Question

Solve $\int \frac{\cot x}{\sqrt{{\cos }^{4}x+{\sin }^{4}x}}dx=......+c$

• A. $\frac{1}{2}\log |{\cot }^{2}x+\sqrt{{\cot }^{4}x+1}|$
• B. $-\frac{1}{2}\log |{\cot }^{2}x+\sqrt{{\cot }^{4}x+1}|$
• C. $\frac{1}{2}\log |{\tan }^{2}x+\sqrt{{\tan }^{4}x+1}|$
• D. $-\frac{1}{2}\log |{\cot }^{}x+\sqrt{{\cot }^{4}x+1}|$

Answer 1

The correct option is B $-\frac{1}{2}\log |{\cot }^{2}x+\sqrt{{\cot }^{4}x+1}|$

$I=\int \frac{\cot x}{\sqrt{{\cos }^{4}x+{\sin }^{4}x}}dx$

$I=\int \frac{\cot x}{\sqrt{{\sin }^{4}x(\frac{{\cos }^{4}x}{{\sin }^{4}x}+1)}}dx$

$I=\int \frac{\cot x}{{\sin }^{2}x\sqrt{{\cot }^{4}x+1}}dx$

$I=\int \frac{\cot xco{\sec }^{2}x}{\sqrt{{\cot }^{4}x+1}}dx$

Now take ${\cot }^{2}x=t$

$\therefore -2\cot xco{\sec }^{2}xdx=dt$

$\therefore \cot xco{\sec }^{2}xdx=-\frac{1}{2}dt$

$\therefore I=-\frac{1}{2}\int \frac{1}{\sqrt{t^2+1}}dt=-\frac{1}{2}\log |t+\sqrt{t^2+1}|+c$

$\therefore I=-\frac{1}{2}\log |{\cot }^{2}x+\sqrt{{\cot }^{4}x+1}|+c$