# Use the table of integrals to find the following integral. \\ \int \frac{ (\sec^2 \theta \tan^2...

## Question:

Use the table of integrals to find the following integral.

{eq}\int \frac{ (\sec^2 \theta \tan^2 \theta)}{\sqrt {4-\tan^2 \theta}}d\theta{/eq}

## Integration by Substitution:

There are two types of integration substitution: substitution of a function and substitution of a trigonometric function.

Substitution of a function: It can use as a u-substitution or v-substitution.

Substitution of a trigonometric function: It can use the trigonometric function for substitution.

For this solution, we use both of the substitution methods.

So, we require the following historical context:

1. Common derivative: {eq}\displaystyle \frac{d}{dx}\left(\tan \left(x\right)\right)=\sec ^2\left(x\right). {/eq}

2. Common derivative: {eq}\displaystyle \frac{d}{dx}\left(\sin \left(x\right)\right)=\cos \left(x\right). {/eq}

3. Take the constant out: {eq}\displaystyle \int c\cdot f\left(v\right)dv=c\cdot \int f\left(v\right)dv. {/eq}

4. Trigonometric identity: {eq}\displaystyle \sin ^2\left(x\right)=\frac{1-\cos \left(2x\right)}{2}. {/eq}

5. The sum rule: {eq}\displaystyle \int f\left(v\right)\pm g\left(v\right)dv=\int f\left(v\right)dv\pm \int g\left(v\right)dv. {/eq}

6. Common integration: {eq}\displaystyle \int \cos \left(w\right)dw=\sin \left(w\right). {/eq}

7. Integration of a constant: {eq}\displaystyle \int adx=ax. {/eq}

We have to solve the integration of $$\displaystyle I = \int \frac{ (\sec^2 \theta \tan^2 \theta)}{\sqrt {4-\tan^2 \theta}}d\theta$$

Apply the substitution for {eq}u=\tan \left(\theta\right) \Rightarrow du = \sec ^2\left(\theta\right) d\theta. {/eq}

$$\displaystyle = \int \frac{u^2}{\sqrt{4-u^2}}du$$

Apply the trigonometric substitution for {eq}u=2\sin \left(v\right) \Rightarrow du = 2 \cos \left(v\right) dv. {/eq}

$$\displaystyle = \int 4\sin ^2\left(v\right)dv$$

Take the constant out.

$$\displaystyle = 4\cdot \int \sin ^2\left(v\right)dv$$

Use the trigonometric identity.

$$\displaystyle = 4\cdot \int \frac{1-\cos \left(2v\right)}{2}dv$$

Take the constant out.

$$\displaystyle = 4\cdot \frac{1}{2}\cdot \int 1-\cos \left(2v\right)dv$$

Apply the sum rule.

$$\displaystyle = 4\cdot \frac{1}{2}\left(\int 1dv-\int \cos \left(2v\right)dv\right)$$

Use the common integration.

$$\displaystyle = 4\cdot \frac{1}{2}\left(v-\frac{1}{2}\sin \left(2v\right)\right)+C$$

Substitute back {eq}v=\arcsin \left(\frac{1}{2}u\right). {/eq}

$$\displaystyle = 4\cdot \frac{1}{2}\left(\arcsin \left(\frac{1}{2}u\right)-\frac{1}{2}\sin \left(2\arcsin \left(\frac{1}{2}u\right)\right)\right)+C$$

Substitute back {eq}u=\tan \left(\theta\right). {/eq}

$$\displaystyle = 4\cdot \frac{1}{2}\left(\arcsin \left(\frac{1}{2}\tan \left(\theta\right)\right)-\frac{1}{2}\sin \left(2\arcsin \left(\frac{1}{2}\tan \left(\theta\right)\right)\right)\right)+C$$

Simplify:

$$\displaystyle = 2\arcsin \left(\frac{1}{2}\tan \left(\theta\right)\right)-\sin \left(2\arcsin \left(\frac{1}{2}\tan \left(\theta\right)\right)\right)+C$$

Where C is constant of the integration.