evaluate integration of x^2sinx with respect to x
Answer (1)
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
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