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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate: $ \int \limits _0 ^{\pi} log(1+\cos x)dx$

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Toolbox:
  • $\int \limits_a^b f(x)dx =F(b)-F(a)$
  • $\int \limits_0^a f(x)dx = \int \limits_0^a f(a-x)dx$
Step 1:
Given $I= \int \limits_0^\pi \log (1+\cos x)dx$-----(1)
By applying the property $\int \limits_0^a f(x)dx = \int \limits_0^a f(a-x)dx$
$I= \int \limits_0^{\pi} \log (1+\cos (\pi-x)dx)$
But $\cos (\pi- x) =-\cos x$
Therefore $I= \int \limits_ 0^{\pi} \log (1-\cos x) dx $-----(2)
Adding equ(1) and equ(2)
$2I= \int \limits_0^{\pi} \bigg[\log (1+\cos x)+\log (1-\cos x)\bigg] dx$
But $\log a+\log b=\log (a \times b)$
Thererfore $ 2I= \int \limits_0^{\pi} \log (1+\cos x) (1-\cos x) dx$
This is of the form $(a+b)(a-b) =a^2-b^2$
Therefore $2I= \int \limits_0^{\pi} \log (1-\cos ^2 x) dx$
We know $1-\cos ^2 x=\sin ^2 x$
Therefore $2I= \int \limits_0^{\pi} \log\; \sin ^2 x \qquad [But \; \log x^2 =2 \log x]$
Step 2:
Therefore $ 2I= \int \limits_0^\pi 2\log \;\sin x dx$
$I= \int \limits_0^\pi \log \sin x dx$
Since $ \int \limits_0^a f(x)dx =\int \limits _0^a f(a-x) dx$
$ 2 \int \limits _0^a f(x)=2 \int \limits_0^a f(2a-x)$
Therefore $ \int \limits_0^\pi \log \; \sin x= 2 \int \limits_0^{\pi/2} \log \; \sin (\pi-x) dx$
$= 2 \int \limits_0^{\pi/2} \log \; \sin x dx$-----(A)
Consider $\int \limits_0^{\pi/2} \log \; \sin x dx$-----(1)
By applying the property $\int \limits_0^a f(x)dx = \int \limits_0^a f(a-x)dx$
$\int \limits_0^{\pi/2} \log \; \sin (\pi/2-x) dx$
But $ \sin (\pi/2-x)=\cos x$
$I=\int \limits_0^{\pi/2} \log \; \sin (\pi/2-x) dx=\int \limits_0^{\pi/2} \cos x dx$------(2)
Step 3:
Adding equ (1) and equ(2)
$2I= \int \limits_0^{\pi/2} \log \; (\sin x+\log \cos x) dx$
But $\log a+\log b=\log (a \times b)$
Hence $ 2I=\int \limits_0^{\pi/2} \log ( \sin x \cos x) dx$
But $ \large\frac{2}{2}$$\sin x \cos x=$$\large\frac{1}{2}$$\sin 2x$
Therefore $ 2I=\int \limits_0^{\pi/2} \log \large \frac{ \sin 2x}{2}$$ dx$
But $ \log \bigg|\frac{a}{b} \bigg|= \log a -\log b$
Therefore $ 2I= \int \limits_0^{\pi/2} (\log \sin 2x -\log 2) dx$
$= \int \limits_0^{\pi/2} \log \; \sin 2x -\int \limits_0^{\pi/2} \log 2 dx$
Let $I_1= \int \limits_0^{\pi/2} \log \sin 2x$
$2dx=dt$
The limits changes as we substitute for t,
When $x=0,t=0$
When $x=\pi/2,t=\pi$
Therefore $I_1=\large\frac{1}{2} $$\int \limits_0^\pi log \sin t.dt$
By property $ \int \limits \limits_0^{2a} f(x) dx= 2 \int \limits_0^a f(x)dx$ if $(2a-x)=f(x) $
Step 4:
Hence $I_1=\large\frac{1}{2}$$ \int \limits_0^\pi \log \sin t dt= \large\frac{1}{2}$$ .2 \int \limits _0^{\pi/2} \log \sin t.dt$
$= \int \limits_0^{\pi/2} \log \sin t.dt$
By property $\int \limits_0^a f(x) dx=\int \limits_0^a f(t) dt$
Hence $I_1= \int \limits_0^{\pi/2} \log \sin x dx$
$=I$
Therefore $2I= I- \int \limits_0^{\pi/2} \log 2 .dx$
$I= - \log \int \limits_0^{\pi/2} \log 2\; dx$
Step 5:
On integrating we get
$I= -\log 2 \int \limits_0^{\pi/2} [dx] = -\log 2 [x]_0^{\pi/2}$
On applying limits we get,
$I= -\log 2 (\pi/2) =-\pi/2 \log 2$
Substituting in equ (A)
$I= 2 \int \limits_0^{\pi/2} \log \sin x dx=2 [ -\pi/2 \log 2]$
$= -\pi \log 2$
But $-\log a =log(1/a)$
Hence $I= \pi \log \large\frac{1}{2}$
answered May 7, 2013 by meena.p
 
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