Question

evalute integration of x sinx dx with limit 0to pi/2​

Answer 1

Step-by-step explanation:

We have,

$\int_{0}^{ \frac{\pi}{2} } x \sin(x) dx$

Applying integration by parts, and then putting the limits,

$[ x \int \sin(x) dx ]_{0}^{ \frac{\pi}{2} } - \int_{0} ^{ \frac{\pi}{2} } ( \frac{d}{dx} (x) \int \sin(x) dx)dx$

$[ - x \cos(x) ]_{0}^{ \frac{\pi}{2} } + \int_{0}^{ \frac{\pi}{2} } \cos(x) dx$

$[ - x \cos(x) ]_{0}^{ \frac{\pi}{2} } + [ \sin(x) ] _{0}^{ \frac{\pi}{2} }$

$= ( - \frac{\pi}{2} \cos( \frac{\pi}{2}) + (0) \cos(0) ) + ( \sin( \frac{\pi}{2} )- \sin(0) )$

$= 1$