Evaluate the following integral int0pi 18 sin 4 x mathrmdx...

Evaluate the following integral. $$ \int_{0}^{\pi} 18 \sin ^{4} x \mathrm{dx} $$ $$ \int_{0}^{\pi} 18 \sin ^{4} x \mathrm{dx} = \square $$ (Type an exact answer, using $$ \pi $$ as needed.)

Transcript

00:01 Section 8 .2, problem number 17, we need to find the definite integral from 0 to pi of 8 sine of the 4th of x d x.

00:12 So the first thing i want to do is just find the antiderivative, then go back, and we'll use fundamental theorem of calculus to figure out the definite integral.

00:19 So i need to figure out then, so let's just see, how do i integrate 8 sine to the 4th of x dx? so we said that the key was to organize this as sine squared.

00:35 So this is the integral of 8.

00:37 This is the sine squared of x squared tx.

00:46 So i can't make a substitution here because i don't have a cosine x dx.

00:49 So i can't do a u substitution.

00:51 What i need to use now is i need to use the identity that the sine squared of x is equal to one -half, one minus the cosine of 2x.

01:05 Well if that's true then when you square both sides of that equation that's where you're going to come up with the sine squared of x squared is equal to one -fourth one minus the cosine of two x squared so that is what we are going to use in this integral right here okay so what we come up with in this point is that the integral 8 sine squared of x squared d x is going to be equal to so i'm just substituting this into the integral so that's going to be 8 times 1 fourth so it's going to be 8 times 1 fourth 1 minus the cosine of 2x squared d x and so this is equal to 2 when you square that you're going to have 1 minus 2 cosine 2x plus cosine squared 2x d x now keep in mind we're going to use the same identity over again that the cosine squared of x is one -half one plus the cosine of 2x therefore the cosine squared of 2x is one -half 1 plus the cosine of 4x.

02:56 So i'm going to make this substitution right here in for this particular part.

03:02 So what i'm going to end up with, let's just copy down everything we have.

03:06 So we've got equals 2 and then times the integral.

03:12 What was it? 1 minus 2 cosine 2x.

03:15 1 minus 2 cosine 2x.

03:23 And then plus the integral of this term that you see here.

03:27 Okay, so this is going to be plus one -half and then plus one -half cosine 4x plus one -half cosine of 4x, all of this, dx.

03:49 Now i can do a little bit of simplification here.

03:53 I'm going to have, this is two times the integral.

03:57 One plus one -half, that's going to be three -half.

03:59 So 3 halves minus 2 cosine 2x plus 1 half cosine 4x dx...

Question

Evaluate the following integral. $$ \int_{0}^{\pi} 18 \sin ^{4} x \mathrm{dx} $$ $$ \int_{0}^{\pi} 18 \sin ^{4} x \mathrm{dx} = \square $$ (Type an exact answer, using $$ \pi $$ as needed.)

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