# Use trigonometric substitution to evaluate the integral. \int_{1/\sqrt{7}}^{\sqrt{2/7}}...

## Question:

Use trigonometric substitution to evaluate the integral.

{eq}\displaystyle \int_{\frac{1}{\sqrt{7}}}^{\sqrt{\frac{2}{7}}} \frac{x^3}{\sqrt{7x^2 - 1}}\,dx {/eq}

## Integration by Trigonometric Substitution:

The integration which makes the effect of changing the function and integration is called integration by substitution. Here we used trigonometric substitution which is defined as the radical function which can be replaced by the trigonometric function is called the trigonometric substitution.

Some fundamental relations we used to solve the integration:

1. An algebraic property: {eq}\displaystyle \int _0^{\frac{\pi }{4}}\sec ^4\left(u\right)du=\int _0^{\frac{\pi }{4}}\sec ^2\left(u\right)\sec ^2\left(u\right)du. {/eq}

2. Eject the constant out: {eq}\displaystyle \int m\cdot f\left(r\right)dr=m\cdot \int f\left(r\right)dr. {/eq}

3. The trigonometric identity: {eq}\displaystyle \sec ^2\left(x\right)=1+\tan ^2\left(x\right). {/eq}

4. The sum rule: {eq}\displaystyle \int f\left(r\right)\pm g\left(r\right)dr=\int f\left(r\right)dr\pm \int g\left(r\right)dr. {/eq}

5. The power rule: {eq}\displaystyle \int r^m dr=\frac{r^{m+1}}{m+1}, \quad m\ne -1. {/eq}

6. integration of a constant: {eq}\displaystyle \int a dr=a r. {/eq}

## Answer and Explanation:

We have to solve the integration of $$\displaystyle I = \int_{\frac{1}{\sqrt{7}}}^{\sqrt{\frac{2}{7}}} \frac{x^3}{\sqrt{7x^2 - 1}}\,dx $$

Apply the trigonometric substitution for {eq}x=\frac{1}{\sqrt{7}}\sec \left(u\right) \Rightarrow dx=\frac{1}{\sqrt{7}}\sec \left(u\right)\tan \left(u\right)du. {/eq}

Limits: {eq}\frac{1}{\sqrt{7}} \rightarrow 0 {/eq} and {eq}\sqrt{\frac{2}{7}} \rightarrow \frac{\pi }{4}. {/eq}

$$\displaystyle = \int _0^{\frac{\pi }{4}}\frac{\sec ^4\left(u\right)}{49}du $$

Eject the constant out.

$$\displaystyle = \frac{1}{49}\cdot \int _0^{\frac{\pi }{4}}\sec ^4\left(u\right)du $$

Use the algebraic property.

$$\displaystyle = \frac{1}{49}\cdot \int _0^{\frac{\pi }{4}}\sec ^2\left(u\right)\sec ^2\left(u\right)du $$

Use the trigonometric identity.

$$\displaystyle = \frac{1}{49}\cdot \int _0^{\frac{\pi }{4}}\left(1+\tan ^2\left(u\right)\right)\sec ^2\left(u\right)du $$

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

Limits: {eq}0 \rightarrow 0 {/eq} and {eq}\frac{\pi }{4} \rightarrow 1. {/eq}

$$\displaystyle = \frac{1}{49}\cdot \int _0^1 1+v^2dv $$

Apply the sum rule.

$$\displaystyle = \frac{1}{49}\left(\int _0^11dv+\int _0^1v^2dv\right) $$

Use the integration of a constant and the power rule.

$$\displaystyle = \frac{1}{49}\left(\left[v\right]^1_0+\left[\frac{v^3}{3}\right]^1_0\right) $$

Compute the boundaries.

$$\begin{align*} \displaystyle &= \frac{1}{49}\left(\left[1-0\right]+\left[\frac{}{3}-0\right]\right)\\ \displaystyle &= \frac{1}{49}\left(1+\frac{1}{3}\right)\\ \displaystyle &= \frac{4}{147}. \end{align*} $$

#### Learn more about this topic:

from Math 104: Calculus

Chapter 13 / Lesson 11