Evaluating ∫₀^(π/8) sin²(x)cos²(x) dx

[opus]

Homework Statement

$\int_0^{π/8}sin^2(x)cos^2(x)$

Homework Equations

The Attempt at a Solution

Please see my attached work to see the train of thought. I've tried this thing about 100 times and still can't get the correct solution. I don't know if it's in the anti derivative evaluations of step (i) or the computation in step (ii)

 

[member 587159]

I'm sorry to inform you that most members won't bother reading attached images. Since the image is hard to read (at least for me it is), all I can do is suggest you an easier appproach:

Observe that $\sin(2x) = 2 \sin(x) \cos(x)$

Use this in your first step and your integral will boil down to (after a substitution) something like $\int \sin^2(x) dx$ which is a standard integral to solve.

 

[opus]

Ok thanks!

 

[opus]

Looks like it was computational. Marking as solved.

 

[member 587159]

Looks like it was computational. Marking as solved.

Did you find the correct answer using your own method/my method?

 

[opus]

What I did was just redo the computations in the evaluation theorem. Still trying to see how to use your suggestion in the first step.

 

[member 587159]

What I did was just redo the computations in the evaluation theorem. Still trying to see how to use your suggestion in the first step.

$\sin^2 (x) \cos^2(x) = 1/4 \sin^2(2x)$

 

[opus]

Are you squaring both sides?

 

[member 587159]

Are you squaring both sides?

$2\sin(x) \cos(x) = \sin(2x) \implies \sin(x) \cos(x) = 1/2 \sin(2x)$

and then indeed I square both sides.

 

[opus]

Are you working to use the power reduction on the sin then?

 

[member 587159]

Are you working to use the power reduction on the sin then?

I use the standard identity $\sin(2x) = 2\sin(x) \sin(x)$

This easily follows from $\sin(a+b) = \sin(a) \cos(b) + \sin(b) \cos(a)$ which is also a standard trig identity.

An easy proof can be given by writing down both sides of

$$e^{i(a+b)} = e^{ia}e^{ib}$$ using $e^{ix} = \cos x + i \sin x$ and comparing imaginary parts.

Knowing your trig identities can save a lot of time in such problems. Worth memorising imo.

 

[opus]

Thats a good idea. I've been mainly just doing what I can to drop the powers and get them into sums or differences.