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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 $$x=\dfrac {\pi}{2}$$
$$\therefore I=\dfrac {1}{2}\int_0^{\frac {\pi}{2}}\log \sin t dt-\dfrac {\pi}{2}\log 2$$
$$\Rightarrow I=\dfrac {1}{2}I-\dfrac {\pi}{2}\log 2$$
$$\Rightarrow \dfrac {I}{2}=-\dfrac {\pi}{2}\log 2$$
$$\Rightarrow =-\pi \log 2$$

Mathematics
RS Agarwal
Standard XII

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