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By using the properties of definite integrals,evaluate the integral\[\int\limits_0^\pi\; log(1+\cos x)\;dx\]

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  • (i)$\int f(x)dx=F(b)-F(a)$
  • (ii)$\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
Given $I=\int\limits_0^\pi log(1+\cos x)\;dx-----(1)$
Applying the property $\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
$I=\int\limits_0^\pi log(1+\cos (\pi-x)\;dx-----(2)$
$=\int\limits_0^\pi log(1-\cos x)\;dx$
Adding equ (1) and equ(2) we get,
$2I=\int\limits_0^\pi log(1+\cos x)\;dx +\int\limits_0^\pi log(1-\cos x)\;dx$
$=\int\limits_0^\pi log(1+\cos x)\;(1-\cos x)dx$
$=\int\limits_0^\pi log(1-\cos ^2x)\;dx$
But $(1-\cos ^2x)=\sin^2 x$
Hence $\int\limits_0^\pi log \sin ^2x\;dx$
$=\int\limits_0^\pi 2 log \sin x\;dx$
$2I=\int\limits_0^\pi log \sin x\;dx$
Therefore $I=\int\limits_0^\pi log \sin x\;dx-----(3)$
Applying the property $\int \limits_0^a f(x)dx=2\int \limits_0^a f(2a-x)dx\; if f(2a-x)=f(x)$
$I=\int\limits_0^\pi log \sin (\pi-x)\;dx$
But $ \sin(\pi-x)=\sin x$
Therefore $=2\int\limits_0^{\pi/2} log \sin (\pi-x)\;dx-----(4)$
Adding equ (3) and equ(4) we get,
$2I=2\int\limits_0^{\pi/2} log \sin x\;dx \qquad => I=\int\limits_0^{\pi/2} log \sin x\;dx-----(5)$
Now applying the property $\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
$I=2 \times \int\limits_0^{\pi/2} log \sin (\pi/2-x)\;dx$
But $ \sin(\frac{\pi}{2}-x)=\cos x$
Therefore $I=2\int\limits_0^{\pi/2} log \cos x\;dx-----(6)$
Now adding equ (5) and equ(6)
$2I=2\int\limits_0^{\pi/2} log \sin x\;dx+2\int\limits_0^{\pi/2} log \cos x\;dx$
$I=\int\limits_0^{\pi/2} log (\sin x+\cos x)\;dx -----(7)$
Add and subtract log 2 in equ(7)
$I=\int\limits_0^{\pi/2} (log \sin x+log \cos x+log 2-log 2)\;dx$
$=\int\limits_0^{\pi/2} (log2 \sin x \cos x-log 2)\;dx$
But $2\sin x\cos x=\sin 2x$
$=\int\limits_0^{\pi/2} log \sin 2x\;dx-\int\limits_0^{\pi/2} log 2 dx$
Let $ 2x=t$ on differentiating w.r.t t,
$2dx=dt \qquad=>dx=dt/2$
When x=0,t=0 and when $x=\frac{\pi}{2},t=\pi$
$I=\frac{1}{2}\int\limits_0^\pi log \sin t.dt -\frac{1}{2} \int \limits_0^{\pi}{2} log 2 dx$
On integrating
$\int \limits_0^{\pi/2} log 2 dx= log 2 [x]_0^{\pi/2}$
On applying limits,
$ log 2(\pi/2-0)=\frac{\pi}{2} log 2$
Therefore $I=\frac{1}{2}\int\limits_0^\pi log \sin t.\;dt-\frac{\pi}{2}log 2$
Consider $ \frac{1}{2}\int \limits_0^\pi log \sin t.dt$
$\frac{1}{2}.2 \int \limits_0^{\pi/2} log \sin t.dt$
This is obtained byapplying the property Applying the property $\int \limits_0^a f(x)dx=2\int \limits_0^a f(2a-x)dx\; if f(2a-x)=f(x)$
Therefore $ \frac{1}{2}.2\int \limits_0^{\pi/2} log \sin x\;dx$=I
Therefore $I=\frac{1}{2}I-\frac{\pi}{2}log 2$
$ =>I-\frac{I}{2}=-\frac{\pi}{2}log 2$
$\frac{I}{2}=-\frac{\pi}{2}log 2$
Therefore $ I=-\pi log 2$



answered Feb 14, 2013 by meena.p

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