Ujjwal Kumawat 13th Apr, 2021

Dear Aspirant

I=∫x^2sinx.dx

Applying integration by part,
I=x^2∫sinx.dx−∫[dxd(x^2)∫sindx]dx
=x^2(−cosx)−2∫x(−cosx).dx
=−x^2cosx+2∫xcosxdx
Again applying integration by parts
I=−x^2cosx+2[x∫cosx.dx−∫dxd(x)∫cosx.dx]
=−x^2cosx+2[xsinx−∫sinx.dx ]
I=−x2cosx+2xsinx+2cosx+c

i hope it will help you

Thank you