Trig Integral Homework: Solving \int_0^{π/8}sin^2(x)cos^2(x)

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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)
 

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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.
 
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Ok thanks!
 
Looks like it was computational. Marking as solved.
 
opus said:
Looks like it was computational. Marking as solved.

Did you find the correct answer using your own method/my method?
 
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.
 
opus said:
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)##
 
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Are you squaring both sides?
 
opus said:
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.
 
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Are you working to use the power reduction on the sin then?
 
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opus said:
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.
 
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  • #12
Thats a good idea. I've been mainly just doing what I can to drop the powers and get them into sums or differences.
 
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