evaluate integration of x^2sinx with respect to x

Question

evaluate integral of 1/sin^2x+cos^x with respect to x

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Answer 1

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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NCERT Study Material

evaluate integration of x^2sinx with respect to x

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

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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