Question

Antiderivative of $\frac{{\sin }^{2}x}{1+{\sin }^{2}x}$ with respect to x is?

• A. $x-\frac{\sqrt{2}}{2}arc\tan \left(\sqrt{2}\tan x\right)+c$
• B. $x-\frac{1}{\sqrt{2}}arc\tan \left(\frac{\tan x}{\sqrt{2}}\right)+c$
• C. $x-\sqrt{2}arc\tan \left(\sqrt{2}\tan x\right)+c$
• D. $x-\sqrt{2}arc\tan \left(\frac{\tan x}{\sqrt{2}}\right)+c$

Answer 1

The correct option is A $x-\frac{\sqrt{2}}{2}arc\tan \left(\sqrt{2}\tan x\right)+c$

$\begin{array}{c}\int \frac{{\sin }^{2}x}{1+{\sin }^{2}x}dx\left[antiderivativesmeansintegeration\right]\\ \int [\frac{(1+{\sin }^{2}x)}{1+{\sin }^2}\frac{-1}{1+{\sin }^{2}x}]dx\\ \int 1dx-\int \begin{array}{c}\frac{dx}{1+{\sin }^{2}x}\\ \end{array}\\ \left(x\right)-\int \frac{dx}{1+{\sin }^{2}x}[dividedby{\cos }^{2}innumeratoranddenomirator\\ \left(x\right)-\int \frac{(\frac{1}{{\cos }^{2}x}dx}{\frac{1}{{\cos }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x}}\\ \left(x\right)-\int \frac{{\sec }^{2}dx}{{\sec }^2+{\tan }^{2}x}\\ \left(x\right)-\int \begin{array}{c}\frac{{\sec }^{2}dx}{1+{\tan }^{2}x{\tan }^{2}x}\\ \end{array}\\ \left(x\right)-\frac{{\sec }^{2}xdx}{1+2{\tan }^{2}x}\\ \left(x\right)-\frac{1}{2}.\int \frac{{\sec }^2}{{\tan }^{2}x+\frac{1}{2}}\\ let{\tan }^{2}x+t\\ {\sec }^{2}xdx=dt\\ dx=\frac{dt}{{\sec }^{2}x}\\ andputting\\ \left(x\right)-\frac{1}{2}.\int \frac{{\sec }^2}{t^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}.\frac{dt}{{\sec }^{2}x}\\ \left(x\right)-\frac{1}{2}.\int \frac{dt}{t^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}\\ x-\frac{1}{2}\left[\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}.{\tan }^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)\right]\\ x-\frac{1}{2}.\left[\sqrt{2{\tan }^{-}\sqrt{2}\tan x}\right]\\ x-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}\tan x\right)\end{array}$