Techniques of Integration

This chapter rigorously examines fundamental techniques for evaluating integrals, focusing on substitution and integration by parts. Mastery of these methods is critical for advanced calculus applications and is a frequently assessed component in CMI examinations, underpinning solutions to a wide range of analytical problems.

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Chapter Contents

|

| Topic |

|---|-------| | 1 | Integration by Substitution | | 2 | Integration by Parts |

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We begin with Integration by Substitution.

Part 1: Integration by Substitution

Integration by substitution is a fundamental technique for simplifying integrals by transforming them into a more manageable form. We use this method to reverse the chain rule, making complex integrands solvable by recognizing a function and its derivative within the expression.

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Core Concepts

1. The Fundamental Substitution Rule for Indefinite Integrals

We define the substitution rule for indefinite integrals to simplify expressions of the form $\int f(g(x))g'(x) \,dx$. By letting $u = g(x)$, we find $du = g'(x) \,dx$, transforming the integral into $\int f(u) \,du$.

Worked Example: Evaluate $\int x \sqrt{x^2+1} \,dx$.

Step 1: Identify $u$ and $du$.

> Let $u = x^2+1$.
> Then, $du = 2x \,dx$.
> We observe that $x \,dx = \frac{1}{2} \,du$.

Step 2: Substitute into the integral.

>

Step 3: Integrate with respect to $u$.

>

>
>

Step 4: Substitute back $u = x^2+1$.

>

Answer: $\frac{1}{3} (x^2+1)^{3/2} + C$

:::question type="MCQ" question="Evaluate the indefinite integral $\int \frac{e^x}{1+e^x} \,dx$." options=["$\ln|1+e^x| + C$","$(1+e^x)^2 + C$","$-\frac{1}{(1+e^x)^2} + C$","$e^x \ln|1+e^x| + C$"] answer="$\ln|1+e^x| + C$" hint="Let $u = 1+e^x$." solution="Step 1: Let $u = 1+e^x$.
> Then $du = e^x \,dx$.

Step 2: Substitute $u$ and $du$ into the integral.
>

Step 3: Integrate with respect to $u$.
>

Step 4: Substitute back $u = 1+e^x$.
>

"
:::

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2. Definite Integrals and Limit Transformation

When performing substitution in definite integrals, we must transform the limits of integration from $x$-values to $u$-values. This avoids the need to substitute back to $x$ at the end.

Worked Example: Evaluate $\int_0^1 x e^{-x^2} \,dx$.

Step 1: Identify $u$, $du$, and transform the limits of integration.

> Let $u = -x^2$.
> Then $du = -2x \,dx$.
> This means $x \,dx = -\frac{1}{2} \,du$.
>
> For the limits:
> When $x=0$, $u = -(0)^2 = 0$.
> When $x=1$, $u = -(1)^2 = -1$.

Step 2: Substitute into the integral with new limits.

>

Step 3: Integrate with respect to $u$.

>

>
>
>

Answer: $\frac{1}{2} \left(1 - \frac{1}{e}\right)$

:::question type="NAT" question="Compute the value of $\int_1^e \frac{\ln x}{x} \,dx$." answer="0.5" hint="Let $u = \ln x$ and transform the limits." solution="Step 1: Let $u = \ln x$.
> Then $du = \frac{1}{x} \,dx$.
>
> For the limits:
> When $x=1$, $u = \ln(1) = 0$.
> When $x=e$, $u = \ln(e) = 1$.

Step 2: Substitute into the integral with new limits.
>

Step 3: Integrate with respect to $u$.
>

Answer: 0.5"
:::

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3. Strategic Choice of Substitution

Identifying the correct substitution $u$ is crucial. We often choose $u$ to be the "inner function" of a composite function, or a term whose derivative is also present (or a constant multiple of it) in the integrand.

Worked Example: Evaluate $\int \frac{1}{x \ln x} \,dx$.

Step 1: Identify $u$ and $du$.

> We observe that the derivative of $\ln x$ is $\frac{1}{x}$.
> Let $u = \ln x$.
> Then $du = \frac{1}{x} \,dx$.

Step 2: Substitute into the integral.

>

Step 3: Integrate with respect to $u$.

>

Step 4: Substitute back $u = \ln x$.

>

Answer: $\ln|\ln x| + C$

:::question type="MCQ" question="Find $\int \frac{\sin(\sqrt{x})}{\sqrt{x}} \,dx$." options=["$-2 \cos(\sqrt{x}) + C$","$\cos(\sqrt{x}) + C$","$\frac{1}{2} \cos(\sqrt{x}) + C$","$- \cos(\sqrt{x}) + C$"] answer="$-2 \cos(\sqrt{x}) + C$" hint="Consider $u = \sqrt{x}$ and find $du$." solution="Step 1: Let $u = \sqrt{x} = x^{1/2}$.
> Then $du = \frac{1}{2} x^{-1/2} \,dx = \frac{1}{2\sqrt{x}} \,dx$.
> This implies $\frac{1}{\sqrt{x}} \,dx = 2 \,du$.

Step 2: Substitute into the integral.
>

Step 3: Integrate with respect to $u$.
>

Step 4: Substitute back $u = \sqrt{x}$.
>

"
:::

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4. Adjusting the Differential and Completing the Square

Sometimes, the substitution requires algebraic manipulation to express $dx$ in terms of $du$, or to express $x$ in terms of $u$. For quadratic denominators, completing the square often precedes substitution to reach a standard integral form.

Worked Example: Evaluate $\int \frac{1}{x^2+2x+5} \,dx$.

Step 1: Complete the square in the denominator.

> The denominator is $x^2+2x+5 = (x^2+2x+1) + 4 = (x+1)^2 + 2^2$.
> The integral becomes $\int \frac{1}{(x+1)^2 + 2^2} \,dx$.

Step 2: Identify $u$ and $du$.

> Let $u = x+1$.
> Then $du = \,dx$.

Step 3: Substitute into the integral.

>

Step 4: Integrate using the standard form $\int \frac{1}{y^2+a^2} \,dy = \frac{1}{a} \arctan\left(\frac{y}{a}\right) + C$.

> Here,$y=u$ and $a=2$.
>

Step 5: Substitute back $u = x+1$.

>

Answer: $\frac{1}{2} \arctan\left(\frac{x+1}{2}\right) + C$

:::question type="MCQ" question="Evaluate $\int \frac{x+1}{\sqrt{4-x^2-2x}} \,dx$." options=["$-\sqrt{4-x^2-2x} + C$","$\arcsin\left(\frac{x+1}{\sqrt{5}}\right) + C$","$\frac{1}{2}\sqrt{4-x^2-2x} + C$","$\sqrt{4-x^2-2x} + C$"] answer="$\arcsin\left(\frac{x+1}{\sqrt{5}}\right) + C$" hint="Complete the square in the denominator and consider $u=x+1$." solution="Step 1: Complete the square in the denominator.
> $4-x^2-2x = 4-(x^2+2x) = 4-(x^2+2x+1-1) = 4-( (x+1)^2 - 1) = 5-(x+1)^2$.
> The integral becomes $\int \frac{x+1}{\sqrt{5-(x+1)^2}} \,dx$.

Step 2: Identify $u$ and $du$.
> Let $u = x+1$.
> Then $du = \,dx$.
> The integral is now $\int \frac{u}{\sqrt{5-u^2}} \,du$.
> This looks like another substitution is needed for $5-u^2$. Let $v = 5-u^2$. Then $dv = -2u \,du$, so $u \,du = -\frac{1}{2} \,dv$.

Step 3: Substitute $v$ into the integral.
>

Step 4: Integrate with respect to $v$.
>

Step 5: Substitute back $v = 5-u^2$ and $u = x+1$.
>

>
Wait, this is incorrect. The derivative of $\arcsin\left(\frac{u}{a}\right)$ is $\frac{1}{\sqrt{a^2-u^2}} du$.
Let's re-evaluate the question and my solution. The question asks for $\int \frac{x+1}{\sqrt{4-x^2-2x}} \,dx$.
My first substitution was $u=x+1$, so $du=dx$. The integral became $\int \frac{u}{\sqrt{5-u^2}} \,du$.
This is NOT $\int \frac{1}{\sqrt{a^2-u^2}} \,du$. It's $\int u(5-u^2)^{-1/2} \,du$.
Let's re-do the example solution:

Step 1: Complete the square in the denominator.
> $4-x^2-2x = 4-(x^2+2x) = 4-(x^2+2x+1-1) = 4-((x+1)^2-1) = 5-(x+1)^2$.
> The integral becomes $\int \frac{x+1}{\sqrt{5-(x+1)^2}} \,dx$.

Step 2: Identify $u$ and $du$.
> Let $u = x+1$.
> Then $du = \,dx$.
> The integral becomes $\int \frac{u}{\sqrt{5-u^2}} \,du$.

Step 3: Perform a second substitution. Let $v = 5-u^2$.
> Then $dv = -2u \,du$, so $u \,du = -\frac{1}{2} \,dv$.
> The integral becomes $\int \frac{1}{\sqrt{v}} \left(-\frac{1}{2}\right) \,dv = -\frac{1}{2} \int v^{-1/2} \,dv$.

Step 4: Integrate with respect to $v$.
>

Step 5: Substitute back $v = 5-u^2$ and $u = x+1$.
>

The provided answer is $\arcsin\left(\frac{x+1}{\sqrt{5}}\right) + C$. This implies the numerator should be $1$ not $x+1$.
Let me check the question itself.
If the question was $\int \frac{1}{\sqrt{4-x^2-2x}} \,dx$, then the answer would be $\arcsin\left(\frac{x+1}{\sqrt{5}}\right) + C$.
The current question $\int \frac{x+1}{\sqrt{4-x^2-2x}} \,dx$ leads to $-\sqrt{4-x^2-2x} + C$.
I need to ensure my question and answer are consistent.
I will change the question to match the $\arcsin$ form. This is a common pattern for substitution.

Corrected Question & Solution:
:::question type="MCQ" question="Evaluate $\int \frac{1}{\sqrt{4-x^2-2x}} \,dx$." options=["$\arcsin\left(\frac{x+1}{\sqrt{5}}\right) + C$","$\ln|x+1 + \sqrt{4-x^2-2x}| + C$","$\frac{1}{2}\ln\left|\frac{\sqrt{5}+x+1}{\sqrt{5}-(x+1)}\right| + C$","$\sqrt{4-x^2-2x} + C$"] answer="$\arcsin\left(\frac{x+1}{\sqrt{5}}\right) + C$" hint="Complete the square in the denominator and use the standard integral form for $\arcsin$." solution="Step 1: Complete the square in the denominator.
> $4-x^2-2x = 4-(x^2+2x) = 4-(x^2+2x+1-1) = 4-((x+1)^2-1) = 5-(x+1)^2$.
> The integral becomes $\int \frac{1}{\sqrt{5-(x+1)^2}} \,dx$.

Step 2: Identify $u$ and $du$.
> Let $u = x+1$.
> Then $du = \,dx$.
> The integral becomes $\int \frac{1}{\sqrt{5-u^2}} \,du$.

Step 3: Integrate using the standard form $\int \frac{1}{\sqrt{a^2-y^2}} \,dy = \arcsin\left(\frac{y}{a}\right) + C$.
> Here, $y=u$ and $a=\sqrt{5}$.
>

Step 4: Substitute back $u = x+1$.
>

"
:::

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Advanced Applications

Substitution can simplify complex integrands that may initially appear intractable. These examples often involve a clever choice of $u$ or lead to one of the inverse trigonometric forms.

Worked Example: Evaluate $\int \frac{e^{2x}}{\sqrt{1-e^{4x}}} \,dx$.

Step 1: Identify a suitable substitution.

> We observe $e^{4x} = (e^{2x})^2$.
> Let $u = e^{2x}$.
> Then $du = 2e^{2x} \,dx$.
> This means $e^{2x} \,dx = \frac{1}{2} \,du$.

Step 2: Substitute into the integral.

>

>

Step 3: Integrate using the standard form $\int \frac{1}{\sqrt{a^2-y^2}} \,dy = \arcsin\left(\frac{y}{a}\right) + C$.

> Here, $y=u$ and $a=1$.
>

Step 4: Substitute back $u = e^{2x}$.

>

Answer: $\frac{1}{2} \arcsin(e^{2x}) + C$

:::question type="NAT" question="Calculate $\int_0^{\pi/4} \frac{\sec^2 x}{1+\tan^2 x} \,dx$." answer="0.785398" hint="Simplify the integrand using a trigonometric identity first, then consider a substitution." solution="Step 1: Simplify the integrand using the identity $1+\tan^2 x = \sec^2 x$.
> The integrand becomes $\frac{\sec^2 x}{\sec^2 x} = 1$.
> So the integral is $\int_0^{\pi/4} 1 \,dx$.
> (Note: This question was intended to be a substitution where $u=\tan x$. Let me re-evaluate to make it a clear substitution question. The current one simplifies too much. I need to make the denominator not simplify to $\sec^2 x$. Let's make it $1+ \tan x$ or something else.)

Corrected Question & Solution:
:::question type="NAT" question="Calculate $\int_0^{\pi/4} \frac{\sec^2 x}{1+\tan x} \,dx$." answer="0.346573" hint="Let $u = 1+\tan x$ and transform the limits." solution="Step 1: Identify $u$, $du$, and transform the limits of integration.
> Let $u = 1+\tan x$.
> Then $du = \sec^2 x \,dx$.
>
> For the limits:
> When $x=0$, $u = 1+\tan(0) = 1+0 = 1$.
> When $x=\pi/4$, $u = 1+\tan(\pi/4) = 1+1 = 2$.

Step 2: Substitute into the integral with new limits.
>

Step 3: Integrate with respect to $u$.
>

>
> Numerically, $\ln(2) \approx 0.693147$.
Ah, the question asks for NAT, so it needs a numerical answer.
Wait, my previous calculation for $\frac{1}{2}(1-1/e)$ was $0.316$.
$\ln(2) \approx 0.693147$. The answer 0.346573 is $\frac{1}{2}\ln(2)$.
Let me check if I made a mistake in the question or the answer.
If the question was $\int_0^{\pi/4} \frac{\sec^2 x}{2(1+\tan x)} \,dx$, then the answer would be $\frac{1}{2}\ln(2)$.
Let me adjust the question to match the intended answer. Or I can just put $\ln(2)$ as the answer. I will put $\ln(2)$ as the answer for now.

Revised Answer for current question:
Answer: 0.693147
"
:::

Final check on the NAT question and solution.
Question: $\int_0^{\pi/4} \frac{\sec^2 x}{1+\tan x} \,dx$.
Solution yields $\ln(2)$. $\ln(2) \approx 0.69314718$.
The given answer in the prompt was 0.346573. This is approximately $\ln(2)/2$.
So, either the question should be $\int_0^{\pi/4} \frac{\sec^2 x}{2(1+\tan x)} \,dx$ or the answer should be $\ln(2)$.
I will use $\ln(2)$ as the answer for the question as stated.

Corrected Answer: 0.693147
(The previous value was incorrect for the stated question.)

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Evaluate $\int \frac{\cos x}{\sin^2 x} \,dx$." options=["$-\frac{1}{\sin x} + C$","$\ln|\sin x| + C$","$\frac{1}{\sin x} + C$","$\tan x + C$"] answer="$-\frac{1}{\sin x} + C$" hint="Let $u = \sin x$." solution="Step 1: Let $u = \sin x$.
> Then $du = \cos x \,dx$.

Step 2: Substitute into the integral.
>

Step 3: Integrate with respect to $u$.
>

Step 4: Substitute back $u = \sin x$.
>

"
:::

:::question type="NAT" question="Compute the definite integral $\int_0^1 (2x+1)^3 \,dx$." answer="10" hint="Use substitution $u=2x+1$ and change the limits." solution="Step 1: Let $u = 2x+1$.
> Then $du = 2 \,dx$, so $dx = \frac{1}{2} \,du$.
>
> For the limits:
> When $x=0$, $u = 2(0)+1 = 1$.
> When $x=1$, $u = 2(1)+1 = 3$.

Step 2: Substitute into the integral with new limits.
>

Step 3: Integrate with respect to $u$.
>

>

"
:::

:::question type="MSQ" question="Which of the following integrals can be solved using the substitution $u=\ln x$?" options=["$\int \frac{\ln x}{x} \,dx$","$\int \frac{1}{x (\ln x)^2} \,dx$","$\int \frac{1}{x \ln x} \,dx$","$\int \frac{x}{\ln x} \,dx$"] answer="$\int \frac{\ln x}{x} \,dx,\int \frac{1}{x (\ln x)^2} \,dx,\int \frac{1}{x \ln x} \,dx$" hint="For $u=\ln x$, we need $du=\frac{1}{x}\,dx$ to be present in the integrand." solution="For $u=\ln x$, we have $du = \frac{1}{x} \,dx$. We need to see if $\frac{1}{x} \,dx$ can be isolated in the integrand.

• Option 1: $\int \frac{\ln x}{x} \,dx$

Can be written as $\int (\ln x) \left(\frac{1}{x} \,dx\right)$. Here,$u=\ln x$ and $du=\frac{1}{x}\,dx$. This works.

• Option 2: $\int \frac{1}{x (\ln x)^2} \,dx$

Can be written as $\int \frac{1}{(\ln x)^2} \left(\frac{1}{x} \,dx\right)$. Here, $u=\ln x$ and $du=\frac{1}{x} \,dx$. This works.

• Option 3: $\int \frac{1}{x \ln x} \,dx$

Can be written as $\int \frac{1}{\ln x} \left(\frac{1}{x} \,dx\right)$. Here, $u=\ln x$ and $du=\frac{1}{x} \,dx$. This works.

• Option 4: $\int \frac{x}{\ln x} \,dx$

If $u=\ln x$, then $x=e^u$ and $dx=e^u\,du$. The integral becomes $\int \frac{e^u}{u} e^u\,du = \int \frac{e^{2u}}{u}\,du$. While a substitution was made, this does not simplify to a standard form and is generally not what is meant by 'solvable using $ u=\ln x

Therefore, options 1, 2, and 3 are correct."
:::

:::question type="MCQ" question="Evaluate $\int \frac{x^2}{\sqrt{1-x^6}} \,dx$." options=["$\frac{1}{3} \arcsin(x^3) + C$","$\frac{1}{2} \arcsin(x^3) + C$","$3 \arcsin(x^3) + C$","$\frac{1}{3} \arctan(x^3) + C$"] answer="$\frac{1}{3} \arcsin(x^3) + C$" hint="Consider $u=x^3$ and the standard integral form for $\arcsin$." solution="Step 1: Observe that $x^6 = (x^3)^2$.
> Let $u = x^3$.
> Then $du = 3x^2 \,dx$, so $x^2 \,dx = \frac{1}{3} \,du$.

Step 2: Substitute into the integral.
>

Step 3: Integrate using the standard form $\int \frac{1}{\sqrt{a^2-y^2}} \,dy = \arcsin\left(\frac{y}{a}\right) + C$.
> Here,$y=u$ and $a=1$.
>

Step 4: Substitute back $u = x^3$.
>

"
:::

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Summary

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What's Next?

in the context of direct simplification. The term $\frac{1}{x}\,dx$ is not readily available.

Therefore, options 1, 2, and 3 are correct."
:::

:::question type="MCQ" question="Evaluate $\int \frac{x^2}{\sqrt{1-x^6}} \,dx$." options=["$\frac{1}{3} \arcsin(x^3) + C$","$\frac{1}{2} \arcsin(x^3) + C$","$3 \arcsin(x^3) + C$","$\frac{1}{3} \arctan(x^3) + C$"] answer="$\frac{1}{3} \arcsin(x^3) + C$" hint="Consider $u=x^3$ and the standard integral form for $\arcsin$." solution="Step 1: Observe that $x^6 = (x^3)^2$.
> Let $u = x^3$.
> Then $du = 3x^2 \,dx$, so $x^2 \,dx = \frac{1}{3} \,du$.

Step 2: Substitute into the integral.
>

Step 3: Integrate using the standard form $\int \frac{1}{\sqrt{a^2-y^2}} \,dy = \arcsin\left(\frac{y}{a}\right) + C$.
> Here, $y=u$ and $a=1$.
>

Step 4: Substitute back $u = x^3$.
>

"
:::

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Summary

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What's Next?

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Part 2: Integration by Parts

Integration by parts is a fundamental technique for integrating products of functions, transforming a complex integral into a potentially simpler one. This method is crucial for solving various integrals encountered in computer science applications, particularly in probability, signal processing, and algorithm analysis.

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Core Concepts

1. The Integration by Parts Formula

We use integration by parts to integrate the product of two functions. The formula is derived from the product rule for differentiation.

Worked Example: Integrate $\int x e^x \, dx$.

Step 1: Choose $u$ and $dv$.

We choose $u = x$ because its derivative, $du = dx$, is simpler. We choose $dv = e^x \, dx$ because it is easily integrable.

> $u = x \implies du = dx$
> $dv = e^x \, dx \implies v = \int e^x \, dx = e^x$

Step 2: Apply the integration by parts formula.

>

> $\int x e^x \, dx = x e^x - \int e^x \, dx

\int x e^x \, dx = x e^x - e^x + C

\int u \, dv = uv - \int v \, du

\int x \cos x \, dx = x \sin x - \int \sin x \, dx

\int x \cos x \, dx = x \sin x - (-\cos x) + C

\int x \cos x \, dx = x \sin x + \cos x + C

\int u \, dv = uv - \int v \, du

\int x \ln x \, dx = (\ln x)\left(\frac{x^2}{2}\right) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx

\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x}{2} \, dx

\int x \ln x \, dx = \frac{x^2}{2} \ln x - \frac{1}{2} \cdot \frac{x^2}{2} + C

\int x \ln x \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C

\int u \, dv = uv - \int v \, du
>

Step 3: Evaluate the remaining integral using a substitution.
Let $w = 1+x^2$, so $dw = 2x \, dx$, or $x \, dx = \frac{1}{2} dw$.
>

> $= \frac{1}{2} \ln(1+x^2) + C \text{ (since } 1+x^2 > 0 \text{)}

\int \arctan x \, dx = x \arctan x - \frac{1}{2} \ln(1+x^2) + C

\int x^2 e^x \, dx = + (x^2)(e^x) - (2x)(e^x) + (2)(e^x) - \int (0)(e^x) \, dx

\int x^2 e^x \, dx = x^2 e^x - 2x e^x + 2e^x + C

\int x^3 \sin x \, dx = +(x^3)(-\cos x) - (3x^2)(-\sin x) + (6x)(\cos x) - (6)(\sin x) + C

\int x^3 \sin x \, dx = -x^3 \cos x + 3x^2 \sin x + 6x \cos x - 6 \sin x + C

I = \frac{1}{2} e^{2x} \cos(3x) - \int \frac{1}{2} e^{2x} (-3 \sin(3x)) \, dx

Step 2: Apply integration by parts a second time to the new integral.

Consider $\int e^{2x} \sin(3x) \, dx$.
Choose $u = \sin(3x)$ and $dv = e^{2x} \, dx$.

> $u = \sin(3x) \implies du = 3 \cos(3x) \, dx$
> $dv = e^{2x} \, dx \implies v = \frac{1}{2} e^{2x}$

>

>

Step 3: Substitute the second result back into the first equation.

Notice that $\int e^{2x} \cos(3x) \, dx$ is the original integral $I$.

>

>

Step 4: Solve for $I$.

>

> \frac{13}{4} I = \frac{1}{4} e^{2x} (2 \cos(3x) + 3 \sin(3x))<div class="math-display"><span class="katex-error" title="ParseError: KaTeX parse error: Can & #x27;t use function & #x27;' in math mode at position 3: > ̲ I = \frac{4}{1…" style="color:#cc0000"> & gt; I = \frac{4}{13} \cdot \frac{1}{4} e^{2x} (2 \cos(3x) + 3 \sin(3x)) + C