Question

$\int \frac{(1-{\tan }^{2}x).\csc 2xdx}{{\sec }^{2}x}$ is equal to

• A. $\frac{\ell n|\sin 2x|}{2}+C$
• B. $\frac{\ell n|\cos 2x|}{2}+C$
• C. $\frac{\ell n|\tan 2x|}{2}+C$
• D. $\frac{\ell n|\cos x+\sin x|2x|}{2}+C$

Answer 1

The correct option is A $\frac{\ell n|\sin 2x|}{2}+C$

$\int \frac{(1-{\tan }^{2}x)\csc x}{\sin 2x{\cos }^{2}x}$

$=\int \frac{(1-{\sec }^2-1){\csc }^{2}x}{{\cos }^{2}x}$

$=-\int \frac{{\cot }^{2}x}{2}{\sec }^{2}x$

$=\frac{\ln |\sin 2x|}{2}+c$