Definite Integration

$ \begin{aligned} & \text { } \int_a^b e^x d x=\lim _{h \rightarrow 0} h\left[e^a+e^{a+h}+e^{a+2 h}+\ldots+e^{a+(n-1) h}\right] \\ & =\lim _{h \rightarrow 0} h e^a\left[1+e^h+e^{2 h}+e^{3 h}+\ldots+e^{(n-1) h}\right] \\ & =\lim _{h \rightarrow 0} h e^a\left\{\frac{\left(e^h\right)^n-1}{e^h-1}\right\} \\ & \quad\left[\because a+a r+\ldots+a r^{n-1}=a\left(\frac{r^n-1}{r-1}\right)\right] \end{aligned} $

$ \begin{aligned} & =\lim _{h \rightarrow 0} h e^a\left(\frac{e^{n h}-1}{e^h-1}\right) \\ & =\lim _{h \rightarrow 0} e^a\left\{\frac{e^{b-a}-1}{\left(\frac{e^h-1}{h}\right)}\right\}=e^a\left(e^{b-a}-1\right)=e^b-e^a \end{aligned} $

We have, $I=\int_0^{\frac{\pi}{2}} \log (\sin x) d x$

Using the property:

$ \int_0^a f(x) d x=\int_0^a f(a-x) d x $

we replace $x$ with $\frac{\pi}{2}-x$, we get

$ \begin{aligned} & I=\int_0^{\frac{\pi}{2}} \log (\cos x) d x \\ \Rightarrow & I+I=\int_0^{\frac{\pi}{2}}[\log (\cos x)+\log (\sin x)] d x \\ = & \int_0^{\frac{\pi}{2}} \log \left(\frac{1}{2} \sin 2 x\right) d x=\int_0^{\frac{\pi}{2}} \log \left(\frac{1}{2}\right) d x+\int_0^{\frac{\pi}{2}} \log (\sin 2 x) d x \\ = & -\frac{\pi}{2} \log 2+\int_0^{\frac{\pi}{2}} \log (\sin x) d x \\ \Rightarrow & 2 I=-\frac{\pi}{2} \log 2+I \\ \Rightarrow & I=-\frac{\pi}{2} \log 2 \end{aligned} $

Let $ I=\int_{0}^{\infty} \frac{1}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x $

$ \begin{array}{l} I=\frac{1}{a^{2}-b^{2}} \int_{0}^{\infty} \frac{\left(x^{2}+a^{2}\right)-\left(x^{2}+b^{2}\right)}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x \\ I=\frac{1}{a^{2}-b^{2}} \int_{0}^{\infty} \frac{1}{\left(x^{2}+b^{2}\right)}-\frac{1}{\left(x^{2}+a^{2}\right)} d x \\ I=\frac{1}{a^{2}-b^{2}}\left[\frac{1}{b} \tan ^{-1} \frac{x}{b}-\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]_{0}^{\infty} \\ I=\frac{1}{a^{2}-b^{2}}\left[\frac{1}{b} \times \frac{\pi}{2}-\frac{1}{a} \times \frac{\pi}{2}\right] \\ I=\frac{1}{(a+b)(a-b)}\left[\frac{a-b}{a b}\right] \times \frac{\pi}{2} \\ I=\frac{\pi}{2 a b(a+b)} \end{array} $

Let $ I=\int_{0}^{\frac{\pi}{2}} \log (\tan x) d x \quad \ldots \text { (i) } $

By applying property,

$ \begin{aligned} \int_{a}^{b} f(x) d x & =\int_{a}^{b} f(a+b-x) d x \\ I & =\int_{0}^{\frac{\pi}{2}} \log \left(\tan \left(\frac{\pi}{2}-x\right)\right) d x \\ I & =\int_{0}^{\frac{\pi}{2}} \log (\cot x) d x \end{aligned} \quad \ldots \text { (ii) } $

Adding Eqs. (i) and (ii),

$ \begin{array}{l} 2 I=\int_{0}^{\frac{\pi}{2}}[\log (\tan x)+\log (\cot x)] d x \\ 2 I=\int_{0}^{\frac{\pi}{2}} \log 1 d x=0 \end{array} $

Let $I = \int_0^1 {{{\log (1 + x)} \over {1 + {x^2}}}dx = \int_0^{\pi /4} {{{\log (1 + \tan \theta )} \over {(1 + {{\tan }^2}\theta )}}{{\sec }^2}\theta d\theta } } $

Let $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta d\theta $

$ = \int_0^{\pi /4} {\log (1 + \tan \theta )d\theta = \int_0^{\pi /4} {\log \left[ {1 + \tan \left( {{\pi \over 4} - \theta } \right)} \right]d\theta } } $

$ = \int_0^{\pi /4} {\log \left( {1 + {{1 - \tan \theta } \over {1 + \tan \theta }}} \right)d\theta } $

$ = \int_0^{\pi /4} {\log \left( {{2 \over {1 + \tan \theta }}} \right)d\theta = \log 2\int_0^{\pi /4} {1d\theta - I} } $

$ \Rightarrow 2I = {\pi \over 4}\log 2 \Rightarrow I = {\pi \over 8}\log 2$

We have,

$I = \int\limits_0^{{\pi \over 2}} {{{dx} \over {\tan x + \cot x + \cos ec\,x + \sec x}}} $

$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{dx} \over {{{\sin x} \over {\cos x}} + {{\cos x} \over {\sin x}} + {1 \over {\sin x}} + {1 \over {\cos x}}}}} $

$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{\sin x\cos xdx} \over {1 + \sin x + \cos x}}} $

$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{2\sin {x \over 2}\cos {x \over 2}\left( {{{\cos }^2}{x \over 2} - {{\sin }^2}{x \over 2}} \right)} \over {2{{\cos }^2}{x \over 2} + 2\sin {x \over 2}\cos {x \over 2}}}} $

$ \Rightarrow I = \int\limits_0^{{\pi \over 2}} {\sin {x \over 2}\left( {\cos {x \over 2} - \sin {x \over 2}} \right)dx} $

$ \Rightarrow I = {1 \over 2}\int\limits_0^{{\pi \over 2}} {(\sin x + \cos x - 1)dx} $

$ \Rightarrow I = {1 \over 2}[ - \cos x + \sin x - x]_0^{\pi /2}$

$ \Rightarrow I = {1 \over 2}\left[ {\left( { - \cos {\pi \over 2} + \sin {\pi \over 2} - {\pi \over 2}} \right) - ( - \cos 0 + \sin 0 - 0} \right]$

$ \Rightarrow I = {1 \over 2}\left[ {0 + 1 - {\pi \over 2} + 1} \right] = 1 - {\pi \over 4}$