Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that: $$ \int (\cos x)^{2n} \, dx = \frac{\sin x (\cos x)^{2n-1}}{2n} + \frac{2n-1}{2n} \int (\cos x)^{2n-2} \, dx $$

share|improve this question
    
Have you tried integration by parts? –  Henry Feb 29 '12 at 9:39
    
Yes, but still can't do it. –  ᴊ ᴀ s ᴏ ɴ Feb 29 '12 at 9:42
add comment

1 Answer 1

up vote 3 down vote accepted

Let $u = (\cos x)^{2n-1}$, $dv = \cos x \, dx$. Then $du = -(2n-1)(\cos x)^{2n-2} \sin x dx$, $v = \sin x$ and we obtain by the integration by parts formula \begin{align} \int (\cos x)^{2n} \ dx & = \sin x (\cos x)^{2n-1} + (2n-1) \int (\cos x)^{2n-2} (\sin x)^2 \ dx \\ & = \sin x (\cos x)^{2n-1} + (2n-1) \int (\cos x)^{2n-2} (1-(\cos x)^2) \ dx \\ & = \sin x (\cos x)^{2n-1} + (2n-1) \int (\cos x)^{2n-2} \ dx - (2n-1) \int (\cos x)^{2n} \ dx \\ \end{align} By putting the last integral on the RHS to the LHS, we obtain $$ 2n \int (\cos x)^{2n} \ dx = \sin x (\cos x)^{2n-1} + (2n-1) \int (\cos x)^{2n-2} \ dx \\\ $$ and by dividing by $2n$ we obtain the desired identity.

It's a very classical trick ; it is repeated several times for powers of $\sin$, $\cos$, $\tan$, $\sec$, $\csc$ and $\cot$ : they are called recursion formulas for integrals. You should really learn to try it out for powers of other trigonometric functions. Also, instead of $2n$, you could've just put $k \ge 2$ ; it also works. I never used in my proof the fact that the exponent of $\cos$ was even, I only used the fact that $2n-2 \ge 0$, or in other words, that $k \ge 2$ ; but the cases $k=1$ and $k=0$ are obvious, so this formula is not needed in those cases.

Hope that helps,

share|improve this answer
1  
I actually saw this during integrating cos(cosx)dx some where, I now understand the process of it completely! Thank you very much! –  ᴊ ᴀ s ᴏ ɴ Feb 29 '12 at 10:04
    
$\cos(\cos(x)) \neq (\cos(x))^2\dots$, watch it! What I've shown is that $\int \underset{2n \text { times}}{\underbrace{\cos(x)\cos(x)\dots \cos(x)}} \, dx = something$. –  Patrick Da Silva May 28 '12 at 3:05
    
I know.____________ –  ᴊ ᴀ s ᴏ ɴ May 28 '12 at 8:57
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.