Integrals — Class 12 Maths Solution

Let $I = \int\limits_0^\pi {\log \left( {1 + \cos x} \right)dx}$

…(i)
$\Rightarrow$ $I = \int\limits_0^\pi {\log \left[ {1 + \cos \left( {\pi - x} \right)} \right]dx} = \int\limits_0^\pi {\log \left( {1 - \cos x} \right)dx}$

….(ii)

Adding (i) and (ii),

we get
$2I = \int\limits_0^\pi {\left[ {\log \left( {1 + \cos x} \right) + \log \left( {1 - \cos x} \right)} \right]dx = \int\limits_0^\pi {\log \left( {1 - {{\cos }^2}x} \right)dx} }$

$= \int\limits_0^\pi {\log {{\sin }^2}xdx} = 2\int\limits_0^\pi {\log \sin xdx}$

$\Rightarrow$ $I = \int\limits_0^\pi {\log \sin xdx} = 2\int\limits_0^\pi {\log \sin xdx} = 2{I_1}$

Where ${I_1} = \int\limits_0^{\pi /2} {\log \sin x\,dx}$

….(iii)
Then, ${I_1} = \int\limits_0^{\pi /2} {\log \sin \left( {\cfrac{\pi }{2} - x} \right)dx}$

$\Rightarrow$ ${I_1} = \int\limits_0^{\pi /2} {{\mathop{\rm logcosx}\nolimits} dx}$
(iv)
Adding (iii) and (iv),

we get
$2{I_1} = \int\limits_0^{\pi /2} {\log \sin xdx} + \int\limits_0^{\pi /2} {{\mathop{\rm logcos}\nolimits} xdx} = \int\limits_0^{\pi /2} {\left( {\log \sin x + \log \cos x} \right)dx}$

$= \int\limits_0^{\pi /2} {\log \left( {\sin x\cos x} \right)dx} = \int\limits_0^{\pi /2} {\log \left( {\cfrac{{2\sin x\cos x}}{2}} \right)dx}$

$= \int\limits_0^{\pi /2} {\log \left( {\cfrac{{\sin 2x}}{2}} \right)dx} = \int\limits_0^{\pi /2} {\log \sin 2xdx} - \int\limits_0^{\pi /2} {\log 2dx}$

$= \int\limits_0^{\pi /2} {\log \sin 2xdx - \left( {\log 2} \right)\left[ x \right]_0^{\pi /2}}$

$= \int\limits_0^{\pi /2} {\log \sin 2xdx - \left( {\log 2} \right)\left( {\cfrac{\pi }{2} - 0} \right)}$

$= \int\limits_0^{\pi /2} {\log \sin 2xdx - \cfrac{\pi }{2}\log 2 = {I_2} - \cfrac{\pi }{2} - \log 2}$

…..(V)
Where ${I_2} = \int\limits_0^{\pi /2} {\log \sin 2x} dx$

Put $2x = t$ $\Rightarrow$ $2dx = dt$

When $x = 0,t = 0$ and when $x = \cfrac{\pi }{2},t = \pi$

$\therefore$ ${I_2} = \cfrac{1}{2}\int\limits_0^\pi {\log \sin tdt} = \cfrac{1}{2} \cdot 2\int\limits_0^{\pi /2} {\log \sin t\,dt}$

$= \int\limits_0^{\pi /2} {\log \,sin\,x\,dx} = {I_1}$ $\therefore$ From (V),

we get
$2{I_1} = {I_1} - \cfrac{\pi }{2}\log 2$ $\Rightarrow$ ${I_1} = - \cfrac{\pi }{2}\log 2$

$\therefore$ $I = 2 \times \left( { - \cfrac{\pi }{2}\log 2} \right) = - \pi \log 2$