Question

The value of \(\rm \int_0^{\pi\over2}{\sec^2x\over2+2\tan^2x}dx\) is:
5. None of these

Answer 1

Correct Answer - Option 4 : \(\rm \pi\over4\)

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• \(\rm∫ {1\over1+ x^2} dx = \tan^{-1}x\)

Calculation:

I = \(\rm \int{\sec^2x\over2+2\tan^2x}dx\)

I = \(\rm \int{\sec^2x\over2(1+\tan^2x)}dx\)

Let tan x = t ⇒ sec² x dx = dt

I = \(\rm \int{1\over2}{dt\over1+t^2}\)

I = \(\rm {1\over2}\tan^{-1}t\)

Putting the value of t

I = \(\rm {1\over2}\tan^{-1}(\tan x)\) = \(\rm {x\over2}\)

Putting the limits 

I = \(\rm \left[{x\over2}\right]_0^{\pi\over2}\)

I = \(\rm {1\over2}\left[{\pi\over2}-0\right]\) = \(\boldsymbol{\pi\over4}\)