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: Identifying the Integrand and Applying Trigonometric Identities

The given integral is: $I = \int \frac{\sin^2 x}{1 + \cos x} dx$

To simplify the integrand, we utilize the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, which implies $\sin^2 x = 1 - \cos^2 x$ [Per the Pythagorean Identity].

Substituting this into the integrand:

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

Step 2: Algebraic Simplification (Difference of Squares)

The numerator $1 - \cos^2 x$ can be expressed as a difference of two squares, $a^2 - b^2 = (a-b)(a+b)$, where $a=1$ and $b=\cos x$ [By Algebraic Factoring].

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

Assuming $\cos x \neq -1$ (i.e., $x \neq (2n+1)\pi$), we can cancel the common term $(1 + \cos x)$ in the numerator and the denominator:

$I = \int (1 - \cos x) dx$

Step 3: Evaluating the Integral

We now apply the linearity property of integrals [Per the Linearity of Integrals], which allows us to split the integral into two parts:

$I = \int 1 dx - \int \cos x dx$

Applying the standard integral formulas:

• $\int 1 dx = x + C_1$
• $\int \cos x dx = \sin x + C_2$

Combining these, where $C = C_1 - C_2$ is the constant of integration:

$I = x - \sin x + C$

Step 4: Visualizing the Function and Geometric Interpretation

The integrand $f(x) = \frac{\sin^2 x}{1 + \cos x} = 1 - \cos x$ represents a shifted cosine wave. Below is a representation of the original function and its transformation.

Step	Transformation	Justification
Identity Substitution	$1 - \cos^2 x$	Trigonometric Identity
Factoring	$(1-\cos x)(1+\cos x)$	Algebraic Difference of Squares
Reduction	$1 - \cos x$	Cancellation Property

Final Solution: $x - \sin x + C$

Step 1: Application of Trigonometric Identities

The integrand is given by:

$I = \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} \, dx$

To simplify the numerator, apply the double-angle identity for cosine: $\cos 2x = 1 - 2\sin^2 x$ [Per the trigonometric identity $\cos 2\theta = 1 - 2\sin^2 \theta$].

Substituting this into the integral expression:

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

Step 2: Algebraic Simplification

Observe the cancellation of the $2\sin^2 x$ terms in the numerator:

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

Utilizing the reciprocal trigonometric identity, $\frac{1}{\cos^2 x} = \sec^2 x$ [By definition of the secant function], the integral simplifies to:

$I = \int \sec^2 x \, dx$

Step 3: Integration and Final Result

Applying the standard integral formula for the square of the secant function [Per the Fundamental Theorem of Calculus/Table of Integrals]:

$\int \sec^2 x \, dx = \tan x + C$

where $C$ represents the constant of integration.

Visual Representation of the Integrand Function

Final Solution: $\tan x + C$

Step 1: Analyzing the Integral Structure

The given integral is $I = \int \frac{e^x(1+x)}{\cos^2(e^x x)} dx$. To solve this, we must identify an appropriate substitution [Per the Method of Substitution for Integration]. We observe that the denominator contains the composite function $\cos^2(e^x x)$. Let $u = e^x x$.

Step 2: Differentiating the Substitution

Differentiate $u$ with respect to $x$ using the Product Rule, which states $\frac{d}{dx}[f(x)g(x)] = f(x)g'(x) + g(x)f'(x)$:

$u = e^x x$

$\frac{du}{dx} = e^x \frac{d}{dx}(x) + x \frac{d}{dx}(e^x)$

$\frac{du}{dx} = e^x(1) + x(e^x)$

$\frac{du}{dx} = e^x(1 + x)$

Therefore, $du = e^x(1 + x) dx$.

Step 3: Transforming the Integral

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

$I = \int \frac{1}{\cos^2(u)} du$

Utilizing the trigonometric identity $\sec^2(u) = \frac{1}{\cos^2(u)}$ [Per Fundamental Trigonometric Identities]:

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

Step 4: Integration and Back-Substitution

The integral of $\sec^2(u)$ is $\tan(u) + C$ [Per the Standard Table of Integrals]:

$I = \tan(u) + C$

Substituting $u = e^x x$ back into the expression:

$I = \tan(e^x x) + C$

Visual Representation of the Transformation

Analytical Summary

Original Form	Transformed Form	Result
$\int \frac{e^x(1+x)}{\cos^2(e^x x)} dx$	$\int \sec^2(u) du$	$\tan(e^x x) + C$

Final Solution: \tan(e^x x) + C

Step 1: Application of Trigonometric Product-to-Sum Identities

To evaluate the integral $\int \sin 3x \cos 4x \, dx$, we must transform the product of trigonometric functions into a sum or difference. We utilize the standard trigonometric identity [Product-to-Sum Formula]:

$\sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)]$

Setting $A = 3x$ and $B = 4x$, we substitute these values into the identity:

$\sin 3x \cos 4x = \frac{1}{2} [\sin(3x + 4x) + \sin(3x - 4x)]$

$\sin 3x \cos 4x = \frac{1}{2} [\sin 7x + \sin(-x)]$

Using the parity property of the sine function [$\sin(-x) = -\sin x$], the expression simplifies to:

$\sin 3x \cos 4x = \frac{1}{2} [\sin 7x - \sin x]$

Step 2: Integration of the Transformed Expression

Substitute the transformed expression back into the integral:

$I = \int \frac{1}{2} (\sin 7x - \sin x) \, dx$

$I = \frac{1}{2} \int \sin 7x \, dx - \frac{1}{2} \int \sin x \, dx$

Step 3: Execution of Integration

Applying the standard integral formula [$\int \sin(ax) \, dx = -\frac{\cos(ax)}{a} + C$]:

For the first term: $\int \sin 7x \, dx = -\frac{\cos 7x}{7}$

For the second term: $\int \sin x \, dx = -\cos x$

Combining these into the expression:

$I = \frac{1}{2} \left[ -\frac{\cos 7x}{7} - (-\cos x) \right] + C$

$I = \frac{1}{2} \left[ \cos x - \frac{\cos 7x}{7} \right] + C$

$I = \frac{\cos x}{2} - \frac{\cos 7x}{14} + C$

Visual Representation of the Integrand

Final Solution: $\frac{\cos x}{2} - \frac{\cos 7x}{14} + C$

Step 1: Application of Trigonometric Product-to-Sum Identities

To evaluate the integral $I = \int \cos 2x \cos 4x \cos 6x \, dx$, we first simplify the product of the trigonometric functions. Using the identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ [Per Trigonometric Identities], we combine two of the terms.

Let us group $\cos 6x$ and $\cos 4x$:

$I = \frac{1}{2} \int (2 \cos 6x \cos 4x) \cos 2x \, dx$

$I = \frac{1}{2} \int [\cos(6x + 4x) + \cos(6x - 4x)] \cos 2x \, dx$

$I = \frac{1}{2} \int (\cos 10x + \cos 2x) \cos 2x \, dx$

Step 2: Expansion of the Integral

Distribute $\cos 2x$ into the parentheses:

$I = \frac{1}{2} \int (\cos 10x \cos 2x + \cos^2 2x) \, dx$

Apply the identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ to $\cos 10x \cos 2x$ and the power-reduction identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ to $\cos^2 2x$:

$I = \frac{1}{2} \int \left[ \frac{1}{2}(\cos 12x + \cos 8x) + \frac{1 + \cos 4x}{2} \right] \, dx$

$I = \frac{1}{4} \int (\cos 12x + \cos 8x + 1 + \cos 4x) \, dx$

Step 3: Integration

Integrate each term individually using the standard integral $\int \cos(ax) \, dx = \frac{\sin(ax)}{a} + C$:

$I = \frac{1}{4} \left[ \int \cos 12x \, dx + \int \cos 8x \, dx + \int 1 \, dx + \int \cos 4x \, dx \right]$

$I = \frac{1}{4} \left[ \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + x + \frac{\sin 4x}{4} \right] + C$

Step 4: Final Algebraic Simplification

Distribute the constant $\frac{1}{4}$ to obtain the final expression:

$I = \frac{\sin 12x}{48} + \frac{\sin 8x}{32} + \frac{x}{4} + \frac{\sin 4x}{16} + C$

Visual Representation of Integral Decomposition

Final Solution: $\frac{\sin 12x}{48} + \frac{\sin 8x}{32} + \frac{\sin 4x}{16} + \frac{x}{4} + C$

Given Variables & Initial Setup

We are required to evaluate the indefinite integral:

$I = \int \tan^3(2x) \sec(2x) \, dx$

Step 1: Decomposition of the Integrand

To facilitate the method of substitution, we utilize the Pythagorean identity $\tan^2(\theta) = \sec^2(\theta) - 1$ [By the fundamental trigonometric identity $\sin^2\theta + \cos^2\theta = 1$, divided by $\cos^2\theta$].

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

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

Step 2: Method of Substitution

Let $u = \sec(2x)$.

Differentiating both sides with respect to $x$ using the Chain Rule [$\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$]:

$\frac{du}{dx} = \sec(2x) \tan(2x) \cdot 2$

$du = 2 \sec(2x) \tan(2x) \, dx$

Therefore, $\sec(2x) \tan(2x) \, dx = \frac{du}{2}$

Step 3: Variable Transformation and Integration

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

$I = \int (u^2 - 1) \frac{du}{2}$

$I = \frac{1}{2} \left[ \int u^2 \, du - \int 1 \, du \right]$

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

$I = \frac{1}{2} \left[ \frac{u^3}{3} - u \right] + C$

$I = \frac{u^3}{6} - \frac{u}{2} + C$

Step 4: Re-substitution

Replace $u$ with $\sec(2x)$ to return to the original variable $x$:

$I = \frac{\sec^3(2x)}{6} - \frac{\sec(2x)}{2} + C$

Visual Representation of the Substitution Logic

Final Solution: $\frac{1}{6} \sec^3(2x) - \frac{1}{2} \sec(2x) + C$

Step 1: Problem Definition and Algebraic Manipulation

The integral to be evaluated is:
$I = \int \frac{1}{\sin x \cos^3 x} \, dx$

To simplify the integrand, we multiply both the numerator and the denominator by $\cos x$. This allows us to convert the trigonometric expressions into $\tan x$ and $\sec x$ terms, which are easier to integrate using the substitution method.

$I = \int \frac{\cos x}{\sin x \cos^4 x} \, dx = \int \frac{1}{\left( \frac{\sin x}{\cos x} \right) \cos^4 x} \, dx$

By the identity $\frac{1}{\cos^4 x} = \sec^4 x$, we have:
$I = \int \frac{\sec^4 x}{\tan x} \, dx$

Step 2: Preparing for Substitution

We know that $\sec^4 x = \sec^2 x \cdot \sec^2 x = (1 + \tan^2 x) \sec^2 x$ [Per the Pythagorean identity $\sec^2 x = 1 + \tan^2 x$]. Substituting this into the integral:

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

Step 3: Variable Substitution

Let $u = \tan x$.
Then, the derivative with respect to $x$ is:
$\frac{du}{dx} = \sec^2 x \implies du = \sec^2 x \, dx$

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

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

Step 4: Evaluating the Integral

Splitting the integrand into partial fractions:

$I = \int \left( \frac{1}{u} + \frac{u^2}{u} \right) \, du = \int \left( \frac{1}{u} + u \right) \, du$

Integrating with respect to $u$ [Using the power rule $\int u^n \, du = \frac{u^{n+1}}{n+1}$ and the logarithmic rule $\int \frac{1}{u} \, du = \ln|u|$]:

$I = \ln|u| + \frac{u^2}{2} + C$

Step 5: Back-Substitution

Replacing $u$ with $\tan x$ to return to the original variable:

$I = \ln|\tan x| + \frac{\tan^2 x}{2} + C$

Visual Representation of Function Behavior

Final Solution: $\ln|\tan x| + \frac{1}{2}\tan^2 x + C$

Step 1: Application of Trigonometric Identities

To integrate the function $f(x) = \frac{1 - \cos x}{1 + \cos x}$, we utilize the half-angle formulas derived from the double-angle identities $\cos(2\theta) = 1 - 2\sin^2\theta$ and $\cos(2\theta) = 2\cos^2\theta - 1$. Specifically:

• $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)$ [Per the identity $\cos x = 1 - 2\sin^2(x/2)$]
• $1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$ [Per the identity $\cos x = 2\cos^2(x/2) - 1$]

Substituting these into the integral:

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

Step 2: Algebraic Simplification

Cancel the constant factor $2$ and express the ratio of sine and cosine as tangent:

$$I = \int \tan^2\left(\frac{x}{2}\right) dx$$

Step 3: Conversion via Pythagorean Identity

Since $\tan^2\theta = \sec^2\theta - 1$ [By the Pythagorean trigonometric identity $\sec^2\theta - \tan^2\theta = 1$], we rewrite the integral:

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

Step 4: Integration and Substitution

Distribute the integral operator:

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

Using the standard integral formula $\int \sec^2(ax) dx = \frac{1}{a}\tan(ax) + C$, where $a = \frac{1}{2}$:

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

Simplifying the coefficient:

$$I = 2\tan\left(\frac{x}{2}\right) - x + C$$

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

Step 1: Application of Trigonometric Identity

To integrate $\int \sin^2(2x+5) \, dx$, we must first utilize the power-reduction formula for the sine function. Per the identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$, we let $\theta = (2x+5)$.

Substituting into the integral:

$\int \sin^2(2x+5) \, dx = \int \frac{1 - \cos(2(2x+5))}{2} \, dx$

Step 2: Algebraic Simplification

Distribute the constant $2$ within the argument of the cosine function and extract the scalar coefficient $\frac{1}{2}$ from the integral:

$\frac{1}{2} \int (1 - \cos(4x + 10)) \, dx$

Step 3: Execution of Integration

Apply the linearity property of integration [Per the Sum/Difference Rule of Integration]:

$\frac{1}{2} \left[ \int 1 \, dx - \int \cos(4x + 10) \, dx \right]$

Integrating the terms individually:

• $\int 1 \, dx = x$
• $\int \cos(4x + 10) \, dx = \frac{\sin(4x + 10)}{4}$ [Per the integration rule $\int \cos(ax+b) \, dx = \frac{\sin(ax+b)}{a} + C$]

Step 4: Final Assembly and Constant of Integration

Multiply the result by the scalar $\frac{1}{2}$ and append the arbitrary constant $C$:

$\frac{1}{2} \left( x - \frac{\sin(4x + 10)}{4} \right) + C$

$\frac{x}{2} - \frac{\sin(4x + 10)}{8} + C$

Final Solution: $\frac{x}{2} - \frac{\sin(4x+10)}{8} + C$

Step 1: Application of Trigonometric Identities

To evaluate the integral $I = \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx$, we utilize the double-angle formula for cosine: $\cos 2\theta = 2\cos^2 \theta - 1$ [Per Trigonometric Double-Angle Identity].

Substituting this identity into the numerator:

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

Simplifying the expression by cancelling the constants:

$I = \int \frac{2\cos^2 x - 2\cos^2 \alpha}{\cos x - \cos \alpha} \, dx$ $I = 2 \int \frac{\cos^2 x - \cos^2 \alpha}{\cos x - \cos \alpha} \, dx$

Step 2: Algebraic Factorization

The numerator $\cos^2 x - \cos^2 \alpha$ follows the algebraic difference of squares identity: $a^2 - b^2 = (a - b)(a + b)$ [By Algebraic Identity].

$I = 2 \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} \, dx$

Assuming $\cos x \neq \cos \alpha$, we cancel the common term $(\cos x - \cos \alpha)$:

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

Step 3: Integration and Linearity

Apply the linearity property of integration $\int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx$ [Per Fundamental Theorem of Calculus].

$I = 2 \left[ \int \cos x \, dx + \int \cos \alpha \, dx \right]$

Note that $\alpha$ is a constant with respect to $x$, therefore $\cos \alpha$ acts as a constant multiplier.

$I = 2 [\sin x + x \cos \alpha] + C$

Distributing the coefficient $2$:

$I = 2 \sin x + 2x \cos \alpha + C$

Visual Representation of the Integrand

Final Solution: 2 \sin x + 2x \cos \alpha + C

Step 1: Application of Trigonometric Identity

To integrate the product of two sine functions, we utilize the product-to-sum trigonometric identity: $\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]$ [By Product-to-Sum Identity].

Given the integral: $I = \int \sin(4x) \sin(8x) \, dx$

Let $A = 8x$ and $B = 4x$ (choosing the larger argument first simplifies the sign): $I = \int \frac{1}{2} [\cos(8x - 4x) - \cos(8x + 4x)] \, dx$

Step 2: Simplification of the Integrand

Substituting the values into the integral: $I = \frac{1}{2} \int [\cos(4x) - \cos(12x)] \, dx$

Step 3: Execution of Term-by-Term Integration

Applying the linearity property of integrals [$\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx$]: $I = \frac{1}{2} \left[ \int \cos(4x) \, dx - \int \cos(12x) \, dx \right]$

Using the standard integral formula $\int \cos(kx) \, dx = \frac{\sin(kx)}{k} + C$:

• For $\int \cos(4x) \, dx$, we get $\frac{\sin(4x)}{4}$
• For $\int \cos(12x) \, dx$, we get $\frac{\sin(12x)}{12}$

Step 4: Algebraic Consolidation

Combining the results: $I = \frac{1}{2} \left[ \frac{\sin(4x)}{4} - \frac{\sin(12x)}{12} \right] + C$

Distributing the scalar $\frac{1}{2}$: $I = \frac{\sin(4x)}{8} - \frac{\sin(12x)}{24} + C$

Visual Representation of Integral Decomposition

Final Solution: $\frac{\sin(4x)}{8} - \frac{\sin(12x)}{24} + C$

Step 1: Analyzing the Denominator and Trigonometric Identities

The integral to be solved is $I = \int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx$.

First, recall the fundamental trigonometric identity: $\sin^2 x + \cos^2 x = 1$ [Pythagorean Identity].

Second, recall the double angle identity: $\sin 2x = 2 \sin x \cos x$.

Substitute these into the denominator expression $1 + \sin 2x$:

$1 + \sin 2x = \cos^2 x + \sin^2 x + 2 \sin x \cos x$

This expression takes the form $a^2 + b^2 + 2ab$, which simplifies to $(a+b)^2$ [Algebraic Expansion Identity].

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

Step 2: Substitution Methodology

The integral becomes:

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

Let $u = \cos x + \sin x$.

Differentiate $u$ with respect to $x$ to find $du$:

$\frac{du}{dx} = -\sin x + \cos x$

$du = (\cos x - \sin x) \, dx$ [Chain Rule/Derivative of Trigonometric Functions].

Step 3: Evaluating the Integral

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

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

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

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

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

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

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

Step 4: Back-Substitution

Replace $u$ with the original substitution $u = \cos x + \sin x$:

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

Visual Representation of Function Relationship

Final Solution: -\frac{1}{\sin x + \cos x} + C

Step 1: Establishing the Integral Expression

The given integral is: $I = \int \sin^3 x \cos^3 x \, dx$.

Step 2: Decomposition of the Integrand

To facilitate the method of substitution, we decompose the power functions while keeping one cosine term for the differential $du$.

$I = \int \sin^3 x \cdot \cos^2 x \cdot \cos x \, dx$ [Using index laws $a^{m+n} = a^m \cdot a^n$]

Utilize the trigonometric identity $\cos^2 x = 1 - \sin^2 x$ [Per the Pythagorean Identity $\sin^2 x + \cos^2 x = 1$]:

$I = \int \sin^3 x (1 - \sin^2 x) \cos x \, dx$

Step 3: Variable Substitution

Let $u = \sin x$.

Differentiating both sides with respect to $x$ gives: $\frac{du}{dx} = \cos x$, which implies $du = \cos x \, dx$ [By the Chain Rule of differentiation].

Step 4: Algebraic Transformation of the Integral

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

$I = \int u^3 (1 - u^2) \, du$

$I = \int (u^3 - u^5) \, du$ [By the distributive property of multiplication over addition/subtraction]

Step 5: Applying the Power Rule for Integration

Apply the Power Rule $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$:

$I = \int u^3 \, du - \int u^5 \, du$

$I = \frac{u^4}{4} - \frac{u^6}{6} + C$ [Where $C$ is the constant of integration]

Step 6: Back-Substitution

Replace $u$ with $\sin x$ to return to the original variable:

$I = \frac{\sin^4 x}{4} - \frac{\sin^6 x}{6} + C$

Analytical Visual Representation of the Substitution Strategy

Final Solution: $\frac{\sin^4 x}{4} - \frac{\sin^6 x}{6} + C$

Given Integral

To evaluate the integral: $I = \int \sin^4 x \, dx$

Step 1: Trigonometric Reduction using Power-Reduction Identities

We utilize the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$ [By Trigonometric Power-Reduction Formula].

Expressing $\sin^4 x$ as $(\sin^2 x)^2$:

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

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

Step 2: Applying the Cosine Power-Reduction Identity

For the term $\cos^2 2x$, use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$. Setting $\theta = 2x$, we get $\cos^2 2x = \frac{1 + \cos 4x}{2}$ [By Trigonometric Identity].

Substituting this into the integral:

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

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

$I = \frac{1}{4} \int \left( \frac{3}{2} - 2\cos 2x + \frac{1}{2}\cos 4x \right) \, dx$

Step 3: Integration Term-by-Term

Distribute the constant and integrate using the rule $\int \cos(ax) dx = \frac{\sin(ax)}{a} + C$ [By Fundamental Theorem of Calculus]:

$I = \frac{1}{4} \left[ \frac{3}{2}x - 2 \cdot \frac{\sin 2x}{2} + \frac{1}{2} \cdot \frac{\sin 4x}{4} \right] + C$

$I = \frac{1}{4} \left[ \frac{3}{2}x - \sin 2x + \frac{1}{8}\sin 4x \right] + C$

Step 4: Final Algebraic Simplification

Multiply the constant $1/4$ through the expression:

$I = \frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C$

Graphical Representation of the Function $\sin^4 x$

Final Solution: $\frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C$

Step 1: Application of Trigonometric Power-Reduction Identities

The goal is to integrate $I = \int \cos^4 2x \, dx$. To reduce the power of the trigonometric function, we utilize the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ [Per the Cosine Double Angle Formula].

Setting $\theta = 2x$, we have:

$\cos^2 2x = \frac{1 + \cos 4x}{2}$

Therefore, the integral becomes:

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

Step 2: Algebraic Expansion

Expanding the squared expression:

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

Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ once more for the $\cos^2 4x$ term, where $\theta = 4x$:

$\cos^2 4x = \frac{1 + \cos 8x}{2}$

Substituting this back into the integral:

$I = \frac{1}{4} \int \left( 1 + 2\cos 4x + \frac{1 + \cos 8x}{2} \right) \, dx$

Step 3: Simplifying the Integrand

Combine the constants and separate the terms:

$I = \frac{1}{4} \int \left( 1 + \frac{1}{2} + 2\cos 4x + \frac{1}{2}\cos 8x \right) \, dx$

$I = \frac{1}{4} \int \left( \frac{3}{2} + 2\cos 4x + \frac{1}{2}\cos 8x \right) \, dx$

Step 4: Integration

Applying the linearity of the integral operator and the standard rule $\int \cos(ax) \, dx = \frac{\sin(ax)}{a} + C$:

$I = \frac{1}{4} \left[ \frac{3}{2}x + 2\left(\frac{\sin 4x}{4}\right) + \frac{1}{2}\left(\frac{\sin 8x}{8}\right) \right] + C$

$I = \frac{1}{4} \left[ \frac{3}{2}x + \frac{1}{2}\sin 4x + \frac{1}{16}\sin 8x \right] + C$

Step 5: Final Simplification

Distribute the constant $1/4$:

$I = \frac{3}{8}x + \frac{1}{8}\sin 4x + \frac{1}{64}\sin 8x + C$

Visual Representation of Function Periodicity

Final Solution: $\frac{3x}{8} + \frac{\sin 4x}{8} + \frac{\sin 8x}{64} + C$

Step 1: Analyzing the Integrand

The given integral is: $I = \int \frac{\cos 2x}{(\cos x + \sin x)^2} \, dx$

Recall the trigonometric identity for the double angle of cosine: $\cos 2x = \cos^2 x - \sin^2 x$ [Per trigonometric identity: $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$].

Step 2: Factoring the Numerator

Substitute the identity into the integral:

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

Apply the algebraic difference of squares formula, $a^2 - b^2 = (a - b)(a + b)$, to the numerator:

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

Step 3: Simplification

Cancel the common term $(\cos x + \sin x)$ from the numerator and the denominator, noting that $\cos x + \sin x \neq 0$ for the integral to be defined:

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

Step 4: Substitution Method

Let $u = \cos x + \sin x$.

Differentiating with respect to $x$ [By the Sum Rule and standard derivatives]:

$\frac{du}{dx} = -\sin x + \cos x = \cos x - \sin x$

Thus, $du = (\cos x - \sin x) \, dx$.

Step 5: Evaluating the Integral

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

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

Applying the standard integral formula $\int \frac{1}{u} \, du = \ln|u| + C$ [Per standard logarithmic integral rule]:

$I = \ln|u| + C$

Step 6: Back-Substitution

Substitute $u = \cos x + \sin x$ back into the result:

$I = \ln|\cos x + \sin x| + C$

Final Solution: $\ln|\sin x + \cos x| + C$

Step 1: Application of Trigonometric Product-to-Sum Identities

To evaluate the integral $I = \int \sin x \sin 2x \sin 3x \, dx$, we first rearrange the terms to utilize the product-to-sum identity. Recall the identity: $\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]$.

Setting $A = 3x$ and $B = x$, we have:

$\sin 3x \sin x = \frac{1}{2} [\cos(3x - x) - \cos(3x + x)] = \frac{1}{2} (\cos 2x - \cos 4x)$ [Trigonometric Identity]

Step 2: Substitution into the Integral

Substituting this back into the original integral expression:

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

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

Step 3: Simplification via Double Angle Identity

We note that $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$ [By the identity $\sin 2\theta = 2\sin \theta \cos \theta$].

For the term $\sin 2x \cos 4x$, we apply the product-to-sum identity again: $\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$.

With $A = 2x$ and $B = 4x$:

$\sin 2x \cos 4x = \frac{1}{2} [\sin(6x) + \sin(-2x)] = \frac{1}{2} (\sin 6x - \sin 2x)$

Step 4: Final Integration

Substituting these into the integral expression:

$I = \frac{1}{2} \int \left( \frac{1}{2} \sin 4x - \frac{1}{2} (\sin 6x - \sin 2x) \right) \, dx$

$I = \frac{1}{4} \int (\sin 4x - \sin 6x + \sin 2x) \, dx$

Performing the integration term by term using $\int \sin(ax) dx = -\frac{\cos(ax)}{a} + C$:

$I = \frac{1}{4} \left[ -\frac{\cos 4x}{4} - \left( -\frac{\cos 6x}{6} \right) + \left( -\frac{\cos 2x}{2} \right) \right] + C$

Step 5: Algebraic Consolidation

$I = \frac{1}{4} \left[ -\frac{\cos 4x}{4} + \frac{\cos 6x}{6} - \frac{\cos 2x}{2} \right] + C$

$I = -\frac{\cos 4x}{16} + \frac{\cos 6x}{24} - \frac{\cos 2x}{8} + C$

Final Solution: $-\frac{\cos 4x}{16} + \frac{\cos 6x}{24} - \frac{\cos 2x}{8} + C$

Step 1: Setting up the Integral

The integral to be evaluated is:

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

To simplify the integrand, we employ a trigonometric manipulation strategy. We introduce a constant factor in the numerator to allow for expansion using the sine addition formula [Per the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$].

Step 2: Algebraic Manipulation

Multiply and divide the integrand by $\sin(a-b)$:

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

Express the numerator $(a-b)$ as $(x-b) - (x-a)$ to match the arguments in the denominator:

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

Step 3: Applying Trigonometric Expansion

Using the expansion identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$, where $A = (x-b)$ and $B = (x-a)$:

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

Split the integral into two separate terms [By the linearity of the integral operator]:

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

Step 4: Simplification and Integration

Cancel the common terms in each fraction:

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

Apply the standard integral formula $\int \tan u \, du = \ln|\sec u| + C$ [Per the fundamental theorem of calculus]:

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

Step 5: Final Logarithmic Simplification

Using the property of logarithms $\ln M - \ln N = \ln(\frac{M}{N})$:

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

Since $\sec \theta = \frac{1}{\cos \theta}$, we have $\frac{\sec(x-b)}{\sec(x-a)} = \frac{\cos(x-a)}{\cos(x-b)}$:

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

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

Step 1: Application of Trigonometric Identity

To integrate $\int \sin^3(2x+1) \, dx$, we utilize the triple-angle identity for sine: $\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)$.

Rearranging for $\sin^3(\theta)$, we obtain: $\sin^3(\theta) = \frac{3\sin(\theta) - \sin(3\theta)}{4}$

Substituting $\theta = (2x+1)$, the integral becomes: $I = \int \frac{3\sin(2x+1) - \sin(3(2x+1))}{4} \, dx$ [Per Trigonometric Identity Substitution]

Step 2: Decomposition and Constant Extraction

Distribute the integral operator and extract the constant factor $\frac{1}{4}$: $I = \frac{1}{4} \left[ 3 \int \sin(2x+1) \, dx - \int \sin(6x+3) \, dx \right]$

Step 3: Integration via Substitution Method

We apply the rule $\int \sin(ax+b) \, dx = -\frac{1}{a}\cos(ax+b) + C$ [By standard integration formula for linear arguments]:

• For the first term: $\int \sin(2x+1) \, dx = -\frac{1}{2}\cos(2x+1)$
• For the second term: $\int \sin(6x+3) \, dx = -\frac{1}{6}\cos(6x+3)$

Step 4: Algebraic Synthesis

Substituting the results back into the expression: $I = \frac{1}{4} \left[ 3 \left( -\frac{1}{2}\cos(2x+1) \right) - \left( -\frac{1}{6}\cos(6x+3) \right) \right] + C$

$I = \frac{1}{4} \left[ -\frac{3}{2}\cos(2x+1) + \frac{1}{6}\cos(6x+3) \right] + C$

Distribute the $\frac{1}{4}$: $I = -\frac{3}{8}\cos(2x+1) + \frac{1}{24}\cos(6x+3) + C$

Visual Representation of Integral Decomposition

Final Solution: -\frac{3}{8}\cos(2x+1) + \frac{1}{24}\cos(6x+3) + C