# What is the integral of Sin^2 x Cos^3 x dx?

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We have to find Int[sin^2 x cos^3 x dx]

Int[sin^2 x cos^3 x dx]

=> Int[sin^2 x (1 - sin^2 x)cos x dx]

let t = sin x , dt = -cos x dx

=> Int[t^2(1 - t^2) dt]

=> Int[t^2 - t^4 dt]

=> t^3/3 - t^5/5

substitute t = sin x

=> (sin x)^3/3 - (sin x)^5/5 + C

**The required integral is (sin x)^3/3 - (sin x)^5/5 + C**

`int sin^2 x cos^3 x dx = int sin^2 x cos^2 x cos x `

`= int sin^2 x (1- sin^2 x) cos x `

put

`therefore int sin^2 x (1- sin^2 x) cos x dx = int [t^2 (1-t^2)] dt `

`= int (t^2 - t^4 ) dt `

`= (t^3)/3 - (t^5)/5 + c `

now plug in `t` value as `sin x`

required integral =`1/3 sin^3 x - 1/5 sin^5 x+ c`

Well

Cos^2 x= 1- SIN^2 X

cos X = sin X dx

so the original integral becomes

integral of sin^2 X (1-sin^2 x)cos X dx

There are a LOT Of sin X's. so lets say sinX = u

it becomes

integral of u^2(1-u^2)du

using reverse power rule

this equals

u^3/3 - u^5/5

= (sin^3 X)/3 - (sin^5 X )/5