Question

# By using the properties of definite integrals, evaluate the integral   $$\displaystyle \int_0^{\pi}\log (1+\cos x)dx$$

Solution

## Let $$I=\int_0^{\pi}\log (1+\cos x)dx$$ ........... (1)$$\Rightarrow I=\int_0^{\pi}\log (1+\cos (\pi-x))dx, \; (\because \int_0^af(x)dx=\int_0^af(a-x)dx)$$$$\Rightarrow I=\int_0^{\pi}\log (1-\cos x)dx$$ ........... (2)Adding (1) and (2), we obtain$$2I=\int_0^{\pi}\left \{\log (1+\cos x)+\log (1-\cos x)\right \}dx$$$$\Rightarrow 2I=\int_0^{\pi}\log (1-\cos^2x)dx$$$$\Rightarrow 2I=\int_0^{\pi}\log \sin^2 x dx$$$$\Rightarrow 2I=2\int_0^{\pi} \log \sin x dx$$$$\Rightarrow I=\int_0^{\pi} \log \sin x dx$$ ........ (3)$$\sin (\pi-x)=\sin x$$$$\therefore I=2\int_0^{\frac {\pi}{2}} \log \sin x dx$$ ............ (4)$$\Rightarrow I=2\int_0^{\frac {\pi}{2}}\log \sin \left (\dfrac {\pi}{2}-x\right )dx=2\int_0^{\frac {\pi}{2}} \log \cos x dx$$ .............. (5)Adding (4) and (5), we obtain$$2I=2\int_0^{\frac {\pi}{2}}(\log \sin x+ \log \cos x)dx$$$$\Rightarrow I=\int_0^{\frac {\pi}{2}} (\log \sin x+ \log \cos x+ \log 2-\log 2)dx$$$$\Rightarrow I=\int_0^{\frac {\pi}{2}} (\log 2 \sin x \cos x-\log 2)dx$$$$\Rightarrow I=\int_0^{\frac {\pi}{2}} \log \sin 2x dx-\int_0^{\frac {\pi}{2}} \log 2 dx$$Let $$2x=t\Rightarrow 2dx=dt$$When $$x=0, t=0$$ and when $$x=\dfrac {\pi}{2}$$$$\therefore I=\dfrac {1}{2}\int_0^{\frac {\pi}{2}}\log \sin t dt-\dfrac {\pi}{2}\log 2$$$$\Rightarrow I=\dfrac {1}{2}I-\dfrac {\pi}{2}\log 2$$$$\Rightarrow \dfrac {I}{2}=-\dfrac {\pi}{2}\log 2$$$$\Rightarrow =-\pi \log 2$$MathematicsRS AgarwalStandard XII

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