$y=\int _{0}^{\pi }\log \left( 1+\cos x \right)dx.....(1)$ - Noon Academy

$y=\int _{0}^{\pi }\log \left( 1+\cos x \right)dx.....(1)$

CBSE | Class 12 | Definite Integrals | Exercise 16C | Maths | RS Aggarwal

$y=\int _{0}^{\pi }\log \left( 1+\cos x \right)dx.....(1)$

Use king theorem of definite integral

$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$

$y=\int _{0}^{\pi }\log (1+\cos (\pi -x))dx$

$y=\int _{0}^{\pi }\log (1-\cos x)dx........(2)$

Adding equation (1) and (2)

$2y=\int _{0}^{\pi }\log \left( 1+\cos x \right)dx+\int _{0}^{\pi }\log (1-\cos x)dx$

$2y=\int _{0}^{\pi }\log {{\sin }^{2}}xdx$

$y=2\int _{0}^{\pi /2}\log \sin xdx......(3)$

Use king theorem of definite integral

$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$

$y=2\int _{0}^{\pi /2}\log \sin \left( \frac{\pi }{2}-x \right)dx$

$y=2\int _{0}^{\pi /2}\log \cos xdx.....(4)$

adding equation (3) and (4)

$2y=2\left( \int _{0}^{\pi /2}\log \sin xdx+\int _{0}^{\pi /2}\log \cos xdx \right)$

$2y=2\left( \int _{0}^{\pi /2}\log \frac{2\sin x\cos x}{2}dx \right)$

$2y=2\left( \int _{0}^{\pi /2}\log \sin 2x-\log 2dx \right)$

Let $2x=t$

$\Rightarrow 2dx=dt$

At $x=0,t=0$

At $x=\pi /2,t=\pi$

$2y=\frac{2}{2}\int _{0}^{\pi /2}\log \sin tdt-\frac{2\pi }{2}\log 2$

$2y=\frac{4}{2}\int _{0}^{\pi /2}\log \sin xdx-\frac{2\pi }{2}\log 2$

$2y=y-\pi \log 2$

$y=-\pi \log 2$

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