Question

How to evaluate this equation?

Answer 1

`int_0^(pi/4) (sin2x)^2*cos2x*dx=1/6`

Explanation:

`int_0^(pi/4) (sin2x)^2*cos2x*dx`

=`1/2int_0^(pi/4) (sin2x)^2*2cos2x*dx`

=`1/2int_0^(pi/4) (sin2x)^2*d(sin2x)`

=`[1/6(sin2x)^3]_0^(pi/4)`

=`1/6*[(sin(pi/2))^3-(sin(0))^3]`

=`1/6*(1-0)`

=`1/6`

Answer 2

`d/dxsin^3(2x)=3sin^2(2x)*d/dxsin(2x)=6sin^2(2x)cos(2x)`, `int_0^(pi/4)sin^2(2x)cos(2x)dx=1/6`

Explanation:

Let's first differentiate `sin^3(2x):`

`d/dxsin^3(2x)=3sin^2(2x)*d/dxsin(2x)=6sin^2(2x)cos(2x)`

By multiple applications of the Chain Rule.

So, we saw that differentiating sine to some power gave us a derivative involving both sine and cosine. This implies that we can evaluate the given integral, which involves both sine and cosine, using `u`-substitution and picking `u` as sine raised to a power.

Well, our integral involves `sin^2(2x),` so we can say:

`u=sin^2(2x)`
Let's calculate our new bounds with `u:`

Upper:
`u=sin^2(pi/2)=1`

Lower:
`u=sin^2(2*0)=0`

`(du)/dx=2sin(2x)*d/dxsin(2x)`

`(du)/dx=4sin(2x)cos(2x)`

`du=4sin(2x)cos(2x)dx`
`(du)/4=sin(2x)cos(2x)dx`

At first, it may not look like `(du)/4` shows up in the integral, but let's rewrite a bit:

`int_0^(pi/4)sin^2(2x)cos(2x)dx=int_0^(pi/4)sin(2x)sin(2x)cos(2x)dx`

So, `(du)/4=sin(2x)cos(2x)dx` does in fact show up in the integral.

Now, even with this substitution, we still have an instance of `sin(2x)` in the integral, but this is no problem after putting in `(du)/4`. Since `u=sin^2(2x), sqrt(u)=sin(2x)`.

Thus, our integral becomes

`1/4int_0^1sqrt(u)du=1/4(2/3(1^(3/2)))=1/4(2/3)=1/6`