INTEGRALS — Maths STD 12 Science — Question

1. $\pi\ln 2$

Solution:

$\int\limits^{\infty}_0\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\frac{1}{1+\text{x}^2}\text{ dx}$

Substitute $\text{x}=\tan\theta$

$\text{dx}=\sec^2\theta\text{d}\theta$

When,

$\text{x}=0\Rightarrow\theta=0$

$\text{x}=\infty\Rightarrow\theta=\frac{\pi}{2}$

$\int\limits^{\frac{\pi}{2}}_0\Big(\tan\theta+\frac{1}{\tan\theta}\Big)\frac{1}{1+\tan^{2}}\times\sec^2\theta\text{d}\theta$

$\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\tan^2\theta+1}{\tan\theta}\Big)\frac{1}{1+\tan^{2}\theta}\times\sec^2\theta\text{d}\theta$

$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\sec^2\theta}{\tan\theta}\Big)\frac{1}{\sec^2\theta}\times\sec^2\theta\text{d}\theta$ $\big[\because1+\tan^2\theta=\sec^2\theta\big]$

$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{\sec^2\theta}{\tan\theta}\Big)\text{d}\theta$

$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{1}{\sin\theta\cdot\cos\theta}\Big)\text{d}\theta$

$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\big(\sin\theta\cos\theta\big)\text{d}\theta$

$\Rightarrow-\int\limits^{\frac{\pi}{2}}_0\big[\log\sin\theta+\log\cos\theta\big]\text{d}\theta$

$\Rightarrow-\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{d}\theta-\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{d}\theta$

Let us conside,

$\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{d}\theta=\text{I}\ ....(\text{i})$

$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\Big(\sin\Big(\frac{\pi}{2}-\theta\Big)\Big)\text{d}\theta$

$\Rightarrow\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{d}\theta\ ...(\text{ii})$

Adding (i) and (ii)

$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta+\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{ d}\theta$

$=\int\limits^{\frac{\pi}{2}}_0\log(\sin\theta\cdot\cos\theta)\text{d}\theta$

$=\int\limits^{\frac{\pi}{2}}_0\log\big(\sin2\theta\big)\text{d}\theta-\int\limits^{\frac{\pi}{2}}_0\log2\text{ d}\theta$

Let us consider $2\theta=\text{t}$

$2\text{d}\theta=\text{dt}$

$2\text{I}=\frac{1}{2}\int\limits^{\pi}_{0}\log(\sin\text{t})\text{dt}-\frac{\pi}{2}\log2$

$2\text{I}=\frac{2}{2}\int\limits^{\pi}_{0}\log(\sin\text{t})\text{dt}-\frac{\pi}{2}\log2$ $\big[\because\sin\theta$ is positive in both 1^st and 2^nd quadrants$\big]$

$2\text{I}=\text{I}-\frac{\pi}{2}\log2$

$2\text{I}-\text{I}=-\frac{\pi}{2}\log2$

$\text{I}=-\frac{\pi}{2}\log2,$ where $\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta$

Now,

$-\int\limits^{\frac{\pi}{2}}_0\log(\sin\theta)\text{ d}\theta-\int\limits^{\frac{\pi}{2}}_0\log\cos\theta\text{ d}\theta$

$-2\int\limits^{\frac{\pi}{2}}_0\log\sin\theta\text{ d}\theta=-2\times\text{I}$

$=-2\times-\frac{\pi}{2}\log2$ $\Big[\text{Where I}=-\frac{\pi}{2}\log2\Big]$

$=\pi\log2$