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

Answer: (C)

$π$

Answer 2

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