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Integrate (sin x)^3 / (1 + cos x)

Integrate (sin x)^3 / (1 + cos x). The previous bit of the question said to write it as a function of u = sinx and v = cosx, but I don't know if it's related to this in any way...

Reply 1

james22

May not be the quickest way but if you let u=cos(x) it becomes a rational function which can be intergrated in the same way as any other.

EDIT: The ways given below are effectively the same, but probably easier.

(edited 10 years ago)

Reply 2

BabyMaths

(\sin x)^3=\sin x (1-\cos x)(1+\cos x)

Reply 3

Felix Felicis

Original post by meowmeowcat

Integrate (sin x)^3 / (1 + cos x). The previous bit of the question said to write it as a function of u = sinx and v = cosx, but I don't know if it's related to this in any way...

\sin^3 x = \sin^2 x \sin x = (1-\cos^2 x) \sin x

.

Difference of two squares on

1-\cos^2 x

Reply 4

meowmeowcat

just realised myself, I'm rusty on integration. Thanks.

Reply 5

user2020user

Original post by BabyMaths

...

Original post by Felix Felicis

...

Out of curiosity, would a Weierstrass substitution work?

I managed to get it down to

\displaystyle \int \dfrac{8t^3}{(1+t^2)^3}\ dt

How would I proceed? I thought of partial fractions but the power of 3 in the denominator is really putting me off.

Reply 6

davros

Original post by Khallil

Out of curiosity, would a Weierstrass substitution work?

I managed to get it down to

\displaystyle \int \dfrac{8t^3}{(1+t^2)^3}\ dt

How would I proceed? I thought of partial fractions but the power of 3 in the denominator is really putting me off.

If you go down that route I think you're going to have to make at least one more substitution like

u = t^2

, but it looks like it could get to be quite a slog!

Reply 7

Felix Felicis

Original post by Khallil

Out of curiosity, would a Weierstrass substitution work?

I managed to get it down to

\displaystyle \int \dfrac{8t^3}{(1+t^2)^3}\ dt

How would I proceed? I thought of partial fractions but the power of 3 in the denominator is really putting me off.

You could go down the partial fractions route with that...

Hmmm

\alpha > 0:

\displaystyle f(\alpha) = \int \frac{dt}{t(1 + \alpha t^2)}

Use partial fractions on that and then consider

4 f''(1)

if you know about differentiation under the integral sign.

Reply 8

Indeterminate

Original post by Khallil

Out of curiosity, would a Weierstrass substitution work?

I managed to get it down to

\displaystyle \int \dfrac{8t^3}{(1+t^2)^3}\ dt

How would I proceed? I thought of partial fractions but the power of 3 in the denominator is really putting me off.

Original post by davros

If you go down that route I think you're going to have to make at least one more substitution like

u = t^2

, but it looks like it could get to be quite a slog!

I got it to

\displaystyle 8\int \left(\dfrac{t}{(1+ t^2)^2} - \dfrac{t}{(1+t^2)^3}\right) dt

which can be split it up into 2 integrals to make substitution easier.

This is to messy a method for me, however. It's much easier to split up the numerator and cancel, as discussed above.