# \int \frac{2x^2}{\sqrt{1-x^2}}dx Basic U-substitution Integrate by Parts Trig.Substitution...

## Question:

c. {eq}\int \frac{2x^2}{\sqrt{1-x^2}}dx {/eq}

Basic U-substitution Integrate by Parts Trig. Substitution Partial functions.

{eq}\int \frac{2x^2}{\sqrt{1-x^2}}dx = {/eq}

d. {eq}\int \frac{2}{(9+x^2)^2} {/eq}

Basic U-substitution Integration by Parts Trig. Substitution Partial functions.

{eq}\int \frac{2}{(9+x^2)^2} = {/eq}

## Integration Using Trigonometric Substitution:

The above question concerns the topic of the integration by substitution method. When an integral cannot be evaluated directly, then we use integration by substitution method. In this question, we have used trigonometric substitution. After substitution, integral becomes simple and can be evaluated directly.

c.

{eq}\begin{align} \displaystyle &\int \frac{2x^2}{\sqrt{1-x^2}}dx---(1)\\ \text{Let us assume that }x=\sin \theta &\rightarrow dx=\cos \theta d\theta\\ &\text{Substituting the values in equation (1), we have}\\ \displaystyle &=\int \frac{2\sin^2 \theta}{\sqrt{1-\sin^2 \theta}} \cos \theta d\theta\\ \displaystyle &=\int \frac{2\sin^2 \theta}{\cos \theta} \cos \theta d\theta &\left [ \text{Using trigonometric identity }\sin^2 \theta+\cos^2 \theta =1\right ]\\ \displaystyle &=\int 2\sin^2 \theta d\theta\\ \displaystyle &=\int \left (1-\cos 2 \theta \right ) d\theta & \left [\text{Using formula }\cos 2\theta=1-2\sin^2 \theta \right ]\\ \displaystyle &= \theta-\frac{\sin 2 \theta }{2 } +C\\ \displaystyle &= \theta-\frac{2\sin \theta \cos \theta }{2 }+C \\ &\text{Replacing the value of }\theta \text{, we have} \\ \displaystyle &= \sin^{-1} x-x\sqrt{1-x^2}+C \end{align} {/eq}

d.

{eq}\begin{align} \displaystyle &\int \frac{2}{(9+x^2)^2}dx ---(2)\\ \text{Let us assume }x=3\tan \theta &\rightarrow dx=3 \sec^2 \theta d\theta\\ &\text{Substituting the values in equation (1), we have}\\ \displaystyle &=\int \frac{2}{(9+ 9\tan^2 \theta)^2}3 \sec^2 \theta d\theta\\ \displaystyle &=\int \frac{2}{81(\sec^4 \theta)}3 \sec^2 \theta d\theta &\left [ \text{Using trigonometric identity }1 +\tan^2 \theta =\sec^2 \theta \right ] \\ \displaystyle &=\int \frac{2}{27(\sec^2 \theta)} d\theta\\ \displaystyle &=\int \frac{2\cos^2 \theta}{27} d\theta &\left [ \text{As }\cos \theta= \frac{1}{\sec \theta} \right ]\\ \displaystyle &=\int \frac{1}{27}\left ( 1+\cos 2\theta \right )d \theta &\left [ \cos 2 \theta =2\cos^2 \theta -1 \right ]\\ \displaystyle &=\frac{1}{27}\left ( \theta + \frac{\sin 2 \theta}{2}\right )+C\\ \displaystyle &=\frac{1}{27}\left ( \theta + \frac{2\sin \theta \cos \theta}{2}\right )+C\\ &\text{Replacing the value of }\theta \text{, we have}\\ \displaystyle &=\frac{1}{27}\left ( \tan^{-1} \left ( \frac{x}{3} \right )+ \frac{3x}{x^2+9} \right ) +C \end{align} {/eq}