# Show that $\int_{0}^{\pi\over 2}\tan{x}\cdot{a+\cos{(x)}\cdot\ln{(\tan{x})}\over 1+\tan{x}}\mathrm dx={a\pi\over 4}$

Can any users show me how to integrate $(1)?$

$$\int_{0}^{\pi\over 2}\tan{x}\cdot{a+\cos{(x)}\cdot\ln{(\tan{x})}\over 1+\tan{x}}\mathrm dx={a\pi\over 4}\tag1$$

$$I_1+I_{2}=\int_{0}^{\pi\over 2}{a\tan{x}\over 1+\tan{x}}\mathrm dx+\int_{0}^{\pi\over 2}{\cos{(x)}\cdot\ln{(\tan{x})}\over 1+\tan{x}}\mathrm dx={a\pi\over 4}\tag2$$

$$I_1=a\int_{0}^{\pi\over2}{\sin{x}\over \sin{x}+\cos{x}}\mathrm dx\tag3$$

$$I_1=a\int_{0}^{\pi\over2}{\sin{x}\cos{x}-\sin^2{x}\over \cos{2x}}\mathrm dx\tag4$$

Not sure what to do next.

• Can we say this new question is the generalization of your last question? – Amin235 Jan 12 '17 at 10:17
• (1) is false: $I_1$ gives $\pi a/4$ and $I_2\neq0$ – tired Jan 12 '17 at 10:43
• @tired The $\cos x$ in $I_2$ should be $\sin x$. – Renascence_5. Jan 12 '17 at 11:05
• @Renascence_5. correct, then the question makes sense! – tired Jan 12 '17 at 11:06

Let $$A=\int_{0}^{\frac{\pi }{2}}\frac{\sin x}{\sin x+\cos x}\, \mathrm{d}x\, \, \, \, \, \, \mathrm{and}\, \, \, \, \, \, B=\int_{0}^{\frac{\pi }{2}}\frac{\cos x}{\sin x+\cos x}\, \mathrm{d}x$$ and $$A+B=\int_{0}^{\frac{\pi }{2}}1\, \mathrm{d}x$$ $$A-B=\int_{0}^{\frac{\pi }{2}}\frac{\sin x-\cos x}{\sin x+\cos x}\, \mathrm{d}x$$ hence, it's easy to see that $$A=\frac{1}{2}\left [ \left ( A+B \right )+\left ( A-B \right ) \right ]=\frac{\pi }{4}$$ So, as tired said, $$I_1=\frac{a\pi }{4}$$
For $I_2$, it should be $$\int_{0}^{\pi\over 2}{\sin{(x)}\cdot\ln{(\tan{x})}\over 1+\tan{x}}\, \mathrm dx$$ let $x\rightarrow \dfrac{\pi }{2}-x$, I got $$I_2=-I_2$$ hence $$I_2=0$$ Now the answer will follow.