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Given that $n$ is a positive integer, show that $$\int_{0}^{\frac{\pi}{2}} \sin^{n}x^ \,dx = \int_{0}^{\frac{\pi}{2}} \cos^{n}x^ \,dx.$$

I know how to use integration by parts but I am not sure how can I apply that to show that this can be solved using that method. Please help me determine the steps on how can I easily prove this. Thank you in advance.

Thomas Andrews
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PRD
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  • If $f$ is any function defined on $[0,1],$ $$\int_{0}^{\pi/2} f(\sin x),dx=\int_0^{\pi/2} f(\cos x),dx.$$ You don’t need to evaluate the integral to prove it. – Thomas Andrews Oct 23 '21 at 14:05

1 Answers1

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Hint:

Note that $\cos x=\sin\left(\frac{\pi}{2}-x\right)$, hence $$\int^{\frac{\pi}{2}}_{0}\cos^n x\,dx=\int^{\frac{\pi}{2}}_{0}\sin^n\left(\frac{\pi}{2}-x\right)\,dx$$

Now, consider the substitution $u=\frac{\pi}{2}-x$.

Kyan Cheung
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  • This is what I got. $-\int^{0}_{\frac{\pi}{2}}\sin^n u,du$ – PRD Oct 23 '21 at 14:15
  • Don't forget that $\frac{du}{dx}=-1$, so you should have a minus sign somewhere, and you shouldn't be getting a $\cos^n$ term. – Kyan Cheung Oct 23 '21 at 14:17
  • Nice, now you should swap bounds which should give you the desired answer. – Kyan Cheung Oct 23 '21 at 14:20
  • Thank you for your guidance, however, the end result was in terms of $u$, not in $x$ – PRD Oct 23 '21 at 14:21
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    You're integrating the same function over the same bounds on both sides of the equality so they are the same. If you remain unconvinced then you could apply the substitution $x=u$ (though if you do that then it feels like overloading a variable...) – Kyan Cheung Oct 23 '21 at 14:24