How do we solve integration problem of function of functions of the kind: $$\int_0^{\pi/2} {\sin (\ln\sqrt {\cos x} )}dx$$
Do we use some substitution e.g., $\sqrt {\cos x} =t$ ? will that work or need another trick?
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What is implicit in that integral? – user251257 Apr 14 '17 at 11:36
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Wolfram alpha says the definite integral: $$ \int \sin(\ln\sqrt{\cos x}) dx = -i2^{-i/2}(e^{-ix}+e^{ix})^{-i/2} (1+e^{2ix})^{-i/2} ((e^{-ix} + e^{ix})^i ,{}_2 F_1( -i/4, -i/2; 1-i/4; e^{-2ix}) + 2^i (1+e^{2ix})^i {}_2 F_1 (i/4, i/2; 1+i/4; -e^{2ix}))$$ and the indefinite integral: $$ \int_0^{\pi /2} \sin(\ln\sqrt{\cos x}) dx = -0.443546$$ – Dando18 Apr 14 '17 at 11:45
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@Dando18 how did you evaluate it? did you use calculator? – user12345 Apr 14 '17 at 11:46
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I put the integral into wolfram alpha. – Dando18 Apr 14 '17 at 11:51
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@user12345 that isn't what people call implicit. You could call it a composite function. – user251257 Apr 14 '17 at 11:51
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It would be rather hard to integrate, so if you're looking to find the indefinite integral you'd be better off using numeric integration. – Dando18 Apr 14 '17 at 11:54
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Well, we have:
$$\mathscr{I}:=\int_0^\frac{\pi}{2}\sin\left(\ln\left(\sqrt{\cos\left(x\right)}\right)\right)\space\text{d}x=\frac{i}{2}\cdot\left\{\int_0^\frac{\pi}{2}\cos^{-\frac{i}{2}}\left(x\right)\space\text{d}x-\int_0^\frac{\pi}{2}\cos^\frac{i}{2}\left(x\right)\space\text{d}x\right\}\tag1$$
Now, use:
$$\int_0^\frac{\pi}{2}\cos^\text{n}\left(x\right)\space\text{d}x=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(\frac{1+\text{n}}{2}\right)}{\Gamma\left(1+\frac{\text{n}}{2}\right)}\tag2$$
When $\Re\left(\text{n}\right)>-1$
So, we get:
$$\mathscr{I}=\frac{i\sqrt{\pi}}{4}\cdot\left\{\frac{\Gamma\left(\frac{1}{2}-\frac{i}{4}\right)}{\Gamma\left(1-\frac{i}{4}\right)}-\frac{\Gamma\left(\frac{1}{2}+\frac{i}{4}\right)}{\Gamma\left(1+\frac{i}{4}\right)}\right\}\tag3$$
Jan Eerland
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@user12345 Well, actually it isn't because I used some very well known properties of the $\ln$ function and the complex representation of the $\sin$ function. – Jan Eerland Apr 14 '17 at 16:51
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@user12345: See also Wallis' integrals, especially the paragraph detailing their relationship to the beta and $\Gamma$ functions. – Lucian Sep 01 '17 at 20:47