Given that:

$$I_n (x) = \int_0^x {\cos^n 2t} \ dt $$

Prove that:

$$nI_n(x) = \frac{1}{2}\sin{2x}\cos^{n-1}{2x} + (n-1)I_{n-2}(x)$$

Now with these sorts of questions I've seen two different ways of starting them.

You could start it by doing integration by parts straight away, so:

let $u = \cos^{n}2t$ and let $\dfrac{dv}{dt} = 1$

$\therefore \dfrac{du}{dt} = n(\cos^{n-1}{2t})(-2\sin{2t})$ and $v = t$

You can then use the formula $uv - \int{v\dfrac{du}{dt}}dt$

However, if I've done it correctly, this results in:

$I_n(x) = x\cos^n2x + 2nx\int_0^x (\cos^{n-1}2t)(\sqrt{1-\cos^2 2t})dt$

And I can't see how this could easily simplify into:

$nI_n(x) = \frac{1}{2}\sin{2x}\cos^{n-1}{2x} + (n-1)I_{n-2}(x)$

However, if you start by expressing $\int_0^x {\cos^n 2t} \space dt $ as $\int_0^x ({\cos 2t})(\cos^{n-1} 2t) \space dt $

and using integration by parts again, saying:

let $u = \cos^{n-1}2t$ and $\dfrac{dv}{dt} = \cos 2t$

$\therefore \dfrac{du}{dx} = (n-1)(\cos^{n-2}2t)(-2\sin2t)$ and $v = \frac{1}{2}\sin2t$

If you then use the formula $uv - \int{v\dfrac{du}{dt}}dt$ this simplifies quite easily into:

$nI_n(x) = \frac{1}{2}\sin{2x}\cos^{n-1}{2x} + (n-1)I_{n-2}(x)$

So for this question, it is best to first express $\int_0^x {\cos^n 2t} \space dt $ as $\int_0^x ({\cos 2t})(\cos^{n-1} 2t) \space dt $ and continue from there. However, I have done questions where this is not the case.

My question is: Can you know before hand which is the best approach to take? Do you just have to do lots of similar questions and build up an intuition for it?

Thank you.

  • 1
    $\begingroup$ My experience is intuition. Sometimes you can realise way before getting stuck that you won't end up with the correct result. The more you do, the better you intuition will be. $\endgroup$
    – user88595
    Mar 5, 2014 at 15:24
  • $\begingroup$ See Wallis' integrals. $\endgroup$
    – Lucian
    Mar 5, 2014 at 15:24
  • $\begingroup$ @user88595 thank you, I better get stuck in then :P $\endgroup$
    – Elise
    Mar 5, 2014 at 15:42
  • $\begingroup$ @Lucian ahh, thank you, this should be useful :D $\endgroup$
    – Elise
    Mar 5, 2014 at 15:44

2 Answers 2


A quick way is via integration by parts as below \begin{align}I_n (x) =& \int_0^x {\cos^n 2t} dt = \frac1{2n}\int_0^x {\cot^{n-1} 2t}\ d(\sin^n2t)\\ \overset{ibp}=& \ \frac{1}{2n}\sin{2x}\cos^{n-1}{2x} + \frac{n-1}n I_{n-2}(x) \end{align}










$\therefore nI_n(x)=\dfrac{1}{2}\sin2x\cos^{n-1}2x+(n-1)I_{n-2}(x)$

  • 1
    $\begingroup$ Thank you, I haven't seen it done like this before. I don't think I can follow what you've done in your third line - where you use integration by parts. Have you let $u = d(\sin2t)$? How does this work? Thank you again :) $\endgroup$
    – Elise
    Mar 5, 2014 at 15:57

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