Question

If $I=\int {\sec }^{2}x{cosec}^{4}xdx=A{\cot }^{3}x+B\tan x+C\cot x+D$ then

• A. $A=-\frac{1}{3}$
• B. $B=2$
• C. $C=-2$
• D. none of these

Answer 1

Answer: (A), (C)

$A=-\frac{1}{3}$
$C=-2$

Given $I=\int {\sec }^{2}x{cosec}^{4}xdx=A{\cot }^{3}x+B\tan x+C\cot x+D$ ....(1)
Consider, $I=\int {\sec }^{2}x{cosec}^{4}xdx$
$=\int \frac{1}{{\cos }^{2}x{\sin }^{4}x}dx$
$=\int \frac{{({\sin }^{2}x+{\cos }^{2}x)}^2}{{\cos }^{2}x{\sin }^{4}x}dx$
$\int \frac{{\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x}{{\cos }^{2}x{\sin }^{4}x}dx$
$=\int ({\sec }^{2}x+2{cosec}^{2}x+\frac{{\cos }^{2}x}{{\sin }^{4}x})dx$
$=\tan x-2\cot x+\int {\cot }^{2}x{cosec}^{2}xdx$
Put $\cot x=t$ in the integral
$-{cosec}^{2}xdx=dt$
$=\tan x-2\cot x-\frac{{\cot }^{3}x}{3}+D$
Put this value in (1) and on comparing we get
$A=-\frac{1}{3},B=1,C=-2$