Evaluate _ 0 ^ x x 1 + ^ 2 x d x.

Question

Evaluate $\int _ { 0 } ^ { \pi } \frac { x \sin x } { 1 + \cos ^ { 2 } x } d x$.

✓

Answer

According to the question , $I =\int _ { 0 } ^ { \pi } \frac { x \sin x } { 1 + \cos ^ { 2 } x } d x$ .......(i)
$\Rightarrow I = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) \sin ( \pi - x ) } { 1 + \cos ^ { 2 } ( \pi - x ) } d x$ $[\because \int _0^a f(x)= \int_0^a f(a-x)]$
$\Rightarrow I = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) \sin x } { 1 + \cos ^ { 2 } x } d x$.....(ii)
On Adding eqs.(i) and (ii),we get
$2 I = \int _ { 0 } ^ { \pi } \frac { \pi \sin x } { \left( 1 + \cos ^ { 2 } x \right) } d x$
Put , $cos x = t$
$-sin x dx = dt$
$\Rightarrow sinx dx = -dt$
Lower limit , when x = 0, then t = 1
Upper limit ,when $x= \pi$, then t = -1
$= - \pi \int _ { 1 } ^ { - 1 } \frac { d t } { 1 + t ^ { 2 } }$
$\Rightarrow 2 I = \pi \int _ { - 1 } ^ { 1 } \frac { d t } { \left( 1 + t ^ { 2 } \right) } $
$= \pi \left[ \tan ^ { - 1 } t \right] _ { - 1 } ^ { 1 }$
$= \pi \left[ \frac { \pi } { 4 } - \left( - \frac { \pi } { 4 } \right) \right] $
$= \frac { \pi ^ { 2 } } { 4 }$
$ \therefore I = \frac { \pi ^ { 2 } } { 4 }$