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Solve the Integral: ∫sinx.sin2xdx

Solve the Integral:
∫sinx.sin2xdx 

Grade:12

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int sinx.sin2xdx
I = \int sinx.(2sinx.cosx)dx
I = \int 2sin^{2}x.cosxdx
sinx = t
cosxdx = dt
I = \int 2t^{2}dt
I = \frac{2t^{3}}{3} + constant
I = \frac{2sin^{3}x}{3} + constant
Sunil Raikwar
askIITians Faculty 45 Points
9 years ago
∫sinx.sin2xdx
=∫sinx.2sinxcosxdx
= 2∫sin2x.cosxdx
let t = sinx
dt = cosxdx
∫ t^2 dx = t^3/3 +c = sin3x/3 +c
where c is a constant


Thanks & Regards
Sunil Raikwar
askIITians Faculty

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