Question

If $\int {\sec }^{2}x{\mathrm{cosec}}^{4}xdx=-\frac{1}{3}{\cot }^{3}x+k\tan x$ $-2\cot x+C$, then $k$ is equal to

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (D)

Let $I=\int {\sec }^{2}x{\mathrm{cosec}}^{4}xdx$
$=\int \frac{1}{{\sin }^{4}x{\cos }^{2}x}dx=\int \frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{4}x{\cos }^{2}x}dx$
$\left(\because 1={\sin }^{2}x+{\cos }^{2}x\right)$
$=\int \frac{dx}{{\sin }^{2}x{\cos }^{2}x}+\int \frac{dx}{{\sin }^{4}x}$
$=\int \frac{\left({\sin }^{2}x+{\cos }^{2}x\right)dx}{{\sin }^{2}x{\cos }^{2}x}+\int {\mathrm{cosec}}^{4}xdx$
$\left(\because 1={\sin }^{2}x+{\cos }^{2}x\right)$
$=\int \left({\sec }^{2}x+{\mathrm{cosec}}^{2}x\right)dx$
$+\int {\mathrm{cosec}}^{2}x\left(1+{\cot }^{2}x\right)dx$
$=\tan x-\cot x+\int {\mathrm{cosec}}^{2}xdx$
$+\int {\mathrm{cosec}}^{2}x{\cot }^{2}xdx$
$=\tan x-\cot x-\cot x-\frac{{\cot }^{3}x}{3}+C$
$=-\frac{1}{3}{\cot }^{3}x+\tan x-2\cot x+C$
But it given that,
$I=-\frac{1}{3}{\cot }^{3}x+k\tan x-2\cot x+C$
$\therefore k=1$