Integrating sin^3 x with Substitution?

[kasse]

What is the best way?

 

[rocomath]

Idk, you didn't show any work.

 

[kasse]

I guess I'll have to use an identity. Maybe sin^2(x) = 1-cos^2(x)?

(1-cos^2(x))*sin(x)

u = 1- cos^2(x)

du/dx = -2cos(x)sin(x)


so that


(1-cos^2(x))*sin(x) dx = (-u sin(x)/2cos(x)) du

Doesn't really help, or?

 

[rocomath]

$$\int\cos x\cos^2 xdx$$

Use a BASIC trig identity to change the 2nd degree cosine function.

 

[kasse]

not cosine, sine

 

[rocomath]

not cosine, sine

What are you talking about?

 

[kasse]

Don't know, I guess I'm too drunk to do maths right now.

 

[rocomath]

Don't know, I guess I'm too drunk to do maths right now.

Try again later :)

 

[kasse]

Oh, I wrote cos^3 x instead of sin^3 x in the headline. That explains my confusion.

sin^3 x

=

sin^2 x*sin x

=

(1 - cos^2 x)sin x

Then substitution?

u = 1-cos^2 x

du/dx = 2cos x*sin*x

so that

sin^3 x dx = - u / 2cos x

Hm...

 

[rocomath]

distribute the sinx and you'll see your solution

 

[HallsofIvy]

Oh, I wrote cos^3 x instead of sin^3 x in the headline. That explains my confusion.

sin^3 x

=

sin^2 x*sin x

=

(1 - cos^2 x)sin x

Then substitution?

u = 1-cos^2 x

du/dx = 2cos x*sin*x

so that

sin^3 x dx = - u / 2cos x

Hm...

So how about just u= cos(x)?

You know what they say "Don't drink and derive"!