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## Calculus 2

### Unit 2: Lesson 3

Integrating using trigonometric identities

# Integral of sin^2(x) cos^3(x)

Another example where u substitution combined with certain trigonometric identities can be used.

## Want to join the conversation?

• I think I'm much more comfortable with the u-substitution as compared to the REVERSE CHAIN RULE. Will that be a problem or is it alright?
• Reverse chain rule is basically doing u substitution in your head, so it would be a bit faster than u-sub but using u-sub won't be a problem, it might even increase accuracy as you right it down
• I got a different answer after substituting the trig identity for sin(x)^2 = 1 - cos(x)^2
So what I have now is ∫(1 - cos(x)^2) * cos(x)^3 * dx
Then after distributing the cos(x)^3 I have, ∫cos(x)^3 - ∫cos(x)^5 * dx
Evaluating this give me: (cos(x)^4)/4 - (cos(x)^6)/6
Is this equivalent to the answer in the video, or where have I gone wrong?
(1 vote)
• Basically you can't integrate the cos(x)^3 and the cos(x)^5. The reason for this is because they represent cos(x)*cos(x)*cos(x) and you cant integrate that without the u substitution. So in the end your only choice is to sub the trig identity for the cos(x)^2
(1 vote)
• I will assume you intend the integrand to be interpreted as `[ (2 - sin 2x) / (1 - cos 2x) ]eᵡ`. To solve the integral, we will first rewrite the sine and cosine terms as follows:

` I) sin(2x) = 2sin(x)cos(x);`
`II) cos(2x) = 2cos²(x) - 1.`

Rewriting yields
`2 - sin(2x)`
`= 2 - 2sin(x)cos(x)`
`= 2[1 - sin(x)cos(x)],`

and
`1 - cos(2x)`
`= 1 - [2cos²(x) - 1]`
`= 2 - 2cos²(x)`
`= 2[1 - cos²(x)]`
`= 2sin²(x).`

Hence
`[ (2 - sin 2x) / (1 - cos 2x) ]eᵡ`
`= ( 2[1 - sin(x)cos(x)] / [ 2sin²(x) ] )eᵡ`
`= [ 1/sin²(x)`` - cos(x)/sin(x) ]eᵡ`
`= eᵡ / sin²(x) - eᵡcot(x).`

Thus `∫ [ (2 - sin 2x) / (1 - cos 2x) ]eᵡ dx = ∫ [ eᵡ / sin²(x) - eᵡcot(x) ] dx`. This may be split up into two integrals as `∫ eᵡ / sin²(x) dx - ∫ eᵡcot(x) dx`. We will first focus on the first of these integrals.

Recall that `d/dx cot(x) = -1 / sin²(x)`. Using integration by parts on the expression `∫ eᵡ / sin²(x) dx` yields

`∫ eᵡ / sin²(x) dx`
`= -eᵡcot(x) + ∫ eᵡcot(x) dx.`

When we plug this into the expression `∫ eᵡ / sin²(x) dx - ∫ eᵡcot(x) dx`, we get

`∫ eᵡ / sin²(x) dx - ∫ eᵡcot(x) dx`
`= -eᵡcot(x) + ∫ eᵡcot(x) dx - ∫ eᵡcot(x) dx`
`= -eᵡcot(x) + C.`

Therefore, the answer is `-eᵡcot(x) + C`, where `C` is an arbitrary real number.
• Can anyone help me with this: ∫ ln(x)/sin(x) dx
• I'm pretty sure there is no standard way to write the antiderivative of that function. The best you can probably do is use a calculator if you needed a definite integral.
• Where does the du go after we solve for the anti derivative of U?
• If you were just solving the integral of x^2 dx, what would happen with the dx? It's just part of the operation and goes away when it is used. Let me know if that doesn't quite make sense.
• why can't ∫sin^2(x)*cos^2(x)*cos(x) be written as ∫1*cos(x)?
after all trig identity says sin^2(x)*cos^2(x)=1
• Is it possible to use U substitution without distributing cosx to (1- sinx^2)? I didn't distribute cos(x) and therefore my solution was different: x - 1/3(sin^3(x)) + C instead of sin(x) - 1/3(sin^3(x)) + C. I feel that when I do distribute my professor does not and ends up finding a different solution than what I would have done and then when I dont distribute, he does. Is there a rule I am missing here?
• In the beginning of learning a concept (like U-sub), it is a good idea to exaggerate the amount of steps it takes to form a solution.

Some little things might be lost if we jump too far between the different steps.

It might be a good idea to control the solutions by deriving the finished antiderivative.

(x - 1/3(sin^3(x)) + C)'=cos^3(x)-cos(x)+1
(sin(x) - 1/3(sin^3(x)) + C)'=cos^3(x)

What could we do to make these derivatives equal eachother?

I hope this was a little helpful!
(1 vote)
• How do we know not to do sin^2(x)-sin^4(x)=-sin^2(x)?
(1 vote)
• Hello, thanks a lot for the video!
I would like to know why does Sal distributes this way ():
`sin² (1 - sin²)`
`(1 - sin²) cos`
Is it an intuitive thing to do or is there any kind of hint for doing this?
Cheers :)
(1 vote)
• He did that to end up with an integrable amount inside the bracket multiplied by its derivatives outside of it so that he could apply u-substitution or the reverse chain rule...
(1 vote)
• let y= (cosx)^4
dy/dx = 4 (cos x) ^3 * (sin x)^2

so my answer is 1/4 { cosx) ^4 } + C, is that correct?