How do I correctly solve the indefinite integral of [sin^3*(13x)*cos^8*(13x)]dx?

[MillerL7]

The integral of [sin^3*(13x)*cos^8*(13x)]dx

I think u=sin^3
so du/dx=cos^3
du=cos^3dx

but then I am really not sure if that is correct, the trig functions confuse me a bit. Please help me, thank you!

 

[RyanSchw]

I don't think that you can simply use a u-substitution here, this appears to be a Trigonometric Integral to me, which has its own special rules.

Since this is in the form of $Sin^n(u)Cos^m(u)$ where m and n are both integers. Where does your book tell you to start?

Hint: focus on m and n, what does that tell you?

 

[Dick]

The chain rule also seems to be confusing you. If u=sin^3(x) then du=3*sin^2(x)*cos(x), NOT cos^3(x). Please review that before even attempting this problem. And the fact that sin^2(x)=1-cos^2(x) may help you a lot. If you play your cards right, it's a pretty simple integral using u substitution.

 

[Dick]

I don't think that you can simply use a u-substitution here, this appears to be a Trigonometric Integral to me, which has its own special rules.

Since this is in the form of $Sin^n(u)Cos^m(u)$ where m and n are both integers. Where does your book tell you to start?

Hint: focus on m and n, what does that tell you?

Why do you say trigonometric integrals have special rules? AFAIK they have the same rules as other integrals. What's special?

 

[RyanSchw]

I guess I phrased that poorly, I simply meant that the integral was in a specific form which can be dealt with easily when you look at the powers of Sin and Cos. In this case he would want to break up $Sin^3(13x)$ using $Sin^2(x) + Cos^2(x) = 1$ because the power of the sine function is odd.

 

[Dick]

I guess I phrased that poorly, I simply meant that the integral was in a specific form which can be dealt with easily when you look at the powers of Sin and Cos. In this case he would want to break up $Sin^3(13x)$ using $Sin^2(x) + Cos^2(x) = 1$ because the power of the sine function is odd.

Right. Sorry, I just jumped on you when you said you can't use a u substitution. Because the answer IS a u substitution. Together with a specific trig rule.

 

[silver-rose]

Hint:

You need to reduce the power of the sin from $$\sin{^3 (13x)}$$ into $$\sin{13x}$$

 

[HallsofIvy]

In general if you have an odd power of a sine or cosine you can factor one of them out to use with the dx, then reduce the remaining even power by sin²x= 1- cos² x or cos²x= 1- sin<sup>2x.

Here you have [sin3(13x) cos8(13x)]dx which has sine to an odd power. Write it as [sin2(13x) cos8(13x)] sin(x)dx= (1- cos2(13x))cos8(13x)] sin(13x) dx and let u= sin(13x). Be careful with the "13x".</sup>

 

[MillerL7]

I tried that formula and got it incorrect...

 

[Dick]

I tried that formula and got it incorrect...

That's not very helpful. How are we supposed to know how you got it wrong?