Question

If $\int {\sin }^{-1}\left(\sqrt{\frac{x}{1+x}}\right)dx=A\left(x\right){\tan }^{-1}\left(\sqrt{x}\right)+B\left(x\right)+C$, where $C$ is a constant of integration, then the ordered pair $\left(A\right(x),B(x\left)\right)$ can be:

• A. $\left(x+1,-\sqrt{x}\right)$
• B. $\left(x-1,-\sqrt{x}\right)$
• C. $\left(x+1,\sqrt{x}\right)$
• D. $\left(x-1,\sqrt{x}\right)$

Answer 1

The correct option is A

$\left(x+1,-\sqrt{x}\right)$

Explanation for the correct option:

Step 1: Simplify $\int {\sin }^{-1}\left(\sqrt{\frac{x}{1+x}}\right)dx$.

Let $x={\tan }^2\theta$, so $dx=2\tan \theta ·{\sec }^2\theta d\theta$. By substituting the value of $x$, we get

$\begin{array}{rcl}\int {\sin }^{-1}\left(\sqrt{\frac{x}{1+x}}\right)dx& =& \int {\sin }^{-1}\left(\sqrt{\frac{{\tan }^2\theta }{1+{\tan }^2\theta }}\right)\left(2\tan \theta ·{\sec }^2\theta d\theta \right)\\ & =& \int {\sin }^{-1}\left(\sqrt{\frac{{\tan }^2\theta }{{\sec }^2\theta }}\right)\left(2\tan \theta ·{\sec }^2\theta \right)d\theta \\ & =& \int {\sin }^{-1}\left(\frac{\tan \theta }{\sec \theta }\right)\left(2\tan \theta ·{\sec }^2\theta \right)d\theta \\ & =& \int {\sin }^{-1}\left(\sin \theta \right)\left(2\tan \theta ·{\sec }^2\theta \right)d\theta \\ & =& 2\int {\sin }^{-1}\left(\sin \theta \right)\left(\frac{\sin \theta }{\cos \theta }·\frac{1}{{\cos }^2\theta }\right)d\theta \\ & =& 2\int \theta \left(\frac{\sin \theta }{{\cos }^3\theta }\right)d\theta \end{array}$

Step 2: Apply integration by parts

We know that $\int uvdx=u\int vdx-\int u\text{'}\left(\int vdx\right)dx$

So,

$2\int \theta \left(\begin{array}{l}\frac{\sin \theta }{{\cos }^3\theta }\end{array}\right)d\theta =2\left[\theta \int \left(\begin{array}{l}\frac{\sin \theta }{{\cos }^3\theta }\end{array}\right)d\theta \right]-\int 1\left(\int \left(\begin{array}{l}\frac{\sin \theta }{{\cos }^3\theta }\end{array}\right)\right)d\theta$ ….. $\left(1\right)$

Let $\cos \theta =u$, so $du=-\sin \theta d\theta$. Now,

$\begin{array}{rcl}\int \begin{array}{l}\frac{\sin \theta }{{\cos }^3\theta }\end{array}d\theta & =& -\int \frac{du}{u^3}\\ & =& -\left(-\frac{1}{2u^2}\right)\\ & =& \frac{1}{2{\cos }^2\theta }\end{array}$

Now, $\left(1\right)$ can be written as

$\begin{array}{rcl}2\left[\theta \left(\begin{array}{l}\frac{1}{2{\cos }^2\theta }\end{array}\right)-\int \left(\frac{1}{2{\cos }^2\theta }\right)d\theta \right]& =& \theta \left(\begin{array}{l}\frac{1}{{\cos }^2\theta }\end{array}\right)-\int \left(\frac{1}{{\cos }^2\theta }\right)d\theta \\ & =& \theta \left(\begin{array}{l}{\sec }^2\theta \end{array}\right)-\int {\sec }^2\theta d\theta \\ & =& \theta \left(\begin{array}{l}{\sec }^2\theta \end{array}\right)-\tan \theta +C\\ & =& {\tan }^{-1}\left(\sqrt{x}\right)\left(1+x\right)-\sqrt{x}+C\left[\mathrm{As}x={\tan }^2\theta \mathrm{and}se{c}^2\theta =1+{\tan }^2\theta \right].....\left(2\right)\end{array}$

Step 3: Find $\left(A\right(x),B(x\left)\right)$.

By comparing $\left(2\right)$, with the equation given in question, we get

Therefore, $A\left(x\right)=1+x\mathrm{and}B\left(x\right)=-\sqrt{x}$.

So, $\left(A\right(x),B(x\left)\right)=\left(x+1,-\sqrt{x}\right)$

Hence, option A is correct.