# Evaluate each integral based on inverse trigonometric functions. (a) \int...

## Question:

Evaluate each integral based on inverse trigonometric functions.

{eq}(a) \int \frac{dx}{\sqrt{1-4x^{2}}}\\(b) \int \frac{e^{t}}{e^{2t}+4}dt {/eq}

## Integration by Trigonometric Substitution:

The radical function which can be replaced by the trigonometric function is called the trigonometric substitution.

Here we describe the problem-solving strategy:

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

2. Remove the constant from the integration: {eq}\displaystyle \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx. {/eq}

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

4. Common integration: {eq}\displaystyle \int \frac{1}{v^2+1}dv=\arctan \left(v\right). {/eq}

## Answer and Explanation:

(a) We have to evaluate the integration of $$\displaystyle I = \int \frac{dx}{\sqrt{1-4x^{2}}} $$

Apply the trigonometric substitution for {eq}x=\frac{1}{2}\sin \left(u\right) \Rightarrow dx = \frac{1}{2}\cos \left(u\right) du. {/eq}

$$\displaystyle = \int \frac{1}{2}du $$

Remove the constant from the integration.

$$\displaystyle = \frac{1}{2}\cdot \int 1du $$

Use the integration of a constant.

$$\displaystyle = \frac{1}{2}\cdot u +C $$

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

$$\displaystyle = \frac{1}{2}\arcsin \left(2x\right)+C $$

Where C is constant of the integration.

(b) We have to evaluate the integration of $$\displaystyle I = \int \frac{e^{t}}{e^{2t}+4}dt $$

Apply the substitution for {eq}u=e^t \Rightarrow du=e^t dt. {/eq}

$$\displaystyle = \int \frac{1}{u^2+4}du $$

Apply the substitution for {eq}u=2v \Rightarrow du=2 dv. {/eq}

$$\displaystyle = \int \frac{1}{2\left(v^2+1\right)}dv $$

Remove the constant from the integration.

$$\displaystyle = \frac{1}{2}\cdot \int \frac{1}{v^2+1}dv $$

Use the common integration.

$$\displaystyle = \frac{1}{2}\arctan \left(v\right) +C $$

Substitute back {eq}v=\frac{u}{2}. {/eq}

$$\displaystyle = \frac{1}{2}\arctan \left(\frac{u}{2}\right) +C $$

Substitute back {eq}u=e^t. {/eq}

$$\displaystyle = \frac{1}{2}\arctan \left(\frac{e^t}{2}\right)+C $$

Where C is constant of the integration.

#### Learn more about this topic:

from Math 104: Calculus

Chapter 13 / Lesson 11