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Evaluate:$\int \limits_0^{\pi/2} log (\cos \theta) d\theta$

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  • (i)$\int \limits _a^b f(x)dx=F(b)-F(a)$
  • (ii) $\int \limits_0^{2a} f(x)dx=2 \int \limits_0^a f(x)dx\;if (2a-x)=f(x)$
  • (iii) $\int \limits_0^a f(x)dx =\int \limits_0^a f(a-x)dx$
Given $=\int \limits_0^{\pi/2} log (\cos \theta) d\theta$-----(1)
By applying the property $\int \limits_0^a f(x)dx =\int \limits_0^a f(a-x)dx$
$I=\int \limits_0^{\pi/2} log \;\cos (\frac{\pi}{2}-\theta) d\theta$
$=\int \limits_0^{\pi/2} log \;\sin \theta d\theta$-----(2)
Adding equ(1) and equ(2)
$2I=\int \limits_0^{\pi/2} (log \sin \theta+log \cos \theta) d\theta$
But log a+log b=log ab
Therefore $2I=\int \limits_0^{\pi/2} log \;\sin \theta \cos \theta. d\theta$
But we know $\sin 2 \theta=2 \sin \theta=2 \sin \theta \cos \theta => \frac{\sin 2 \theta}{2}=\sin \theta \cos \theta$
Therefore $2I=\int \limits_0^{\pi/2} log (\frac{\sin 2 \theta}{2}) d\theta$
But $log (\frac{a}{b})=log a-log b$
Hence $2I=\int \limits_0^{\pi/2} log \;\sin 2\theta d\theta-\int \limits_0^{\pi/2} log \;2 d\theta$
Let $x_1=\int \limits_0^{\pi/2} log \;\sin 2\theta d\theta$
Put $2 \theta=t$ on differentiating w.r.t $\theta$, we get ,
$2d\theta=dt =>$ The limits also change when we substitute for $\theta$
When $ \theta=0,t=0$
$\theta =\frac{\pi}{2}\;t=\pi$
Therefore $I_1=\int \limits_0^{\pi} log \sin t. \frac{dt}{2}$
$=\frac{1}{2} \int \limits_0^{\pi} log \sin t.dt$
Here $f(t)=\log \sin t=> log \sin (\pi-t)= log \sin t=f(t)$
By applying the property,
$\Sigma \int \limits_0^{2a} f(x)dx=2 \int \limits_0^a f(x)dx\;if\;(2a-x)=f(x)$
$I_1=\frac{1}{2} \times 2 \int \limits_0^{\pi/2} log \sin t.dt$
$=\int \limits_0^{\pi/2} log \sin t.dt$
By the property $\int \limits_0^a f(x)dx=\int \limits_0^a f(t)dt$
Therefore $I_1=\int \limits_0^{\pi/2} log \sin \theta d\theta$
But this is equal to I from equ(2)
Therefore $2I=I-\int \limits_0^{\pi/2} log 2 d\theta$
=>$I=-\int \limits_0^{\pi/2} log2\; d \theta$
On integrating we get
On applying limits we get,
$I=-log 2.\frac{\pi}{2}-0$
$=-\frac{\pi}{2} log 2$
answered Apr 1, 2013 by meena.p
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