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

In summary: Thats a good idea. I've been mainly just doing what I can to drop the powers and get them into sums or differences.
  • #1
opus
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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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  • #2
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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  • #3
Ok thanks!
 
  • #4
Looks like it was computational. Marking as solved.
 
  • #5
opus said:
Looks like it was computational. Marking as solved.

Did you find the correct answer using your own method/my method?
 
  • #6
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.
 
  • #7
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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  • #8
Are you squaring both sides?
 
  • #9
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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  • #10
Are you working to use the power reduction on the sin then?
 
  • #11
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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Related to Trig Integral Homework: Solving \int_0^{π/8}sin^2(x)cos^2(x)

1. What is the purpose of solving \int_0^{π/8}sin^2(x)cos^2(x)?

The purpose of solving this trigonometric integral is to find the definite integral of the given function within the given limits of integration. This is a common type of problem in calculus and is important for understanding the relationship between trigonometric functions and their integrals.

2. What is the first step in solving \int_0^{π/8}sin^2(x)cos^2(x)?

The first step is to use the trigonometric identity sin^2(x) = (1-cos(2x))/2 to rewrite the integral as \int_0^{π/8}(1-cos(2x))/2 * cos^2(x). This simplifies the integral and makes it easier to solve.

3. How do you solve \int_0^{π/8}sin^2(x)cos^2(x)?

After rewriting the integral using the trigonometric identity, you can use the power-reducing formula cos^2(x) = (1+cos(2x))/2 to further simplify the integral to \int_0^{π/8}(1-cos(2x))^2/4. From there, you can use the substitution u = 2x to transform the integral into a more manageable form.

4. Is there a shortcut method for solving \int_0^{π/8}sin^2(x)cos^2(x)?

Yes, there is a shortcut method known as the double angle formula. This formula states that sin^2(x)cos^2(x) = (1/4)sin^2(2x). By using this formula, you can simplify the integral to \int_0^{π/8}(1/4)sin^2(2x). From there, you can use the power-reducing formula again to solve the integral.

5. What is the final answer for \int_0^{π/8}sin^2(x)cos^2(x)?

The final answer for this integral is 1/8 - 1/32π, which can be obtained by evaluating the integral using the substitution u = 2x and then plugging in the limits of integration. It is always important to check your answer by taking the derivative of the antiderivative to ensure that it matches the original function.

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