Question

The value of the integral \(∫^{ \frac{π}{2}}_{ \frac{-π}{2}} \frac{dx}{(1+e^x)(sin^6x+cos^6x)}\)  is equal to

• A. \(2π\)
• B. 0
• C. \(π\)
• D. \(\frac{π}{2}\)

Answer 1

The Correct Option is C

\(I\) = \(∫^{ \frac{π}{2}}_{ \frac{-π}{2}} \frac{dx}{(1+e^x)(sin^6x+cos^6x)}\).....(i)

\(I\) = \(∫^{ \frac{π}{2}}_{ \frac{-π}{2}} \frac{dx}{(1+e^{-x})(sin^6x+cos^6x)}\).....(ii)

From equation (i) & (ii)

\(2I\) = \(∫ ^{\frac{π}{2}} _{\frac{-π}{2}} \frac{dx}{(sin^6x+cos^6x) }\) = \(∫ ^{\frac{π}{2}}_0 \frac{ dx}{(1-\frac{3}{4}sin^22x)}\)

\(⇒ I = ∫ ^{\frac{π}{2}}_ 0 \frac{4sec^22xdx}{(4+tan^22x)} = 2 ∫^{\frac{π}{4}} _0\frac{ 4sec^22x}{4+tan^22x }dx\)

Now, 

\(tan2x = t\) and \(2sec^22xdx = dt\)

At  \(x = 0, t = 0\)

is  \(x = \frac{π}{4}, t →∞\)

\(∴ I = 2 ∫^∞_0 \frac{2dt}{4+t^2} = 2(tan^{-1} \frac{t}{2})^∞_0\)

= \(2 \frac{π}{2} = π\)

Hence, the correct option is (C): \(\pi\)