Question

$I=\int \frac{\sin x\cos x}{1-{\sin }^{4}x}dx$ is equal to

• A. $I=\frac{1}{4}\log ({\sec }^{2}x+{\tan }^{2}x)+C$
• B. $I=\frac{1}{4}\log ({\sec }^{2}x-{\tan }^{2}x)+C$
• C. $I=\frac{1}{2}\log ({\sec }^{2}x+{\tan }^{2}x)+C$
• D. $I=\frac{1}{2}\log ({\sec }^{2}x-{\tan }^{2}x)+C$

Answer 1

Answer: (A)

$I=\frac{1}{4}\log ({\sec }^{2}x+{\tan }^{2}x)+C$

Consider the following Equation.

$I=\int \frac{\sin x\cos x}{1-{\sin }^{4}x}dx$

Let $t={\sin }^{2}x$

$\frac{1}{2\sin x\cos x}dt=dx$

Therefore,

$=\int \frac{\sin x\cos x}{1-{\sin }^{4}x}\times \frac{1}{2\sin x\cos x}dt$

$=\frac{1}{2}\underset{}{\overset{}{\int }}\frac{1}{1-t^2}dt$

$=\frac{1}{2}\left[\frac{1}{2}\log \left(\frac{1+t}{1-t}\right)\right]+C$

$=\frac{1}{4}\log \left(\frac{1+t}{1-t}\right)+C$

On putting the value of $t$, we get

$=\frac{1}{4}\log \left(\frac{1+{\sin }^{2}x}{1-{\sin }^{2}x}\right)+C$

$=\frac{1}{4}\log \left(\frac{1+{\sin }^{2}x}{{\cos }^{2}x}\right)+C$

$=\frac{1}{4}\log ({\sec }^{2}x+{\tan }^{2}x)+C$

Hence, this is the correct answer.