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Assume that $f$ is even, integrable, and such that $\int_0^1 f(x) dx = \pi$. Determine the definite integral $$\int_\frac{-\pi}{2}^\frac{\pi}{2}\cos(x) f(\sin x) dx.$$

Since f is even it follows that the given integral is equal to $$2 \int_0^\frac{\pi}{2}\cos(x) f(\sin(x)) dx.$$ I see that it's possible to write $\cos(x)f(\sin(x))$ as $g'(x)f(g(x))$ but not sure if that's of any use moving forward.

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  • $\begingroup$ You did most of the work, just make a change of variable. $\endgroup$
    – nicomezi
    Commented Oct 13, 2017 at 15:37

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Yes, you are correct and you are almost done. If $f$ is even then $g(x):=\cos(x) f(\sin(x)) $ is even too $$g(-x)=\cos(-x) f(\sin(-x))=\cos(x) f(-\sin(x))=\cos(x) f(\sin(x))=g(x),$$ and $$\int_\frac{-\pi}{2}^\frac{\pi}{2} \cos(x) f(\sin(x)) dx= 2\int_0^\frac{\pi}{2} \cos(x) f(\sin(x)) dx=2\int_0^1 f(t) dt=2\pi$$ where we made the change of variable $t=\sin(x)$ with $dt=\cos(x) dx.$ Note that as $x$ goes from $0$ to $\pi/2$ then $t=\sin(x)$ goes from $0$ to $1$.

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  • $\begingroup$ Ah the thing I couldn't figure out was how to get the $\frac{π}{2}$ to a $1$ $\endgroup$
    – JohnDoe
    Commented Oct 13, 2017 at 15:47
  • $\begingroup$ @JohnDoe Is it clear now? $\endgroup$
    – Robert Z
    Commented Oct 13, 2017 at 15:48
  • $\begingroup$ yes thanks a lot! $\endgroup$
    – JohnDoe
    Commented Oct 13, 2017 at 15:48

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