Finding indefinite integral ∫ x.sinx ^ 2. Cosx ^ 2DX
∫x.sinx^2.cosx^2dx
=(1/2)∫xsin2x^2dx
Let u = 2x ^ 2
du=4x
The original formula = (1 / 8) ∫ sinudu
=-(1/8)cosu+C
=-(1/8)cos2x^2+C
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