NIMCET 2020 Question Paper with solutions

Question : 25 of 120

Marks: +1, -0

∫ \sec^2 x \cosec ^4 xdx=-1/3\cot^3 x+k \tan x-2\cot x+c

, the value of k is

Solution:

I=∫1/{\cos^2 x{\sin^4 x}/{\cos^4 x}\cos^4 x}

I=∫{\sec^4 x}/{\tan^4 x}\sec^2 xdx

\tan x =t

\sec^2 dx=dt

I=∫{(1+t^2)^2}/t^4dt

I=t^{-4}+2t^{-2}+1

=-1/3t^{-3}-2t^{-1}+t

=-1/3\cot^3x-2\cot x+\tan x

k=1

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