Finding antiderivative of cos^5(x)...

Limits, differentiation, related rates, integration, trig integrals, etc.

Finding antiderivative of cos^5(x)...

Hey can anyone help me with question 34 of exercise 9C. I've done it about 10 times and cant get it right.

FInd the antiderivative of:

cos ^5x or (cosx)^5

Ive been doing:

=(cosx)(cos^2x)^2
=(cosx)((cos2x+1)/2)^2
=cosx(1/4 + cos2x/2 + cos^2(2x)/4)

This is the point where I get stuck.
I've gone
= cosx/4 + cos3x/2 + cos^2(3x)/4
I know its not right. Can anyone help me get to the solution?
Answer: y= sinx -2/3sin^3(x) + 1/5sin^5(x)

Cheers! Been bugging me all yesterday because all the next questions are similar.

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Joined: Fri May 29, 2009 2:56 am

Re: Finding antiderivative of cos^5(x)...

nuadre wrote:Hey can anyone help me with question 34 of exercise 9C. I've done it about 10 times and cant get it right.

FInd the antiderivative of:

cos ^5x or (cosx)^5

Ive been doing:

=(cosx)(cos^2x)^2
=(cosx)((cos2x+1)/2)^2
=cosx(1/4 + cos2x/2 + cos^2(2x)/4)

This is the point where I get stuck.
I've gone
= cosx/4 + cos3x/2 + cos^2(3x)/4
I know its not right. Can anyone help me get to the solution?
Answer: y= sinx -2/3sin^3(x) + 1/5sin^5(x)

Cheers! Been bugging me all yesterday because all the next questions are similar.

$\int \cos^5(x)dx=\int\cos(x)\cos^4(x)dx=\int\cos(x)(\cos^2(x))^2dx$

$=\int\cos(x)(1-\sin^2(x))^2dx$

now let $u=sin(x)$

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA