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How do I go about solving the following integral: $$\int_0^{\pi/2} (\sin x)^{\cos x} (\cos x \cot x - \log (\sin x)^{\sin x})\, dx$$

Giving a first try doesn't help as the starting point cannot be grasped that is to say it baffles that how and where to start. The hint given is substitute $u =(\sin x)^{\cos x}$ then what remains to be computed is $\frac {d}{dx} (\sin x)^{\cos x}$ that is a problem in itself as it is differentiating exponential function of a function. Any thoughts:

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Hint:Take $u={\sin x}^{\cos x}$ and find $du$

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    $\begingroup$ @ResearchEngineer yes $\endgroup$ Apr 5 '17 at 19:10
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Note that $$d\left( (\sin x)^{ \cos x } \right) =(\sin x)^{ \cos x }(\cos x\cot x-\log (\sin x)^{ \sin x })dx$$ (you can check it) so from Newton- Leibniz formula we can conclude that $$\\ \\ \\ \int _{ 0 }^{ \pi /2 } \left( \sin x \right) ^{ \cos x }\left( \cos x\cot x-\log \left( \sin x \right) ^{ \sin x } \right) dx=\left( \sin { \frac { \pi }{ 2 } } \right) ^{ \cos { \frac { \pi }{ 2 } } }-{ \left( \sin { 0 } \right) }^{ \cos { 0 } }=1-0=1\\ $$

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    $\begingroup$ There's an extra $dx$ in there. Either $df(x) = f'(x) \, dx$ or $\dfrac d{dx} f(x) = f'(x)$, but not $\dfrac d{dx} f(x) = f'(x) \, dx$. $\endgroup$
    – user307169
    Apr 5 '17 at 18:58
  • $\begingroup$ @tilper, thank you for correction $\endgroup$
    – haqnatural
    Apr 5 '17 at 19:02

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