Question

Question #5ea5f

Answer 1

I found: `1/2[x-sin(x)cos(x)]+c`

Explanation:

Try this:

Answer 2

Alternatively, you could make use of trig identities to find the same result: `intsin^2xdx=1/2(x-sinxcosx)+C`

Explanation:

In addition to Gio's method, there is another way of doing this integral, using trig identities. (If you don't like trig or math in general, I wouldn't blame you for disregarding this answer - but sometimes the use of trig is unavoidable in problems).

The identity we will be using is: `sin^2x=1/2(1-cos2x)`.

We can therefore rewrite the integral like so:
`int1/2(1-cos2x)dx`
`=1/2int1-cos2x`

Using the sum rule we get:
`1/2(int1dx-intcos2xdx)`

The first integral simply evaluates to `x`. The second integral is a little more challenging. We know that the integral of `cosx` is `sinx` (because `d/dxsinx=cosx`), but what about `cos2x`? We'll need to adjust for the chain rule by multiplying by `1/2`, so as to balance the `2x`:
`d/dx1/2sin2x=2*1/2cos2x=cos2x`

So `intcos2xdx=1/2sin2x+C` (don't forget the integration constant!) Using that info, plus the fact that `int1dx=x+C`, we have:
`1/2(color(red)(int1dx)-color(blue)(intcos2xdx))=1/2(color(red)(x)-color(blue)(1/2sin2x))+C`

Use the identity `sin2x=2sinxcosx`, we find:
`1/2(x-1/2sin2x)+C=1/2(x-1/2(2sinxcosx))+C`
`=1/2(x-sinxcosx)+C`

And that is the answer Gio found using the integration by parts method.