I am currently working on a practice problem on trig substitution.

I have no problem solving it my way but I don’t see how i can use the below stated identity and partial integration. (I am sorry i am not using mathjax, im looking into it but it is quite confusing for me)

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And here is my solution. enter image description here

I tried rewriting as 1- sin^2(theta), but when i attempt to use partial integration it just does not solve, and I have no idea how to rewrite it.

Many thanks for your help



Integration by parts: $u=cos\theta,\ dv=cos\theta d\theta$, therefore $du=-sin\theta d\theta,\ v=sin\theta$. Result, $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}cos^2\theta d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sin^2\theta d\theta$, since $cos{-\frac{\pi}{2}}=cos{\frac{\pi}{2}}=0$

$2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}cos^2\theta d\theta=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sin^2\theta d\theta$. The two integrals sum to $2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta=2\pi$.

Therefore the integral in question $=\pi$.

  • $\begingroup$ Many thanks that the exercise should contain integration by parts $\endgroup$ – Ang Sep 11 '18 at 17:48
  • $\begingroup$ @Petra I revised answer to include integration by parts. $\endgroup$ – herb steinberg Sep 13 '18 at 16:20
  • $\begingroup$ wow many thanks that you took the time I would have never seen this $\endgroup$ – Ang Sep 13 '18 at 16:49

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