Question

How can I integrate `sin^2 xcosx`??

Answer 1

The integral is equal to `1/3sin^3x + C`

Explanation:

Let `u = sinx`. Then `du = cosx dx` and `dx = (du)/cosx`.

`I = int u^2 cosx * (du)/cosx`

`I = int u^2 du`

`I = 1/3u^3 + C`

`I = 1/3sin^3x + C`

Hopefully this helps!

Answer 2

`1/3 sin^3x + C`

Explanation:

Given: `int sin^2x cos x dx`

Use `u`-substitution.
Let `u = sin x; " "du = cos x dx; " "dx = (du)/(cos x)`

`int sin^2x cos x dx = int u^2 cancel(cos x) (du)/(cancel(cos x)) = int u^2 du = 1/3 u^3 +C`

`int sin^2x cos x dx = 1/3 sin^3x + C`

Answer 3

Is an inmediate integral. See below

Explanation:

`intsin^2xcosxdx=1/3sin^3x+C` because the derivative of `sin^3x` is `3sin^2xcosx`. To remove `3` we need to insert `1/3` before

Answer 4

`1/3sin^3 x + c`

Explanation:

`int sin^2 x cosx dx`

Make an appropriate u sub:

`u = sinx`

`du = cosx dx`

`=> int u^2 du`

`=> 1/3u^3 +c`

`=> 1/3sin^3 x + c`