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: Reformulation of the Integrand

Given the integral:

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

We simplify the integrand by distributing the division by $x$ across the term $(x+1)$:

$I = \int \left( \frac{x}{x} + \frac{1}{x} \right) (x + \log x)^2 \, dx$

Which reduces to:

$I = \int \left( 1 + \frac{1}{x} \right) (x + \log x)^2 \, dx$ [Algebraic simplification via the distributive property]

Step 2: Substitution Methodology

To evaluate the integral, we perform a change of variables. Let $u$ be the base of the squared term:

$u = x + \log x$

Now, compute the differential $du$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(\log x)$ $\frac{du}{dx} = 1 + \frac{1}{x}$ [By the Sum Rule of differentiation and the Derivative of $\log_e x$]

Rearranging gives:

$du = \left( 1 + \frac{1}{x} \right) \, dx$

Step 3: Evaluating the Transformed Integral

Substitute $u$ and $du$ into the integral expression derived in Step 1:

$I = \int (u)^2 \, du$

Applying the Power Rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, where $n=2$:

$I = \frac{u^{2+1}}{2+1} + C$ $I = \frac{u^3}{3} + C$ [By the Fundamental Theorem of Calculus]

Step 4: Reversion to Original Variable

Substitute $u = x + \log x$ back into the result to obtain the final expression in terms of $x$:

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

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

Step 1: Identifying the Mathematical Framework

To evaluate the definite integral $I = \int_{0}^{\frac{\pi}{2}} \cos 2x \, dx$, we employ the Fundamental Theorem of Calculus (Part II). This theorem states that if $F(x)$ is the antiderivative of $f(x)$, then $\int_{a}^{b} f(x) \, dx = F(b) - F(a)$. [Fundamental Theorem of Calculus].

Step 2: Finding the Antiderivative

We identify the integrand $f(x) = \cos 2x$. To find the antiderivative $F(x)$, we apply the method of integration by substitution or the reverse chain rule for trigonometric functions. Since $\frac{d}{dx}(\sin(kx)) = k \cos(kx)$, it follows that $\int \cos(kx) \, dx = \frac{1}{k} \sin(kx) + C$.

Setting $k = 2$: $F(x) = \int \cos 2x \, dx = \frac{\sin 2x}{2}$

Step 3: Applying Limits of Integration

Substituting the upper limit $b = \frac{\pi}{2}$ and the lower limit $a = 0$ into the antiderivative:

$I = \left[ \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}}$

$I = \frac{1}{2} \left[ \sin\left(2 \cdot \frac{\pi}{2}\right) - \sin(2 \cdot 0) \right]$

Step 4: Algebraic Evaluation

Evaluate the trigonometric terms:

• $\sin(2 \cdot \frac{\pi}{2}) = \sin(\pi) = 0$ [Trigonometric Identity]
• $\sin(0) = 0$

$I = \frac{1}{2} [0 - 0] = 0$

Visual Representation of the Integrand

The integral represents the net signed area between the curve $y = \cos 2x$ and the $x$-axis from $0$ to $\frac{\pi}{2}$. Because the area from $0$ to $\frac{\pi}{4}$ (positive) exactly cancels the area from $\frac{\pi}{4}$ to $\frac{\pi}{2}$ (negative), the result is zero.

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

Step 1: Decomposition of the Definite Integral

Given the definite integral: $I = \int_{0}^{\frac{\pi}{4}} (2\sec^2 x + x^3 + 2) \, dx$.
By the Linearity Property of Integration [$\int_{a}^{b} [f(x) + g(x)] dx = \int_{a}^{b} f(x) dx + \int_{a}^{b} g(x) dx$], we decompose the expression into three distinct integrable terms:

$I = 2 \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx + \int_{0}^{\frac{\pi}{4}} x^3 \, dx + 2 \int_{0}^{\frac{\pi}{4}} 1 \, dx$

Step 2: Identifying Fundamental Antiderivatives

We apply the Fundamental Theorem of Calculus Part II: $\int_{a}^{b} f(x) dx = F(b) - F(a)$, where $F'(x) = f(x)$.

• For $\int \sec^2 x \, dx$, the antiderivative is $\tan x$ [Standard Trigonometric Integral].
• For $\int x^3 \, dx$, the antiderivative is $\frac{x^4}{4}$ [Power Rule: $\int x^n dx = \frac{x^{n+1}}{n+1}$].
• For $\int 1 \, dx$, the antiderivative is $x$.

Step 3: Evaluating the Antiderivatives

Applying the limits $[0, \frac{\pi}{4}]$ to each term:

$I = 2[\tan x]_{0}^{\frac{\pi}{4}} + \left[ \frac{x^4}{4} \right]_{0}^{\frac{\pi}{4}} + 2[x]_{0}^{\frac{\pi}{4}}$

Calculating the values:

1. Term 1: $2(\tan(\frac{\pi}{4}) - \tan(0)) = 2(1 - 0) = 2$
2. Term 2: $\left( \frac{(\frac{\pi}{4})^4}{4} - 0 \right) = \frac{\pi^4}{4^4 \cdot 4} = \frac{\pi^4}{256 \cdot 4} = \frac{\pi^4}{1024}$
3. Term 3: $2(\frac{\pi}{4} - 0) = \frac{2\pi}{4} = \frac{\pi}{2}$

Step 4: Algebraic Summation

Summing the results obtained in Step 3:

$I = 2 + \frac{\pi^4}{1024} + \frac{\pi}{2}$

Visual Representation of Area under the curve $f(x) = 2\sec^2 x + x^3 + 2$

Final Solution: $2 + \frac{\pi}{2} + \frac{\pi^4}{1024}$

Step 1: Formal Definition of the Integral

Let the given definite integral be defined as $I$:

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

Step 2: Applying the King's Property of Definite Integrals

We utilize the property: $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$ [Per the King's Property of Definite Integrals].

Substituting $x$ with $(a-x)$ in Equation 1:

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

$I = \int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} \, dx$ --- (Equation 2)

Step 3: Summation of Equations

Adding Equation 1 and Equation 2:

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

$2I = \int_{0}^{a} \left( \frac{\sqrt{x} + \sqrt{a-x}}{\sqrt{x} + \sqrt{a-x}} \right) \, dx$

Step 4: Algebraic Simplification and Evaluation

Since $\frac{\sqrt{x} + \sqrt{a-x}}{\sqrt{x} + \sqrt{a-x}} = 1$ for all $x \in (0, a)$:

$2I = \int_{0}^{a} 1 \, dx$

$2I = [x]_{0}^{a}$ [By the Fundamental Theorem of Calculus]

$2I = a - 0$

$2I = a$

Step 5: Solving for I

$I = \frac{a}{2}$

Visual Representation of the Area Under the Curve

Final Solution: \frac{a}{2}

Step 1: Theoretical Foundation

To evaluate the definite integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x \, dx$, we utilize the standard integral formula for the cosecant function. The antiderivative of $\csc x$ is given by:

$\int \csc x \, dx = \ln |\csc x - \cot x| + C$ [Standard Integral Calculus Identity]

Step 2: Applying the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus, Part II, if $F(x)$ is the antiderivative of $f(x)$, then:

$\int_{a}^{b} f(x) \, dx = F(b) - F(a)$

Substituting our limits $a = \frac{\pi}{6}$ and $b = \frac{\pi}{4}$:

$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x \, dx = \left[ \ln |\csc x - \cot x| \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}$

Step 3: Evaluating at the Boundaries

Substitute the upper limit $x = \frac{\pi}{4}$:

$F(\frac{\pi}{4}) = \ln |\csc(\frac{\pi}{4}) - \cot(\frac{\pi}{4})| = \ln |\sqrt{2} - 1|$

[Using trigonometric ratios: $\csc(\frac{\pi}{4}) = \sqrt{2}$, $\cot(\frac{\pi}{4}) = 1$]

Substitute the lower limit $x = \frac{\pi}{6}$:

$F(\frac{\pi}{6}) = \ln |\csc(\frac{\pi}{6}) - \cot(\frac{\pi}{6})| = \ln |2 - \sqrt{3}|$

[Using trigonometric ratios: $\csc(\frac{\pi}{6}) = 2$, $\cot(\frac{\pi}{6}) = \sqrt{3}$]

Step 4: Final Algebraic Simplification

Subtracting the values:

Result $= \ln |\sqrt{2} - 1| - \ln |2 - \sqrt{3}|$

Using the logarithmic quotient rule $\ln(a) - \ln(b) = \ln(\frac{a}{b})$:

Result $= \ln \left| \frac{\sqrt{2} - 1}{2 - \sqrt{3}} \right|$

Rationalize the denominator to simplify the expression:

$\frac{\sqrt{2} - 1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2\sqrt{2} + \sqrt{6} - 2 - \sqrt{3}}{4 - 3} = 2\sqrt{2} + \sqrt{6} - 2 - \sqrt{3}$

Visual Representation of the Integration Area

Final Solution: \ln \left| \frac{\sqrt{2} - 1}{2 - \sqrt{3}} \right| \text{ or } \ln(2\sqrt{2} + \sqrt{6} - 2 - \sqrt{3})

Step 1: Identifying the Derivative Relationship

To find the anti-derivative (integral) of the function $f(x) = \cos 3x$ by the method of inspection, we must identify a function $F(x)$ such that its derivative yields the original function:

$\frac{d}{dx} [F(x)] = \cos 3x$ [Definition of Anti-derivative]

Step 2: Selecting the Trial Function

We know from the standard trigonometric derivatives that the derivative of the sine function results in a cosine function:

$\frac{d}{dx} (\sin ax) = a \cos ax$ [Chain Rule of Differentiation]

In our case, $a = 3$. Therefore, let us consider the function $\sin 3x$.

Step 3: Calculating the Derivative of the Trial Function

Differentiating $\sin 3x$ with respect to $x$:

$\frac{d}{dx} (\sin 3x) = \cos 3x \cdot \frac{d}{dx} (3x)$ [Applying the Chain Rule]

$\frac{d}{dx} (\sin 3x) = \cos 3x \cdot 3$

$\frac{d}{dx} (\sin 3x) = 3 \cos 3x$

Step 4: Isolating the Target Function

To isolate $\cos 3x$ as requested by the inspection method, divide both sides of the equation by $3$:

$\frac{1}{3} \cdot \frac{d}{dx} (\sin 3x) = \cos 3x$

Using the linearity property of derivatives, $\frac{d}{dx} [k \cdot f(x)] = k \cdot \frac{d}{dx} [f(x)]$, we can move the constant inside the derivative operator:

$\frac{d}{dx} \left( \frac{1}{3} \sin 3x \right) = \cos 3x$

Step 5: Visualizing the Relationship

Final Solution: The anti-derivative of $\cos 3x$ is $\frac{1}{3} \sin 3x$.

Step 1: Substitution for Simplification

To integrate the function $I = \int \frac{2x}{(x^2+1)(x^2+3)} dx$, we observe that the derivative of $x^2$ is $2x$. Let us use the method of substitution.

Let $t = x^2$. [By the Method of Substitution]

Then, the differential $dt = 2x \, dx$.

The integral transforms into:

$I = \int \frac{1}{(t+1)(t+3)} dt$

Step 2: Partial Fraction Decomposition

We express the integrand as a sum of partial fractions:

$\frac{1}{(t+1)(t+3)} = \frac{A}{t+1} + \frac{B}{t+3}$

Multiplying both sides by $(t+1)(t+3)$, we obtain the identity:

$1 = A(t+3) + B(t+1)$

To find the constants $A$ and $B$, we use the method of equating coefficients or substituting critical values: [By Partial Fraction Theorem]

• Set $t = -1$: $1 = A(-1+3) \implies 1 = 2A \implies A = \frac{1}{2}$
• Set $t = -3$: $1 = B(-3+1) \implies 1 = -2B \implies B = -\frac{1}{2}$

Step 3: Evaluating the Integral

Substitute the constants back into the integral form:

$I = \int \left( \frac{1/2}{t+1} - \frac{1/2}{t+3} \right) dt$ $I = \frac{1}{2} \int \frac{1}{t+1} dt - \frac{1}{2} \int \frac{1}{t+3} dt$

Using the standard integral $\int \frac{1}{x} dx = \ln|x| + C$:

$I = \frac{1}{2} \ln|t+1| - \frac{1}{2} \ln|t+3| + C$ $I = \frac{1}{2} \ln\left| \frac{t+1}{t+3} \right| + C$

Step 4: Re-substitution and Simplification

Substituting $t = x^2$ back into the result:

$I = \frac{1}{2} \ln\left| \frac{x^2+1}{x^2+3} \right| + C$

Since $x^2+1 > 0$ and $x^2+3 > 0$ for all real $x$, the absolute value bars are redundant.

Final Solution: \frac{1}{2} \ln\left( \frac{x^2+1}{x^2+3} \right) + C

Step 1: Problem Definition and Algebraic Rearrangement

We are tasked to evaluate the following indefinite integral:

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

Utilizing the trigonometric identity $\frac{1}{\cos^2 x} = \sec^2 x$ [Per the Reciprocal Identity of Trigonometry], we rewrite the integrand to facilitate substitution:

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

Step 2: Method of Substitution

Let $u = 1 - \tan x$. To find the differential $du$, we differentiate $u$ with respect to $x$ [By the Chain Rule]:

$\frac{du}{dx} = \frac{d}{dx}(1 - \tan x) = 0 - \sec^2 x$

$du = -\sec^2 x \, dx$

Consequently, $\sec^2 x \, dx = -du$.

Step 3: Variable Transformation and Integration

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

$I = \int \frac{-du}{u^2} = -\int u^{-2} \, du$

Applying the Power Rule for Integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ [Where $n \neq -1$]:

$I = - \left( \frac{u^{-2+1}}{-2+1} \right) + C$

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

$I = \frac{1}{u} + C$

Step 4: Back-Substitution

Reverting the variable $u$ to the original expression $1 - \tan x$:

$I = \frac{1}{1 - \tan x} + C$

Visual Representation of the Substitution Logic

Analytical Verification

Expression	Result
Integral term	$\int (1-\tan x)^{-2} \sec^2 x \, dx$
Resulting primitive	$(1-\tan x)^{-1} + C$
Derivative check	$-1(1-\tan x)^{-2} \cdot (-\sec^2 x) = \frac{\sec^2 x}{(1-\tan x)^2}$

Final Solution: $\frac{1}{1 - \tan x} + C$

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 and Identifying the Substitution

We are tasked with evaluating the indefinite integral:

$I = \int \frac{x^3 \sin(\tan^{-1} x^4)}{1+x^8} \, dx$

To simplify the integrand, we observe the composition of functions. Let $u = \tan^{-1}(x^4)$. [Per the chain rule of differentiation, $\frac{d}{dx} \tan^{-1}(f(x)) = \frac{1}{1+(f(x))^2} \cdot f'(x)$]

Step 2: Performing the Differential Substitution

Let $u = \tan^{-1}(x^4)$.

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = \frac{1}{1 + (x^4)^2} \cdot \frac{d}{dx}(x^4)$

$\frac{du}{dx} = \frac{1}{1 + x^8} \cdot 4x^3$

Rearranging the terms to solve for the differential $dx$:

$du = \frac{4x^3}{1+x^8} \, dx \implies \frac{x^3}{1+x^8} \, dx = \frac{1}{4} \, du$

Step 3: Transforming the Integral

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

$I = \int \sin(u) \cdot \frac{1}{4} \, du$

$I = \frac{1}{4} \int \sin(u) \, du$

Step 4: Integration and Back-Substitution

Using the fundamental integration formula $\int \sin(u) \, du = -\cos(u) + C$, where $C$ is the constant of integration:

$I = \frac{1}{4} (-\cos(u)) + C$

$I = -\frac{1}{4} \cos(u) + C$

Substituting the original variable $u = \tan^{-1}(x^4)$ back into the result:

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

Step 5: Geometric Interpretation of $\cos(\tan^{-1} x^4)$

Let $\theta = \tan^{-1}(x^4)$, which implies $\tan \theta = \frac{x^4}{1}$. By considering a right-angled triangle with opposite side $x^4$ and adjacent side $1$, the hypotenuse $h$ is given by $\sqrt{(x^4)^2 + 1^2} = \sqrt{x^8 + 1}$.

Therefore, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^8 + 1}}$.

Replacing the cosine term with its algebraic equivalent:

$I = -\frac{1}{4} \left( \frac{1}{\sqrt{x^8 + 1}} \right) + C$

Final Solution: $-\frac{1}{4\sqrt{x^8+1}} + C$

Step 1: Application of Integration by Parts

To evaluate the integral $I = \int x \cos^{-1} x \, dx$, we employ the method of Integration by Parts, which states: $\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \cdot \int v \, dx \right) dx$ [Per the Integration by Parts Theorem].

According to the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), we assign:

• $u = \cos^{-1} x$
• $dv = x \, dx$

Differentiating $u$ and integrating $dv$ yields:

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

Step 2: Substitution into the Formula

Applying the components to the integration formula:

$I = \left( \cos^{-1} x \right) \left( \frac{x^2}{2} \right) - \int \left( -\frac{1}{\sqrt{1-x^2}} \right) \left( \frac{x^2}{2} \right) dx$

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

Step 3: Evaluating the Remaining Integral

To integrate $\int \frac{x^2}{\sqrt{1-x^2}} \, dx$, we manipulate the numerator to create a term that matches the denominator:

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

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

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

Step 4: Applying Standard Integral Formulas

We use the following standard results:

1. $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$
2. $\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right)$

Thus, $\int \frac{x^2}{\sqrt{1-x^2}} \, dx = -\left( \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x \right) + \sin^{-1} x$

$= -\frac{x}{2}\sqrt{1-x^2} - \frac{1}{2}\sin^{-1} x + \sin^{-1} x = -\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x$

Step 5: Final Assembly

Substitute the result back into the expression for $I$:

$I = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \left[ -\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x \right] + C$

$I = \frac{x^2}{2} \cos^{-1} x - \frac{x}{4}\sqrt{1-x^2} + \frac{1}{4}\sin^{-1} x + C$

Final Solution: $\frac{x^2}{2} \cos^{-1} x - \frac{x}{4}\sqrt{1-x^2} + \frac{1}{4}\sin^{-1} x + C$

Step 1: Decomposition of the Integrand

To integrate $I = \int \tan^4 x \, dx$, we employ the trigonometric identity $\tan^2 x = \sec^2 x - 1$ [By Pythagorean Identity]. We rewrite the integrand as follows:

$I = \int \tan^2 x \cdot \tan^2 x \, dx$

$I = \int \tan^2 x (\sec^2 x - 1) \, dx$

$I = \int \tan^2 x \sec^2 x \, dx - \int \tan^2 x \, dx$ [By Linearity of Integration]

Step 2: Solving the Individual Integrals

Let $I = I_1 - I_2$, where $I_1 = \int \tan^2 x \sec^2 x \, dx$ and $I_2 = \int \tan^2 x \, dx$.

Evaluating $I_1$:

For $I_1 = \int \tan^2 x \sec^2 x \, dx$, we use the method of substitution. Let $u = \tan x$. Then, $du = \sec^2 x \, dx$ [By Chain Rule].

$I_1 = \int u^2 \, du = \frac{u^3}{3} + C_1$ [By Power Rule of Integration: $\int x^n dx = \frac{x^{n+1}}{n+1}$]

$I_1 = \frac{\tan^3 x}{3} + C_1$

Evaluating $I_2$:

For $I_2 = \int \tan^2 x \, dx$, we again use the identity $\tan^2 x = \sec^2 x - 1$.

$I_2 = \int (\sec^2 x - 1) \, dx$

$I_2 = \int \sec^2 x \, dx - \int 1 \, dx$

$I_2 = \tan x - x + C_2$ [Since $\frac{d}{dx}(\tan x) = \sec^2 x$]

Step 3: Synthesis of Results

Combining $I_1$ and $I_2$ into the expression $I = I_1 - I_2$:

$I = \left( \frac{\tan^3 x}{3} + C_1 \right) - (\tan x - x + C_2)$

$I = \frac{\tan^3 x}{3} - \tan x + x + (C_1 - C_2)$

Setting the combined integration constant $C = C_1 - C_2$, we reach the final form.

Visual Representation of the Integrand Geometry

Final Solution: $\frac{1}{3}\tan^3 x - \tan x + x + C$

Step 1: Problem Definition and Algebraic Manipulation

We are required to evaluate the integral:

$I = \int \frac{x}{\sqrt{x+4}} \, dx$

Given the structure of the integrand, we utilize the algebraic substitution method to simplify the radical expression [Per the Method of Substitution for Integration].

Step 2: Substitution Method

Let $u = x + 4$. Then, differentiating both sides with respect to $x$ gives:

$\frac{du}{dx} = 1 \implies du = dx$

From the substitution equation $u = x + 4$, we isolate $x$ to express the numerator in terms of $u$:

$x = u - 4$

Step 3: Transforming the Integral

Substitute $x = u - 4$, $dx = du$, and $\sqrt{x+4} = \sqrt{u}$ into the integral expression:

$I = \int \frac{u - 4}{\sqrt{u}} \, du$

$I = \int \left( \frac{u}{\sqrt{u}} - \frac{4}{\sqrt{u}} \right) \, du$

$I = \int \left( u^{1 - 1/2} - 4u^{-1/2} \right) \, du$

$I = \int \left( u^{1/2} - 4u^{-1/2} \right) \, du$

Step 4: Applying Power Rule of Integration

Applying the power rule $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$ [Per the Fundamental Theorem of Calculus]:

$I = \frac{u^{1/2 + 1}}{1/2 + 1} - 4 \cdot \frac{u^{-1/2 + 1}}{-1/2 + 1} + C$

$I = \frac{u^{3/2}}{3/2} - 4 \cdot \frac{u^{1/2}}{1/2} + C$

$I = \frac{2}{3}u^{3/2} - 8u^{1/2} + C$

Step 5: Back-Substitution

Reverting the variable $u$ back to $x+4$:

$I = \frac{2}{3}(x+4)^{3/2} - 8(x+4)^{1/2} + C$

To simplify further, factor out $2(x+4)^{1/2}$:

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

$I = 2\sqrt{x+4} \left[ \frac{x+4 - 12}{3} \right] + C$

$I = 2\sqrt{x+4} \left[ \frac{x-8}{3} \right] + C$

$I = \frac{2}{3}(x-8)\sqrt{x+4} + C$

Final Solution: $\frac{2}{3}(x-8)\sqrt{x+4} + C$

Step 1: Formal Representation of the Integral

To evaluate the integral of the function $I = \int \frac{1}{\sqrt{9-25x^2}} \, dx$, we first identify the standard form required for integration involving quadratic radicals.

Step 2: Algebraic Manipulation and Factorization

We aim to bring the denominator into the standard form $\sqrt{a^2 - u^2}$. To achieve this, we factor out $25$ from the expression inside the square root:

$9 - 25x^2 = 25\left(\frac{9}{25} - x^2\right)$

Substituting this back into the integral:

$I = \int \frac{1}{\sqrt{25\left(\left(\frac{3}{5}\right)^2 - x^2\right)}} \, dx$

[By the property of radicals $\sqrt{A \cdot B} = \sqrt{A} \cdot \sqrt{B}$]

$I = \int \frac{1}{5\sqrt{\left(\frac{3}{5}\right)^2 - x^2}} \, dx$

$I = \frac{1}{5} \int \frac{1}{\sqrt{\left(\frac{3}{5}\right)^2 - x^2}} \, dx$

Step 3: Application of the Standard Integration Formula

We utilize the standard integration formula for the form $\int \frac{1}{\sqrt{a^2 - x^2}} \, dx$:

$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C$

[Per the standard integral table for inverse trigonometric functions]

Here, $a = \frac{3}{5}$. Substituting these values into the expression:

$I = \frac{1}{5} \left( \arcsin\left(\frac{x}{3/5}\right) \right) + C$

Step 4: Simplification of the Result

Simplify the fraction within the inverse sine function:

$I = \frac{1}{5} \arcsin\left(\frac{5x}{3}\right) + C$

Visual Representation of the Integration Geometry

Final Solution: $\frac{1}{5} \arcsin\left(\frac{5x}{3}\right) + C$

Step 1: Analyzing the Integrand and Property Application

The integral to evaluate is $I = \int_{0}^{4} |x-1| \, dx$. The integrand involves an absolute value function $|f(x)|$, which changes its definition at the point where the argument is zero. Specifically, $|x-1| = 0$ at $x = 1$.

We apply the property of definite integrals: $\int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx$ where $a < c < b$. [Property: Additivity of Interval of Integration]

Since the interval is $[0, 4]$, we split the integral at $c = 1$:

$I = \int_{0}^{1} |x-1| \, dx + \int_{1}^{4} |x-1| \, dx$

Step 2: Defining the Piecewise Function

Based on the definition of absolute value:

• For $x \in [0, 1]$, $x-1 \le 0$, thus $|x-1| = -(x-1) = 1-x$.
• For $x \in [1, 4]$, $x-1 \ge 0$, thus $|x-1| = x-1$.

Substituting these into the split integrals:

$I = \int_{0}^{1} (1-x) \, dx + \int_{1}^{4} (x-1) \, dx$

Step 3: Integration and Evaluation

Applying the Fundamental Theorem of Calculus: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$.

For the first part:

$\int_{0}^{1} (1-x) \, dx = \left[ x - \frac{x^2}{2} \right]_{0}^{1} = \left( 1 - \frac{1^2}{2} \right) - (0 - 0) = 1 - 0.5 = 0.5$

For the second part:

$\int_{1}^{4} (x-1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{4} = \left( \frac{4^2}{2} - 4 \right) - \left( \frac{1^2}{2} - 1 \right)$ $= (8 - 4) - (0.5 - 1) = 4 - (-0.5) = 4 + 0.5 = 4.5$

Step 4: Summation

$I = 0.5 + 4.5 = 5.0$

Visual Representation of the Area

Final Solution: The value of the integral $\int_{0}^{4} |x-1| \, dx$ is $5$.

Step 1: Identifying the Integral and Substitution Strategy

We are tasked to evaluate the indefinite integral:

$I = \int \frac{x}{9 - 4x^2} \, dx$

To solve this, we employ the method of integration by substitution. We observe that the derivative of the denominator $9 - 4x^2$ is proportional to the numerator $x$.

[Justification: The power rule for differentiation states $\frac{d}{dx}(x^n) = nx^{n-1}$. Here, $\frac{d}{dx}(9 - 4x^2) = -8x$.]

Step 2: Performing the Change of Variables

Let $u = 9 - 4x^2$.

Differentiating both sides with respect to $x$:

$\frac{du}{dx} = -8x \implies du = -8x \, dx$

Rearranging to isolate the terms present in the integral:

$x \, dx = -\frac{1}{8} du$

Step 3: Transforming the Integral

Substitute $u$ and $du$ into the original integral $I$:

$I = \int \frac{1}{u} \cdot \left(-\frac{1}{8}\right) du$

$I = -\frac{1}{8} \int \frac{1}{u} \, du$

[Justification: By the Constant Multiple Rule for integration, $\int k f(u) du = k \int f(u) du$.]

Step 4: Evaluating the Integral

Using the standard integration formula for the reciprocal function:

$\int \frac{1}{u} \, du = \ln|u| + C$

[Justification: The Fundamental Theorem of Calculus implies that the antiderivative of $\frac{1}{x}$ is $\ln|x|$.]

$I = -\frac{1}{8} \ln|u| + C$

Step 5: Back-Substitution

Substitute $u = 9 - 4x^2$ back into the expression:

$I = -\frac{1}{8} \ln|9 - 4x^2| + C$

Visual Representation of the integrand

Final Solution: $-\frac{1}{8} \ln|9 - 4x^2| + C$

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

Given Integral and Property Setup

The objective is to evaluate the definite integral:

$I = \int_{2}^{8} |x-5| dx$

[By the property of definite integrals: $\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$, where $a < c < b$]

Step 1: Analyzing the Modulus Function

The integrand $f(x) = |x-5|$ changes its definition based on the value of $x$ relative to the critical point $x = 5$:

• $|x-5| = -(x-5)$ for $x < 5$
• $|x-5| = (x-5)$ for $x \ge 5$

Since the interval of integration is $[2, 8]$ and the point $5$ lies within this interval, we split the integral at $x = 5$:

$I = \int_{2}^{5} -(x-5) dx + \int_{5}^{8} (x-5) dx$

Step 2: Evaluating the Split Integrals

Part A: Evaluating $\int_{2}^{5} -(x-5) dx$

$= -\left[ \frac{x^2}{2} - 5x \right]_{2}^{5}$

$= -\left[ (\frac{25}{2} - 25) - (\frac{4}{2} - 10) \right]$

$= -\left[ -12.5 - (-8) \right] = -(-4.5) = 4.5$

Part B: Evaluating $\int_{5}^{8} (x-5) dx$

$= \left[ \frac{x^2}{2} - 5x \right]_{5}^{8}$

$= \left[ (\frac{64}{2} - 40) - (\frac{25}{2} - 25) \right]$

$= \left[ (32 - 40) - (-12.5) \right] = -8 + 12.5 = 4.5$

Step 3: Calculating Total Sum

$I = 4.5 + 4.5 = 9$

Geometric Representation

Final Solution: 9

Step 1: Algebraic Simplification of the Integrand

Given the integral: $I = \int (1 - x)\sqrt{x} \, dx$

To facilitate integration, we apply the distributive property of multiplication over subtraction [By the Distributive Law: $a(b-c) = ab - ac$]. First, express the radical as a fractional exponent: $\sqrt{x} = x^{1/2}$.

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

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

Using the Laws of Exponents ($x^a \cdot x^b = x^{a+b}$), we compute the power for the second term:

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

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

Step 2: Application of Linearity and the Power Rule

By the linearity property of integration [$\int [f(x) \pm g(x)] \, dx = \int f(x) \, dx \pm \int g(x) \, dx$], we split the integral:

$I = \int x^{1/2} \, dx - \int x^{3/2} \, dx$

Apply the Power Rule for integration, which states $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$:

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

Step 3: Arithmetic Simplification

Calculate the exponents and the denominators:

For the first term: $\frac{1}{2} + 1 = \frac{3}{2}$

For the second term: $\frac{3}{2} + 1 = \frac{5}{2}$

$I = \frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2} + C$

Multiply by the reciprocal of the fractions:

$I = \frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2} + C$

Visualization of the Function

Final Solution: $\frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2} + C$