Question

If $\int {\sec }^{2}x\cdot co{\sec }^{4}x\cdot dx=-\frac{1}{3}{\cot }^{3}x+k\tan x-2\cot x+c$, then $k$ is equal to

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

$\int \frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{4}x{\cos }^{2}x}dx=\int \frac{1}{{\sin }^{2}x{\cos }^{2}x}dx+\int \frac{1}{{\sin }^{4}x}dx$
$=\int \frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx+\int \frac{1}{{\sin }^{4}x}dx$
$=\int {\sec }^{2}xdx+\int cose{c}^{2}xdx+\int cose{c}^{2}x[1+{\cot }^{2}x]dx$

$=\tan x-\cot x-\cot x-\frac{{\cot }^{3}x}{3}$

$K=1$