Question

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

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (D)

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