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I was stuck on evaluating this integral :$$I=\int_{\pi/4}^{3\pi/4}\dfrac{\sin x\; \rm dx}{4^{\pi/4}+4^{\tan^{-1}{\left(\frac{4^x}{2^{\pi}}\right)}}}\tag{1}\label{eq1}$$

My attempt:

I used a property of Definite integration, which says, $\int_a^b f (x) \, \rm dx=\int_a^b f (a+b-x) \, \rm dx$. This converted $I$ to $$I=\int_{\pi/4}^{3\pi/4}\dfrac{\sin x\; \rm dx}{4^{\pi/4}+4^{\cot^{-1}{\left(\frac{4^x}{2^{\pi}}\right)}}}\tag{2}\label{eq2}$$ Adding \eqref{eq1} and \eqref{eq2} won't give result in any simplifactions. What else could be done?

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2 Answers 2

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The second expression for $I$ can be written as $$I= \int_{\pi/4}^{3\pi/4} \frac{\sin x}{4^{\pi/4} +4^{\pi/2 -\tan^{-1} (\frac{4^x}{\pi}) }}dx\\ \implies 4^{\pi/4} I= \int_{\pi/4}^{3\pi/4} \frac{4^{\tan^{-1}(4^x/\pi )}\sin x}{4^{\pi/4} +4^{\tan^{-1} (4^x/\pi)}}dx$$ But also, $$4^{\pi/4} I = \int_{\pi/4}^{3\pi/4} \frac{4^{\pi/4} \sin x}{4^{\pi/4} +4^{\tan^{-1} (4^x/\pi)}} dx $$ Adding the two gives $$2\cdot 4^{\pi/4} I =\int_{\pi/4}^{3\pi/4} \sin x dx =\sqrt 2 \\ \implies I =2^{ -\left(\frac{\pi+1}{2} \right)}$$

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Hint:

Note the identity: $$\cot^{-1}(x)=\frac {\pi}{2}-\tan^{-1}(x)$$ Apply it to the changed integral you obtained, multiply it with $4^{\frac {\pi}{4}}$, and add to the original integral. Can you proceed now?

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