Question

If $\int _ { 0 } ^ { \pi } x f \left( \cos ^ { 2 } x + \tan ^ { 4 } x \right) d x$ $k$ is

• A. $\frac { \pi } { 2 }$
• B. $\pi$
• C. $- \frac { \pi } { 2 }$
• D. None of these

Answer 1

Answer: (B)

$\pi$

Answer 2

$\int _ { 0 } ^ { \pi } x f \left( \cos ^ { 2 } x + \tan ^ { 4 } x \right) d x = k \int _ { 0 } ^ { \pi / 2 } f \left( \cos ^ { 2 } x + \tan ^ { 4 } x \right) d x$

By the property of definite integral

$I = \int _ { 0 } ^ { \pi } x f \left( \cos ^ { 2 } x + \tan ^ { 4 } x \right) d x$ …..(i)

$= \int _ { 0 } ^ { \pi } ( \pi - x ) f \left( \cos ^ { 2 } x + \tan ^ { 4 } x \right) d x$ …..(ii)

Adding (i) and (ii), we have

$2 I = \pi \int _ { 0 } ^ { \pi } f \left( \cos ^ { 2 } x + \tan ^ { 4 } x \right) d x$

⇒ $2 I = 2 \pi \int _ { 0 } ^ { \pi / 2 } f \left( \cos ^ { 2 } x + \tan ^ { 4 } x \right) d x$

⇒ $I = \pi \int _ { 0 } ^ { \pi / 2 } f \left( \cos ^ { 2 } x + \tan ^ { 4 } x \right) d x$

On comparing with given integral, we get

$k = \pi$.