Question

Integration of x²sinx

Answer 1

Hii friend!!

$\int\limits {x^{2}} {sin x} dx \\ \\ = (-cos x) x^{2} - (-cos x) . 2xdx \\ \\ = -2(-cos x)xd x - x^{2} cos x \\ \\ = -2 . (-x cos xd x) - x^{2} cos x\\ \\ = -2 (-sinxx + sin xdx) - x^{2} cos x \\ \\ = -2 ( -cos x - x sin x ) - x^{2}cosx \\ \\ = -x^{2}cos x - 2(-xsinx - cosx) \\ \\ = -x^{2} cos x + 2x sin x + 2cosx + C$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\huge\boxed{\sf{\int \:x^2sinx\:\:dx}}$

♣ ᴀɴꜱᴡᴇʀ :

$\large\boxed{\sf{\int \:x^2\sin \left(x\right)dx=-x^2\cos \left(x\right)+2\left(x\sin \left(x\right)+\cos \left(x\right)\right)+C}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\sf{\text { Apply Integration By Parts: } u=x^{2}, v^{\:\prime}=\sin (x)}$

$\sf{=-x^2\cos \left(x\right)-\int \:-2x\cos \left(x\right)dx}$

$\sf{\int-2 x \cos (x) d x=-2(x \sin (x)+\cos (x))}$

$\sf{=-x^2\cos \left(x\right)-\left(-2\left(x\sin \left(x\right)+\cos \left(x\right)\right)\right)}$

$\mathrm{Simplify}$

$\sf{=-x^2\cos \left(x\right)+2\left(x\sin \left(x\right)+\cos \left(x\right)\right)}$

$\sf{Add\:a\:constant\:to\:the\:solution}$

$\Large\boxed{\sf{=-x^2\cos \left(x\right)+2\left(x\sin \left(x\right)+\cos \left(x\right)\right)+C}}$