Question

${\int }_{\frac{\pi }{4}}^{\frac{\pi }{4}}log(sinx+cosx)dx$

Answer 1

Let $I={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}log(sinx+cosx)dx$ ...(i)

$I={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}log\{sin(\frac{\pi }{4}-\frac{\pi }{4}-x)+cos(\frac{\pi }{4}-\frac{\pi }{4})\}dx={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}log\left\{sin\right(-x)+cos(-x\left)\right\}dxandI={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}log(cosx-sinx)dx...\left(ii\right)$

From Eqs. (i) and (ii),

$2I={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}logcos2xdx2I={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}logcos2xdx...\left(iii\right)[\because {\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x),iff(-x)=f(x\left)\right]Put2x=t\Rightarrow dx=\frac{dt}{2}Asx\to 0,thent\to 0andx\to \frac{\pi }{4},thent\to \frac{\pi }{2}2I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}logcostdt...\left(iv\right)2I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}logcos(\frac{\pi }{2}-t)dt[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]\Rightarrow 2I={\int }_0^{\frac{\pi }{4}}logsintdx...\left(v\right)$

On adding Eqs. (iv) and (v) , we get

$4I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}logsintcostdt\Rightarrow 4I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}log\frac{sin2t}{2}dt\Rightarrow 4I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}logsin2xdx-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}log2dx\Rightarrow 4I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}logsin(\frac{\pi }{2}-2x)dx-log2.\frac{\pi }{4}\Rightarrow 4I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}logcos2xdx-\frac{\pi }{4}log2\Rightarrow 4I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}logcos2xdx-\frac{\pi }{4}log2[\because {\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x\left)dx\right]\Rightarrow 4I=2I-\frac{\pi }{4}log2[fromEq.(iii\left)\right]\therefore I=-\frac{\pi }{8}log2=\frac{\pi }{8}log\left(\frac{1}{2}\right)$