# Which of the following integrals would be substitution? A) \int sin(\theta)e^{\theta}d{\theta} B)...

## Question:

Which of the following integrals would be substitution?

A) {eq}\int sin(\theta)e^{\theta}d{\theta} {/eq}

B){eq}\int \frac{3}{\sqrt{x^2-7}}dx {/eq}

C){eq}\int (\frac{\sqrt{x^3}+7x^2+x}{x})dx {/eq}

D) {eq}\int sin^2(x)cos^3(x)dx {/eq}

## Indefinite Integration:

Indefinite integral can be solve by directly by formula or can solve by substitution method.

Some Important Formula shown below:

{eq}* \int secx dx= \ln |secx +tanx|+C {/eq}

{eq}\int x^n dx=\frac{x^{n+1}}{n+1}+C {/eq}

{eq}\int \frac{dx}{\sqrt{x^2-a^2}}= \ln \left[\frac{x}{a}+\frac{1}{a}{\sqrt{x^2-a^2}} \right]+C {/eq}

Integration by-parts: let u and v be a function of x.

{eq}* \int u \cdot v dx=u \cdot \int v dx -\int \left [ \frac{du}{dx} \int v dx \right]dx {/eq}

## Answer and Explanation:

A) {eq}\displaystyle \int \sin(\theta)e^{\theta}d{\theta} {/eq}

Use integration by-parts to solve:

let {eq}\displaystyle I=\int \sin(\theta) e^{\theta} d(\theta)\\ \displaystyle I= \sin(\theta) e^{\theta} -\int \cos(\theta) e^{\theta}d(\theta) \qquad \text{let,} \enspace u=\sin(\theta), v =e^{\theta}\\ \displaystyle I= \sin(\theta) e^{\theta} - \cos(\theta) e^{\theta} -\int \sin(\theta) e^{\theta}d(\theta) \qquad \text{again use integration by-parts and let,} \enspace u=cos(\theta), v =e^{\theta}\\ \displaystyle 2I = \sin (\theta) e^{\theta} - \cos(\theta) e^{\theta}\\ \displaystyle 2I=e^{\theta}(\sin (\theta) - \cos(\theta))\\ \displaystyle \color{blue}{I=\int \sin (\theta)e^{\theta}d{\theta} = \frac{1}{2}e^{\theta}(\sin(\theta) - \cos(\theta)) } {/eq}

B) {eq}\displaystyle I= \int \frac{3}{\sqrt{x^2-7}}dx {/eq}

Here, use substitution method of integration:

let, {eq}\displaystyle x=\sqrt 7 \sec t \Rightarrow dx=\sqrt 7 \sec t \tan t dt\\ \displaystyle I=3\int \frac{\sqrt 7 \sec t \tan t}{\sqrt{7\sec^2t-7}}dt\\ \displaystyle I=\frac{3\sqrt 7}{\sqrt 7 } \int \frac{\sec t \cdot tant}{tant}dt\\ \displaystyle I=3\int \sec t dt\\ \displaystyle I=3\ln|\sec t + \tan t |+C \qquad \text{use standard formula} \int \sec x dx= \ln |\sec x + \tan x|+C\\ \displaystyle I=3\ln \left |\frac{x}{\sqrt 7} + \tan \left ( \sec^{-1}\left ( \frac{x}{\sqrt 7} \right ) \right ) \right |+C\\ \displaystyle \color{blue}{I=3 \ln \left| \frac{x}{\sqrt 7}+ \sqrt{ \left( \frac{x^2-7}{7}\right )}\right|+C}\\ {/eq}

Step 3:

C){eq}\displaystyle \int (\frac{\sqrt{x^3}+7x^2+x}{x})dx {/eq}

We can solve this by direct formula:

{eq}\displaystyle \int \left ( x^{\frac{3}{2}-1}+7x+1 \right) dx\\ \displaystyle \int \left( x^{\frac{1}{2}+7x+1}\right)dx\\ \qquad \displaystyle \int (\frac{\sqrt{x^3}+7x^2+x}{x})dx = \frac{2}{3}x^{\frac{3}{2}}+\frac{7}{2}x^2+x+C \qquad \text{use the formula } \enspace \int x^n dx=\frac{x^{n+1}}{n+1}+C\\ {/eq}

Step 4:

D) {eq}\displaystyle \int \sin^2 (x) \cos^3(x)dx {/eq}

Use substitution method to solve the above integral :

let, {eq}\displaystyle \int \sin^2x \cdot \cos^2x \cdot \cos x dx\\ \displaystyle \int \sin^2x \cdot (1-sin^2x) \cdot \cos x dx\\ \displaystyle \int (\sin^2x - \sin^4x) \cdot \cos x dx\\ \sin x=t \Rightarrow \cos x dx=dt\\ \displaystyle \Rightarrow \int (t^2-t^4) dt\\ \displaystyle =\frac{t^3}{3}-\frac{t^5}{5}+C\\ \displaystyle \displaystyle \int \sin^2(x)\cos^3(x)dx=\frac{\sin^3x}{3}-\frac{\sin^5x}{5}+C {/eq}