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Evaluate the integral. 1. \int \sqrt { 1 - 25 x ^ { 2 } } d x 2. \int \frac { x ^ { 2 } } { \sqrt...

Question:

Evaluate the integral.

1. {eq}\int \sqrt { 1 - 25 x ^ { 2 } } d x {/eq}

2. {eq}\int \frac { x ^ { 2 } } { \sqrt { 5 + x ^ { 2 } } } d x {/eq}

Integration by Trigonometric Substitution:

Take any function {eq}\cos (x) {/eq}, which is a trigonometric function, and its derivative is {eq}-\sin (x) {/eq}. Therefore, the integration of trigonometric function is given by:

{eq}\int -\sin \left(x\right)dx=\cos \left(x\right)+C. {/eq}

We will use some important properties to solve the integration:

1. Move the constant out from the integration: {eq}\int b\cdot p\left(x\right)dx=b\cdot \int p\left(x\right)dx. {/eq}

2. Common differentiation: {eq}\frac{d}{dy}\left(\sin \left(y\right)\right)=\cos \left(y\right). {/eq}

3. The sum rule: {eq}\displaystyle \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx. {/eq}

4. The power rule: {eq}\displaystyle \int x^adx=\frac{x^{a+1}}{a+1}, \quad a\ne -1. {/eq}

5. Common integration: {eq}\displaystyle \int \cos \left(v\right)dv=\sin \left(v\right). {/eq}

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

7. Trigonometric identity: {eq}\displaystyle \sec ^2\left(x\right)-\tan ^2\left(x\right)=1. {/eq}

8. Common integration: {eq}\displaystyle \int \sec \left(u\right)du=\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|. {/eq}

9. Reduction formula: {eq}\displaystyle \int \sec ^n\left(x\right)dx=\frac{\sec ^{n-1}\left(x\right)\sin \left(x\right)}{n-1}+\frac{n-2}{n-1}\int \sec ^{n-2}\left(x\right)dx. {/eq}

10. Trigonometric identity: {eq}\displaystyle \sin \left(\arctan \left(x\right)\right)=\frac{x}{\sqrt{1+x^2}}. {/eq}

11. Trigonometric identity: {eq}\displaystyle \tan \left(\arctan \left(x\right)\right)=x. {/eq}

12. Trigonometric identity: {eq}\displaystyle \sec \left(\arctan \left(x\right)\right)=\sqrt{1+x^2}. {/eq}

Answer and Explanation:

1. We have to solve the integration of $$\displaystyle I = \int \sqrt { 1 - 25 x ^ { 2 } } d x $$

Apply trigonometric substitution for {eq}x=\frac{1}{5}\sin \left(u\right) \Rightarrow dx = \frac{1}{5}\cos \left(u\right) du. {/eq}

$$\displaystyle = \int \frac{1}{5}\cos ^2\left(u\right)du $$

Move the constant out from the integration.

$$\displaystyle = \frac{1}{5}\cdot \int \cos ^2\left(u\right)du $$

Use the trigonometric identity.

$$\displaystyle = \frac{1}{5}\cdot \int \frac{1+\cos \left(2u\right)}{2}du $$

Move the constant out from the integration.

$$\displaystyle = \frac{1}{5}\cdot \frac{1}{2}\cdot \int 1+\cos \left(2u\right)du $$

Apply the sum rule.

$$\displaystyle = \frac{1}{5}\cdot \frac{1}{2}\left(\int 1du+\int \cos \left(2u\right)du\right) $$

Use the common integration.

$$\displaystyle = \frac{1}{5}\cdot \frac{1}{2}\left(u+\frac{1}{2}\sin \left(2u\right)\right)+C $$

Substitute back {eq}u=\arcsin \left(5x\right). {/eq}

$$\displaystyle = \frac{1}{5}\cdot \frac{1}{2}\left(\arcsin \left(5x\right)+\frac{1}{2}\sin \left(2\arcsin \left(5x\right)\right)\right)+C $$

Simplify:

$$\displaystyle = \frac{1}{10}\left(\arcsin \left(5x\right)+\frac{1}{2}\sin \left(2\arcsin \left(5x\right)\right)\right)+C $$

Where C is constant of the integration.



2. We have to solve the integration of $$\displaystyle I = \int \frac { x ^ { 2 } } { \sqrt { 5 + x ^ { 2 } } } d x $$

Apply trigonometric substitution for {eq}x=\sqrt{5}\tan \left(u\right) \Rightarrow dx =\sqrt{5}\sec ^2 \left(u\right) du. {/eq}

$$\displaystyle = \int \frac{5\sqrt{5}\tan ^2\left(u\right)\sec ^2\left(u\right)}{\sqrt{5+5\tan ^2\left(u\right)}}du $$

Move the constant out from the integration.

$$\displaystyle = 5\sqrt{5}\cdot \int \frac{\tan ^2\left(u\right)\sec ^2\left(u\right)}{\sqrt{5+5\tan ^2\left(u\right)}}du $$

Use the trigonometric identity.

$$\displaystyle = 5\sqrt{5}\cdot \int \frac{\sec ^2\left(u\right)\tan ^2\left(u\right)}{\sqrt{\sec ^2\left(u\right)}\sqrt{5}}du $$

Simplify:

$$\displaystyle = 5\sqrt{5}\cdot \int \frac{\sec \left(u\right)\tan ^2\left(u\right)}{\sqrt{5}}du $$

Move the constant out from the integration.

$$\displaystyle = 5\sqrt{5}\frac{1}{\sqrt{5}}\cdot \int \sec \left(u\right)\tan ^2\left(u\right)du $$

Use the trigonometric identity.

$$\displaystyle = 5\sqrt{5}\frac{1}{\sqrt{5}}\cdot \int \left(-1+\sec ^2\left(u\right)\right)\sec \left(u\right)du $$

Expand:

$$\displaystyle = 5\sqrt{5}\frac{1}{\sqrt{5}}\cdot \int \:-\sec \left(u\right)+\sec ^3\left(u\right)du $$

Apply the sum rule.

$$\displaystyle = 5\sqrt{5}\frac{1}{\sqrt{5}}\left(-\int \sec \left(u\right)du+\int \sec ^3\left(u\right)du\right) $$

Apply reduction formula.

$$\displaystyle = 5\sqrt{5}\frac{1}{\sqrt{5}}\left(-\int \sec \left(u\right)du+\frac{\sec ^2\left(u\right)\sin \left(u\right)}{2}+\frac{1}{2}\cdot \int \sec \left(u\right)du\right) $$

Use the common integration.

$$\displaystyle = 5\sqrt{5}\frac{1}{\sqrt{5}}\left(-\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|+\frac{\sec ^2\left(u\right)\sin \left(u\right)}{2}+\frac{1}{2}\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|\right)+C $$

Substitute back {eq}u=\arctan \left(\frac{1}{\sqrt{5}}x\right). {/eq}

$$\displaystyle = 5\sqrt{5}\frac{1}{\sqrt{5}}\left(-\ln \left|\tan \left(\arctan \left(\frac{1}{\sqrt{5}}x\right)\right)+\sec \left(\arctan \left(\frac{1}{\sqrt{5}}x\right)\right)\right|+\frac{\sec ^2\left(\arctan \left(\frac{1}{\sqrt{5}}x\right)\right)\sin \left(\arctan \left(\frac{1}{\sqrt{5}}x\right)\right)}{2}+\frac{1}{2}\ln \left|\tan \left(\arctan \left(\frac{1}{\sqrt{5}}x\right)\right)+\sec \left(\arctan \left(\frac{1}{\sqrt{5}}x\right)\right)\right|\right)+C $$

Use the trigonometric identity.

$$\begin{align*} \displaystyle &= 5\sqrt 5 \frac{1}{{\sqrt 5 }}\left( { - \ln \left| {\frac{1}{{\sqrt 5 }}x + \sqrt {1 + {{\left( {\frac{1}{{\sqrt 5 }}x} \right)}^2}} } \right| + \frac{{{{\left( {\sqrt {1 + {{\left( {\frac{1}{{\sqrt 5 }}x} \right)}^2}} } \right)}^2}\frac{{\frac{1}{{\sqrt 5 }}x}}{{\sqrt {1 + {{\left( {\frac{1}{{\sqrt 5 }}x} \right)}^2}} }}}}{2} + \frac{1}{2}\ln \left| {\frac{1}{{\sqrt 5 }}x + \sqrt {1 + {{\left( {\frac{1}{{\sqrt 5 }}x} \right)}^2}} } \right|} \right) + C\\ \displaystyle &= 1\cdot \:5\left(\frac{\frac{\frac{1}{\sqrt{5}}x}{\sqrt{\left(\frac{1}{\sqrt{5}}x\right)^2+1}}\left(\sqrt{\left(\frac{1}{\sqrt{5}}x\right)^2+1}\right)^2}{2}-\frac{1}{2}\ln \left|\frac{1}{\sqrt{5}}x+\sqrt{\left(\frac{1}{\sqrt{5}}x\right)^2+1}\right|\right)+C\\ \displaystyle &= 5\left(-\frac{1}{2}\ln \left|\frac{1}{\sqrt{5}}x+\sqrt{\frac{1}{5}\left(x^2+5\right)}\right|+\frac{1}{10}x\sqrt{x^2+5}\right)+C \end{align*} $$

Where C is constant.


Learn more about this topic:

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How to Use Trigonometric Substitution to Solve Integrals

from Math 104: Calculus

Chapter 13 / Lesson 11
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