Integrating sin^11(x) using substitution

[cango91]

Homework Statement

$$\int sin^{11}x.dx$$

Homework Equations

The Attempt at a Solution

$$\int (sin^{2}x)^{5}.sinx.dx$$
$$\int (1-cos^{2}x)^{5}.sinx.dx$$

let cosx be u, statement became

$$- \int (1-u^{2})^{5}.du$$

and I'm stuck here. Any help is appreciated,
thank you

 

[Dick]

Why don't you just multiply the expression (1-u^2)^5 out to get just powers of u?

 

[cango91]

This is potentially one of the 10 questions to be asked in tomorrow's 45 minutes exam, there should be a quicker way...

Plus expanding the whole expression would be too long and impractical..

Thanks anyways,
any other suggestions?

 

[Dick]

(1+a)^5=1+5a+10a^2+10a^3+5a^4+a^5. Put a=-u^2. It's pretty easy if you remember the binomial theorem. I can't think of anything easier.

 

[natives]

No quicker way,EXPAND it!

 

[dirk_mec1]

$$\int {\sin ^k xdx} = \int {\sin ^{k - 1} x\sin xdx}$$

IBP:
$$<br /> - \cos x\sin ^{k - 1} x - \int { - \cos x\left( {k - 1} \right)\sin ^{k - 2} x} \cos xdx<br />$$

rewrite:$$<br /> - \cos x\sin ^{k - 1} x + \left( {k - 1} \right)\int {\cos ^2 x\sin ^{k - 2} x} dx<br />$$

Use the identity:

$$<br /> \cos ^2 x\sin ^{k - 2} x = \left( {1 - \sin ^2 x} \right)\sin ^{k - 2} x = \sin ^{k - 2} x - \sin ^k x<br />$$

And so:

$$<br /> - \cos x\sin ^{k - 1} x + \left( {k - 1} \right)\int {\sin ^{k - 2} x} dx - \left( {k - 1} \right)\int {\sin ^k x} dx<br />$$

$$<br /> k\int {\sin ^k xdx} = - \cos x\sin ^{k - 1} x + \left( {k - 1} \right)\int {\sin ^{k - 2} x} dx<br />$$

Conclusion:
$$<br /> \int {\sin ^k xdx} = - \frac{{\cos x\sin ^{k - 1} x}}{k} + \frac{{k - 1}}{k}\int {\sin ^{k - 2} x} dx<br />$$