

$I = \int \frac{1}{x^2 \cdot x^3 \cdot (1 + x^{-4})^{\frac{3}{4}}} dx$

$I = \int \frac{1}{x^5 (1 + x^{-4})^{\frac{3}{4}}} dx$

Step 2: Substitution Method

Let $u = 1 + x^{-4}$.

Differentiating with respect to $x$:

$\frac{du}{dx} = 0 + (-4)x^{-5} = -\frac{4}{x^5}$

$dx = -\frac{x^5}{4} du$

[By Chain Rule and Power Rule for differentiation]

Step 3: Transforming the Integral

Substitute $u$ and $dx$ into the integral expression:

$I = \int \frac{1}{x^5 \cdot u^{\frac{3}{4}}} \cdot \left( -\frac{x^5}{4} \right) du$

$I = -\frac{1}{4} \int u^{-\frac{3}{4}} du$

Step 4: Integration and Back-Substitution

Applying the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C$:

$I = -\frac{1}{4} \left[ \frac{u^{-\frac{3}{4} + 1}}{-\frac{3}{4} + 1} \right] + C$

$I = -\frac{1}{4} \left[ \frac{u^{\frac{1}{4}}}{\frac{1}{4}} \right] + C$

$I = -u^{\frac{1}{4}} + C$

Re-substituting $u = 1 + x^{-4} = 1 + \frac{1}{x^4} = \frac{x^4+1}{x^4}$:

$I = -\left( \frac{x^4+1}{x^4} \right)^{\frac{1}{4}} + C$

$I = -\frac{(x^4+1)^{\frac{1}{4}}}{(x^4)^{\frac{1}{4}}} + C$

$I = -\frac{(x^4+1)^{\frac{1}{4}}}{x} + C$

Final Solution: $-\frac{(x^4+1)^{\frac{1}{4}}}{x} + C$

Step 1: Simplifying the Integrand

The given integral is $I = \int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}$. To resolve the radical difference in the denominator, we employ the method of rationalization by multiplying the numerator and denominator by the conjugate of the expression [Per the algebraic identity $(a-b)(a+b) = a^2 - b^2$].

$I = \int_{0}^{1} \frac{1}{\sqrt{1+x}-\sqrt{x}} \times \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}} \, dx$

$I = \int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{(1+x) - x} \, dx$

$I = \int_{0}^{1} (\sqrt{1+x} + \sqrt{x}) \, dx$

Step 2: Applying the Power Rule for Integration

We apply the Fundamental Theorem of Calculus. Recall that $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ [Per the Power Rule for Integration].

$I = \int_{0}^{1} (1+x)^{1/2} \, dx + \int_{0}^{1} x^{1/2} \, dx$

$I = \left[ \frac{(1+x)^{3/2}}{3/2} \right]_{0}^{1} + \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1}$

$I = \frac{2}{3} \left[ (1+x)^{3/2} \right]_{0}^{1} + \frac{2}{3} \left[ x^{3/2} \right]_{0}^{1}$

Step 3: Evaluating the Definite Limits

Substituting the upper and lower limits $[F(b) - F(a)]$:

$I = \frac{2}{3} \left( (1+1)^{3/2} - (1+0)^{3/2} \right) + \frac{2}{3} \left( (1)^{3/2} - (0)^{3/2} \right)$

$I = \frac{2}{3} \left( 2^{3/2} - 1 \right) + \frac{2}{3} \left( 1 - 0 \right)$

$I = \frac{2}{3} (2\sqrt{2} - 1) + \frac{2}{3} (1)$

$I = \frac{4\sqrt{2}}{3} - \frac{2}{3} + \frac{2}{3}$

$I = \frac{4\sqrt{2}}{3}$

Visual Representation of the Area under the curve $f(x) = \frac{1}{\sqrt{1+x}-\sqrt{x}}$

Final Solution: $\frac{4\sqrt{2}}{3}$

Step 1: Setting up the Integral

To evaluate the definite integral $I = \int_{0}^{1} \sin^{-1} x \, dx$, we employ the technique of Integration by Parts. The formula for integration by parts is defined as:

$\int u \, dv = uv - \int v \, du$ [By the Product Rule of Differentiation in reverse]

Let $u = \sin^{-1} x$ and $dv = dx$.

Step 2: Differentiating and Integrating the Components

We derive the components as follows:

• $du = \frac{d}{dx}(\sin^{-1} x) \, dx = \frac{1}{\sqrt{1 - x^2}} \, dx$
• $v = \int 1 \, dx = x$

Step 3: Applying Integration by Parts

Substituting these into the formula:

$I = [x \sin^{-1} x]_{0}^{1} - \int_{0}^{1} x \cdot \frac{1}{\sqrt{1 - x^2}} \, dx$

$I = [x \sin^{-1} x]_{0}^{1} - \int_{0}^{1} \frac{x}{\sqrt{1 - x^2}} \, dx$

Step 4: Evaluating the First Term and Substitution for the Second Term

Evaluate the boundary term: $[x \sin^{-1} x]_{0}^{1} = (1 \cdot \sin^{-1}(1)) - (0 \cdot \sin^{-1}(0)) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

For the integral $\int_{0}^{1} \frac{x}{\sqrt{1 - x^2}} \, dx$, let $t = 1 - x^2$. Then $dt = -2x \, dx$, which implies $x \, dx = -\frac{1}{2} dt$.

Adjusting the limits: When $x=0, t=1$. When $x=1, t=0$.

The integral becomes: $-\frac{1}{2} \int_{1}^{0} t^{-1/2} \, dt = \frac{1}{2} \int_{0}^{1} t^{-1/2} \, dt$.

Step 5: Final Computation

Integrating $t^{-1/2}$ with respect to $t$ yields $2t^{1/2}$:

$\frac{1}{2} [2 \sqrt{t}]_{0}^{1} = [\sqrt{t}]_{0}^{1} = 1 - 0 = 1$.

Substituting this result back into the expression for $I$:

$I = \frac{\pi}{2} - 1$.

Final Solution: $\int_{0}^{1} \sin^{-1} x \, dx = \frac{\pi}{2} - 1$

Step 1: Formal Expression Analysis

The integral to be evaluated is: $$I = \int \frac{\cos x}{\sqrt{4 - \sin^2 x}} \, dx$$ [By the Principle of Substitution, we observe that the numerator $\cos x \, dx$ is the derivative of $\sin x$, suggesting an immediate change of variables.]

Step 2: Substitution and Differential Transformation

Let $u = \sin x$. Then, differentiating both sides with respect to $x$: $$\frac{du}{dx} = \cos x \implies du = \cos x \, dx$$ [Per the Chain Rule of calculus, this substitution effectively simplifies the integrand from a trigonometric form to an algebraic form.]

Step 3: Reformulation of the Integral

Substituting $u$ and $du$ into the original integral: $$I = \int \frac{du}{\sqrt{4 - u^2}}$$ We recognize this expression in the standard form: $$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$$ [By the Table of Standard Integrals for inverse trigonometric functions.]

Step 4: Calculation and Integration

Here, $a^2 = 4$, which implies $a = 2$. Applying the formula: $$I = \sin^{-1}\left(\frac{u}{2}\right) + C$$ Substituting back the original variable $u = \sin x$: $$I = \sin^{-1}\left(\frac{\sin x}{2}\right) + C$$

Geometric Visualization of the Inverse Relationship

Final Solution: $\sin^{-1}\left(\frac{\sin x}{2}\right) + C$

Step 1: Analyzing the Integral Structure

We are required to evaluate the indefinite integral:

$I = \int \frac{\sin x}{\sin(x-a)} \, dx$

The integrand features a denominator term $\sin(x-a)$. To simplify the numerator, we apply the technique of substitution by algebraic manipulation, forcing the argument of the numerator to match the denominator [By Algebraic Substitution Principle].

Step 2: Decomposition of the Numerator

Rewrite $x$ in the numerator as $(x - a) + a$. This allows for the application of the trigonometric addition formula [Per $\sin(A+B) = \sin A \cos B + \cos A \sin B$]:

$\sin x = \sin((x - a) + a)$

$\sin x = \sin(x - a) \cos a + \cos(x - a) \sin a$

Step 3: Substitution into the Integral

Substitute this expansion back into the original integral expression:

$I = \int \frac{\sin(x - a) \cos a + \cos(x - a) \sin a}{\sin(x - a)} \, dx$

Distribute the denominator across the two terms in the numerator:

$I = \int \left( \frac{\sin(x - a) \cos a}{\sin(x - a)} + \frac{\cos(x - a) \sin a}{\sin(x - a)} \right) \, dx$

Step 4: Integration and Simplification

Simplify the expression using basic trigonometric identities [Per $\frac{\cos \theta}{\sin \theta} = \cot \theta$]:

$I = \int (\cos a + \sin a \cdot \cot(x - a)) \, dx$

Since $a$ is a constant, $\cos a$ and $\sin a$ are treated as constant coefficients during the integration process with respect to $x$:

$I = \int \cos a \, dx + \int \sin a \cdot \cot(x - a) \, dx$

$I = x \cos a + \sin a \int \cot(x - a) \, dx$

Step 5: Final Integration

Apply the standard integration formula for the cotangent function [Per $\int \cot u \, du = \ln |\sin u| + C$]:

Let $u = x - a$, then $du = dx$.

$\int \cot(x - a) \, dx = \ln |\sin(x - a)|$

Combining all parts:

$I = x \cos a + \sin a \cdot \ln |\sin(x - a)| + C$

Visual Representation of the Integration Path

Final Solution: $x \cos a + \sin a \cdot \ln |\sin(x - a)| + C$

Step 1: Formal Definition of the Integral

To evaluate the integral $I = \int f'(ax+b)[f(ax+b)]^n \, dx$, we utilize the method of substitution, which is predicated on the Chain Rule of differentiation [Fundamental Theorem of Calculus].

Step 2: Applying the Method of Substitution

Let $u = f(ax+b)$.

To determine the differential $du$, we differentiate $u$ with respect to $x$ using the Chain Rule:

$\frac{du}{dx} = \frac{d}{dx} [f(ax+b)] = f'(ax+b) \cdot \frac{d}{dx}(ax+b)$ [Chain Rule for Differentiation]

Since $\frac{d}{dx}(ax+b) = a$, we have:

$\frac{du}{dx} = a \cdot f'(ax+b)$

Rearranging the terms to solve for $f'(ax+b) \, dx$:

$f'(ax+b) \, dx = \frac{1}{a} \, du$

Step 3: Transforming the Integral

Substitute the variables $u$ and $du$ into the original integral expression:

$I = \int [f(ax+b)]^n \cdot f'(ax+b) \, dx$

$I = \int u^n \cdot \left( \frac{1}{a} \right) du$

$I = \frac{1}{a} \int u^n \, du$

Step 4: Evaluating the Power Integral

Applying the standard power rule for integration $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, provided $n \neq -1$ [Power Rule of Integration]:

$I = \frac{1}{a} \cdot \left( \frac{u^{n+1}}{n+1} \right) + C$

$I = \frac{u^{n+1}}{a(n+1)} + C$

Step 5: Back-Substitution

Replace $u$ with the original function $f(ax+b)$:

$I = \frac{[f(ax+b)]^{n+1}}{a(n+1)} + C$

Visual Representation of Substitution Logic

Final Solution: $\frac{[f(ax+b)]^{n+1}}{a(n+1)} + C$, where $n \neq -1$

Step 1: Decomposition via Partial Fractions

To evaluate the integral $I = \int_{1}^{3} \frac{dx}{x^2(x+1)}$, we express the integrand as a sum of partial fractions [Per the Method of Partial Fractions]:

$\frac{1}{x^2(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$

Multiplying both sides by $x^2(x+1)$, we obtain the identity:

$1 = Ax(x+1) + B(x+1) + Cx^2$

$1 = A(x^2 + x) + Bx + B + Cx^2$

$1 = (A+C)x^2 + (A+B)x + B$

Step 2: Solving for Coefficients

Equating coefficients of like powers of $x$ on both sides:

• Constant term: $B = 1$
• Coefficient of $x$: $A + B = 0 \implies A + 1 = 0 \implies A = -1$
• Coefficient of $x^2$: $A + C = 0 \implies -1 + C = 0 \implies C = 1$

Thus, the integrand is: $\frac{1}{x^2(x+1)} = -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1}$

Step 3: Integration

We now evaluate the definite integral using the Fundamental Theorem of Calculus:

$I = \int_{1}^{3} \left( -\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x+1} \right) dx$

$I = \left[ -\log|x| - \frac{1}{x} + \log|x+1| \right]_{1}^{3}$

$I = \left[ \log\left| \frac{x+1}{x} \right| - \frac{1}{x} \right]_{1}^{3}$

Step 4: Applying Limits

Substitute the upper and lower limits into the antiderivative:

$I = \left( \log\left| \frac{3+1}{3} \right| - \frac{1}{3} \right) - \left( \log\left| \frac{1+1}{1} \right| - \frac{1}{1} \right)$

$I = \left( \log\frac{4}{3} - \frac{1}{3} \right) - \left( \log 2 - 1 \right)$

$I = \log\frac{4}{3} - \log 2 - \frac{1}{3} + 1$

$I = \log\left( \frac{4/3}{2} \right) + \frac{2}{3}$

$I = \log\left( \frac{2}{3} \right) + \frac{2}{3}$

Final Solution: \frac{2}{3} + \log \frac{2}{3}

Step 1: Analyzing the Integrand and Critical Points

The integral to evaluate is $I = \int_{1}^{4} (|x-1| + |x-2| + |x-3|) \, dx$. The absolute value functions $f(x) = |x-a|$ change their definition at the critical point $x = a$. Thus, we must partition the interval of integration $[1, 4]$ based on the points $x=1, x=2,$ and $x=3$.

The defined intervals are: $[1, 2]$, $[2, 3]$, and $[3, 4]$.

Step 2: Defining the Piecewise Function

We evaluate the behavior of each modulus term within the sub-intervals:

Interval	$|x-1|$	$|x-2|$	$|x-3|$	Sum $f(x)$
$1 \le x < 2$	$x-1$	$-(x-2)$	$-(x-3)$	$x - 1 - x + 2 - x + 3 = -x + 4$
$2 \le x < 3$	$x-1$	$x-2$	$-(x-3)$	$x - 1 + x - 2 - x + 3 = x$
$3 \le x \le 4$	$x-1$	$x-2$	$x-3$	$x - 1 + x - 2 + x - 3 = 3x - 6$

Step 3: Applying the Additive Property of Definite Integrals

[By the property $\int_{a}^{c} f(x) dx = \int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx$], we partition the integral $I$:

$I = \int_{1}^{2} (-x+4) dx + \int_{2}^{3} (x) dx + \int_{3}^{4} (3x-6) dx$

Step 4: Evaluating the Definite Integrals

Part A: $\int_{1}^{2} (-x+4) dx = \left[ -\frac{x^2}{2} + 4x \right]_1^2 = (-2+8) - (-\frac{1}{2}+4) = 6 - 3.5 = 2.5$

Part B: $\int_{2}^{3} (x) dx = \left[ \frac{x^2}{2} \right]_2^3 = \frac{9}{2} - \frac{4}{2} = \frac{5}{2} = 2.5$

Part C: $\int_{3}^{4} (3x-6) dx = \left[ \frac{3x^2}{2} - 6x \right]_3^4 = (\frac{48}{2} - 24) - (\frac{27}{2} - 18) = (24-24) - (13.5-18) = 0 - (-4.5) = 4.5$

Step 5: Final Summation

$I = 2.5 + 2.5 + 4.5 = 9.5$

Final Solution: The value of the integral is $\frac{19}{2}$ or $9.5$.

Step 1: Simplification of Logarithmic Expressions

Given the integrand: $I = \int \frac{e^{5\log x} - e^{4\log x}}{e^{3\log x} - e^{2\log x}} \, dx$

We utilize the fundamental property of logarithms: $a\log b = \log(b^a)$ [By the Power Rule of Logarithms].

Applying this to each term:

• $e^{5\log x} = e^{\log(x^5)} = x^5$ [Since $e^{\log_e f(x)} = f(x)$]
• $e^{4\log x} = e^{\log(x^4)} = x^4$
• $e^{3\log x} = e^{\log(x^3)} = x^3$
• $e^{2\log x} = e^{\log(x^2)} = x^2$

Step 2: Algebraic Reduction

Substitute these values back into the integral expression:

$I = \int \frac{x^5 - x^4}{x^3 - x^2} \, dx$

Factor out the greatest common factors from the numerator and the denominator:

$I = \int \frac{x^4(x - 1)}{x^2(x - 1)} \, dx$

Assuming $x \neq 1$ and $x \neq 0$ for the expression to be defined, we cancel the common factor $(x-1)$ and simplify the power terms:

$I = \int \frac{x^4}{x^2} \, dx = \int x^2 \, dx$

Step 3: Integration and Final Computation

We apply the Power Rule for integration, which states $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$ [By the Fundamental Theorem of Calculus].

$I = \frac{x^{2+1}}{2+1} + C$

$I = \frac{x^3}{3} + C$

Final Solution: \frac{x^3}{3} + C

Step 1: Decomposition and Substitution

Given the integral: $I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \tan^{-1}(\sin x) \, dx$.

Utilizing the double-angle identity $\sin(2x) = 2\sin x \cos x$ [By Trigonometric Identity], the integral becomes:

$I = \int_{0}^{\frac{\pi}{2}} 2\sin x \cos x \tan^{-1}(\sin x) \, dx$

Let $u = \sin x$. Then, the differential is $du = \cos x \, dx$ [By Method of Substitution].

Adjusting the limits of integration:

• When $x = 0$, $u = \sin(0) = 0$.
• When $x = \frac{\pi}{2}$, $u = \sin(\frac{\pi}{2}) = 1$.

Substituting these into the integral yields:

$I = 2 \int_{0}^{1} u \tan^{-1}(u) \, du$

Step 2: Integration by Parts

Applying the Integration by Parts formula $\int f(u)g'(u) \, du = f(u)G(u) - \int f'(u)G(u) \, du$ [By Integration by Parts], where $f(u) = \tan^{-1}(u)$ and $g'(u) = u$.

Let $f(u) = \tan^{-1}(u) \implies f'(u) = \frac{1}{1+u^2}$.

Let $g'(u) = u \implies g(u) = \frac{u^2}{2}$.

Substituting into the formula:

$I = 2 \left[ \left( \frac{u^2}{2} \tan^{-1}(u) \right)_0^1 - \int_{0}^{1} \frac{u^2}{2(1+u^2)} \, du \right]$

$I = \left[ u^2 \tan^{-1}(u) \right]_0^1 - \int_{0}^{1} \frac{u^2}{1+u^2} \, du$

Step 3: Algebraic Manipulation and Final Evaluation

Evaluating the first term:

$[1^2 \tan^{-1}(1) - 0^2 \tan^{-1}(0)] = 1 \cdot \frac{\pi}{4} = \frac{\pi}{4}$.

For the remaining integral, perform algebraic division [By Polynomial Long Division]:

$\frac{u^2}{1+u^2} = \frac{u^2+1-1}{1+u^2} = 1 - \frac{1}{1+u^2}$.

The integral becomes:

$I = \frac{\pi}{4} - \int_{0}^{1} \left( 1 - \frac{1}{1+u^2} \right) \, du$

$I = \frac{\pi}{4} - \left[ u - \tan^{-1}(u) \right]_0^1$

$I = \frac{\pi}{4} - \left[ (1 - \tan^{-1}(1)) - (0 - \tan^{-1}(0)) \right]$

$I = \frac{\pi}{4} - \left( 1 - \frac{\pi}{4} \right)$

$I = \frac{\pi}{4} - 1 + \frac{\pi}{4}$

$I = \frac{\pi}{2} - 1$

Visual Representation of the Domain

Final Solution: \frac{\pi}{2} - 1

Step 1: Decomposition of the Integrand

The given integral is $I = \int \frac{1}{x - x^3} \, dx$. First, factor the denominator: $x - x^3 = x(1 - x^2) = x(1 - x)(1 + x)$. [By Algebraic Factorization]

We express the integrand as a sum of partial fractions: $\frac{1}{x(1 - x)(1 + x)} = \frac{A}{x} + \frac{B}{1 - x} + \frac{C}{1 + x}$

Step 2: Solving for Coefficients

Multiply both sides by $x(1 - x)(1 + x)$ to obtain the identity: $1 = A(1 - x)(1 + x) + Bx(1 + x) + Cx(1 - x)$ $1 = A(1 - x^2) + B(x + x^2) + C(x - x^2)$ $1 = (B - C - A)x^2 + (B + C)x + A$

• Comparing constants: $A = 1$.
• Comparing coefficients of $x$: $B + C = 0 \implies B = -C$.
• Comparing coefficients of $x^2$: $B - C - A = 0 \implies B - C = 1$.

Substituting $B = -C$ into $B - C = 1$: $-C - C = 1 \implies -2C = 1 \implies C = -\frac{1}{2}$ $B = -(-\frac{1}{2}) = \frac{1}{2}$

Step 3: Integration

Now, substitute the partial fractions back into the integral: $I = \int \left( \frac{1}{x} + \frac{1/2}{1 - x} - \frac{1/2}{1 + x} \right) \, dx$

Apply the linearity property of integrals [Per Linearity of Integration]: $I = \int \frac{1}{x} \, dx + \frac{1}{2} \int \frac{1}{1 - x} \, dx - \frac{1}{2} \int \frac{1}{1 + x} \, dx$

Integrating term by term: $\int \frac{1}{x} \, dx = \ln|x|$ $\int \frac{1}{1 - x} \, dx = -\ln|1 - x|$ [Using substitution $u = 1-x, du = -dx$] $\int \frac{1}{1 + x} \, dx = \ln|1 + x|$

Combining the results: $I = \ln|x| - \frac{1}{2} \ln|1 - x| - \frac{1}{2} \ln|1 + x| + C$

Step 4: Algebraic Simplification

Using logarithm properties: $I = \ln|x| - \frac{1}{2} (\ln|1 - x| + \ln|1 + x|) + C$ $I = \ln|x| - \frac{1}{2} \ln|(1 - x)(1 + x)| + C$ $I = \ln|x| - \frac{1}{2} \ln|1 - x^2| + C$ $I = \ln|x| - \ln|1 - x^2|^{1/2} + C$ $I = \ln\left| \frac{x}{\sqrt{1 - x^2}} \right| + C$

Final Solution: $\ln\left| \frac{x}{\sqrt{1 - x^2}} \right| + C$

Step 1: Decomposition and Trigonometric Simplification

Given the integral: $I = \int_{\frac{\pi}{2}}^{\pi} e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$.

Utilize trigonometric identities to simplify the integrand:

• $\sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)$ [By Double Angle Formula]
• $1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)$ [By Half-Angle Identity]

Substituting these into the expression: $\frac{1 - \sin x}{1 - \cos x} = \frac{1 - 2 \sin(\frac{x}{2}) \cos(\frac{x}{2})}{2 \sin^2(\frac{x}{2})} = \frac{1}{2 \sin^2(\frac{x}{2})} - \frac{2 \sin(\frac{x}{2}) \cos(\frac{x}{2})}{2 \sin^2(\frac{x}{2})}$

$= \frac{1}{2} \csc^2\left(\frac{x}{2}\right) - \cot\left(\frac{x}{2}\right)$

Step 2: Identifying the Form of the Integral

The integral now takes the form: $I = \int_{\frac{\pi}{2}}^{\pi} e^x \left[ -\cot\left(\frac{x}{2}\right) + \frac{1}{2} \csc^2\left(\frac{x}{2}\right) \right] dx$.

Let $f(x) = -\cot\left(\frac{x}{2}\right)$. Then, differentiating with respect to $x$ using the Chain Rule: $f'(x) = -\left( -\csc^2\left(\frac{x}{2}\right) \cdot \frac{1}{2} \right) = \frac{1}{2} \csc^2\left(\frac{x}{2}\right)$.

The integral matches the standard form: $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$ [By the Fundamental Theorem of Calculus/Integration by Parts corollary].

Step 3: Evaluation of Definite Integral

Applying the formula: $I = \left[ e^x \cdot (-\cot\left(\frac{x}{2}\right)) \right]_{\frac{\pi}{2}}^{\pi}$

Evaluating at limits: Upper limit ($\pi$): $e^{\pi} (-\cot(\frac{\pi}{2})) = e^{\pi} (0) = 0$.

Lower limit ($\frac{\pi}{2}$): $e^{\frac{\pi}{2}} (-\cot(\frac{\pi}{4})) = e^{\frac{\pi}{2}} (-1) = -e^{\frac{\pi}{2}}$.

Step 4: Final Calculation

$I = [0] - [-e^{\frac{\pi}{2}}] = e^{\frac{\pi}{2}}$.

Final Solution: $e^{\frac{\pi}{2}}$