শর্তমতে,
\(u=u(x)\) এবং \(w=w(x)\)
ধরি,
\(v=\frac{d}{dx}(w)\)
\(\Rightarrow \int{vdx}=\int{\frac{d}{dx}(w)dx}\) ➜Note উভয় পার্শে \(x\)-এর সাপেক্ষে যোগজীকরণ করে।
\(\Rightarrow \int{vdx}=\int{dw}\)
\(\Rightarrow \int{vdx}=w\) ➜Note যোগজীকরণ ধ্রুবক \((c)\) উপেক্ষা করে ।
\(\Rightarrow u\int{vdx}=uw\) ➜Note উভয় পার্শে \(u\) গুণ করে ।
\(\Rightarrow \frac{d}{dx}\left(u\int{vdx}\right)=\frac{d}{dx}(uw)\) ➜Note উভয় পার্শে \(x\)-এর সাপেক্ষে অন্তরীকরণ করে।
\(\Rightarrow \frac{d}{dx}\left(u\int{vdx}\right)=u\frac{d}{dx}(w)+w\frac{d}{dx}(u)\) ➜Note \(\because \frac{d}{dx}(uv)=u\frac{d}{dx}(v)+v\frac{d}{dx}(u)\)
\(\Rightarrow \frac{d}{dx}\left(u\int{vdx}\right)=uv+\int{vdx}.\frac{d}{dx}(u)\) ➜Note \(\because \frac{d}{dx}(w)=v, w=\int{vdx}\)
\(\Rightarrow \frac{d}{dx}\left(u\int{vdx}\right)=uv+\frac{d}{dx}(u)\int{vdx}\)
\(\Rightarrow \int{\frac{d}{dx}\left(u\int{vdx}\right)dx}=\int{(uv)dx}+\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\) ➜Note উভয় পার্শে \(x\)-এর সাপেক্ষে যোগজীকরণ করে।
\(\Rightarrow \int{d\left(u\int{vdx}\right)}=\int{(uv)dx}+\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\) \(\Rightarrow u\int{vdx}=\int{(uv)dx}+\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\) ➜Note \(\because \int{dx}=x\)
\(\Rightarrow \int{(uv)dx}+\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}=u\int{vdx}\)
\(\therefore \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)

\(\int{\ln{|x|}dx}\)
\(=\int{\ln{|x|}.1dx}\)
\(=\ln{|x|}\int{1dx}-\int{\left\{\frac{d}{dx}(\ln{|x|})\int{1dx}\right\}dx}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\ln{|x|}.x-\int{\left\{\frac{1}{x}.x\right\}dx}\) ➜Note \(\because \int{1dx}=x, \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(=x\ln{|x|}-\int{1dx}\)
\(=x\ln{|x|}-x+c\) ➜Note \(\because \int{1dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x(\ln{|x|}-1)+c\)

\(\int{\log{|x|}dx}\)
\(=\int{\log_ae\times\log_e{|x|}dx}\) ➜Note \(\because \log_p{|x|}=\log_pe\times\log_e{|x|}\)
\(=\log_ae\int{\ln{|x|}dx}\)
\(=\log_ae\int{\ln{|x|}.1dx}\)
\(=\log_ae\left[\ln{|x|}\int{1dx}-\int{\left\{\frac{d}{dx}(\ln{|x|})\int{1dx}\right\}dx}\right]\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\log_ae\left[\ln{|x|}.x-\int{\left\{\frac{1}{x}.x\right\}dx}\right]\) ➜Note \(\because \int{1dx}=x, \frac{d}{dx}(\ln{|x|})=\frac{1}{x}\)
\(=\log_ae\left(x\ln{|x|}-\int{1dx}\right)\)
\(=\log_ae\left(x\ln{|x|}-x\right)+c\) ➜Note \(\because \int{1dx}=x\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=\log_a{e}\times x(\ln{|x|}-1)+c\)

\(\int{\sin^{-1}{x}dx}\)
\(=\int{\sin^{-1}{x}.1dx}\)
\(=\sin^{-1}{x}\int{1dx}-\int{\left\{\frac{d}{dx}(\sin^{-1}{x})\int{1dx}\right\}dx}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\sin^{-1}{x}.x-\int{\left\{\frac{1}{\sqrt{1-x^2}}.x\right\}dx}\) ➜Note \(\because \int{1dx}=x, \frac{d}{dx}(\sin^{-1}{x})=\frac{1}{\sqrt{1-x^2}}\)
\(=x\sin^{-1}{x}-\int{\frac{x}{\sqrt{1-x^2}}dx}\)
\(=x\sin^{-1}{x}+\frac{1}{2}\int{\frac{-2xdx}{\sqrt{1-x^2}}}\)
\(=x\sin^{-1}{x}+\frac{1}{2}\int{\frac{dt}{\sqrt{t}}}\)
\(=x\sin^{-1}{x}+\frac{1}{2}.2\sqrt{t}+c\) ➜Note \(\because \int{\frac{dx}{\sqrt{x}}}=2\sqrt{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x\sin^{-1}{x}+\sqrt{t}+c\)
\(=x\sin^{-1}{x}+\sqrt{1-x^2}+c\) ➜Note \(\because t=1-x^2\)

\(\int{\cos^{-1}{x}dx}\)
\(=\int{\cos^{-1}{x}.1dx}\)
\(=\cos^{-1}{x}\int{1dx}-\int{\left\{\frac{d}{dx}(\cos^{-1}{x})\int{1dx}\right\}dx}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\cos^{-1}{x}.x-\int{\left\{-\frac{1}{\sqrt{1-x^2}}.x\right\}dx}\) ➜Note \(\because \int{1dx}=x, \frac{d}{dx}(\cos^{-1}{x})=-\frac{1}{\sqrt{1-x^2}}\)
\(=x\cos^{-1}{x}+\int{\frac{x}{\sqrt{1-x^2}}dx}\)
\(=x\cos^{-1}{x}-\frac{1}{2}\int{\frac{-2xdx}{\sqrt{1-x^2}}}\)
\(=x\cos^{-1}{x}-\frac{1}{2}\int{\frac{dt}{\sqrt{t}}}\)
\(=x\cos^{-1}{x}-\frac{1}{2}.2\sqrt{t}+c\) ➜Note \(\because \int{\frac{dx}{\sqrt{x}}}=2\sqrt{x}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x\cos^{-1}{x}-\sqrt{t}+c\)
\(=x\cos^{-1}{x}-\sqrt{1-x^2}+c\) ➜Note \(\because t=1-x^2\)

\(\int{\tan^{-1}{x}dx}\)
\(=\int{\tan^{-1}{x}.1dx}\)
\(=\tan^{-1}{x}\int{1dx}-\int{\left\{\frac{d}{dx}(\tan^{-1}{x})\int{1dx}\right\}dx}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\tan^{-1}{x}.x-\int{\left\{\frac{1}{1+x^2}.x\right\}dx}\) ➜Note \(\because \int{1dx}=x, \frac{d}{dx}(\tan^{-1}{x})=\frac{1}{1+x^2}\)
\(=x\tan^{-1}{x}-\int{\frac{1}{1+x^2}xdx}\)
\(=x\tan^{-1}{x}-\int{\frac{1}{t}\times{\frac{1}{2}}dt}\)
\(=x\tan^{-1}{x}-\frac{1}{2}\int{\frac{1}{t}dt}\)
\(=x\tan^{-1}{x}-\frac{1}{2}\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x\tan^{-1}{x}-\frac{1}{2}\ln{|1+x^2|}+c\) ➜Note \(\because t=1+x^2\)

\(\int{cosec^{-1}{x}dx}\)
\(=\int{cosec^{-1}{x}.1dx}\)
\(=cosec^{-1}{x}\int{1dx}-\int{\left\{\frac{d}{dx}(cosec^{-1}{x})\int{1dx}\right\}dx}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=cosec^{-1}{x}.x-\int{\left\{-\frac{1}{x\sqrt{x^2-1}}.x\right\}dx}\) ➜Note \(\because \int{1dx}=x, \frac{d}{dx}(cosec^{-1}{x})=-\frac{1}{x\sqrt{x^2-1}}\)
\(=x \ cosec^{-1}{x}+\int{\frac{1}{\sqrt{x^2-1}}dx}\)
\(=x \ cosec^{-1}{x}+\int{\frac{1}{\sqrt{x^2-1^2}}dx}\)
\(=x \ cosec^{-1}{x}+\ln{|x+\sqrt{x^2-1^2}|}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{x^2-a^2}}dx}=\ln{|x+\sqrt{x^2-a^2}|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x \ cosec^{-1}{x}+\ln{|x+\sqrt{x^2-1}|}+c\)

\(\int{\sec^{-1}{x}dx}\)
\(=\int{\sec^{-1}{x}.1dx}\)
\(=\sec^{-1}{x}\int{1dx}-\int{\left\{\frac{d}{dx}(\sec^{-1}{x})\int{1dx}\right\}dx}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\sec^{-1}{x}.x-\int{\left\{\frac{1}{x\sqrt{x^2-1}}.x\right\}dx}\) ➜Note \(\because \int{1dx}=x, \frac{d}{dx}(\sec^{-1}{x})=\frac{1}{x\sqrt{x^2-1}}\)
\(=x\sec^{-1}{x}-\int{\frac{1}{\sqrt{x^2-1}}dx}\)
\(=x\sec^{-1}{x}-\int{\frac{1}{\sqrt{x^2-1^2}}dx}\)
\(=x\sec^{-1}{x}-\ln{|x+\sqrt{x^2-1^2}|}+c\) ➜Note \(\because \int{\frac{1}{\sqrt{x^2-a^2}}dx}=\ln{|x+\sqrt{x^2-a^2}|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x\sec^{-1}{x}-\ln{|x+\sqrt{x^2-1}|}+c\)

\(\int{\cot^{-1}{x}dx}\)
\(=\int{\cot^{-1}{x}.1dx}\)
\(=\cot^{-1}{x}\int{1dx}-\int{\left\{\frac{d}{dx}(\cot^{-1}{x})\int{1dx}\right\}dx}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\cot^{-1}{x}.x-\int{\left\{-\frac{1}{1+x^2}.x\right\}dx}\) ➜Note \(\because \int{1dx}=x, \frac{d}{dx}(\cot^{-1}{x})=-\frac{1}{1+x^2}\)
\(=x\cot^{-1}{x}+\int{\frac{1}{1+x^2}xdx}\)
\(=x\cot^{-1}{x}+\int{\frac{1}{t}\times{\frac{1}{2}}dt}\)
\(=x\cot^{-1}{x}+\frac{1}{2}\int{\frac{1}{t}dt}\)
\(=x\cot^{-1}{x}+\frac{1}{2}\ln{|t|}+c\) ➜Note \(\because \int{\frac{1}{x}dx}=\ln{|x|}\) এবং \(c\) যোগজীকরণ ধ্রুবক।
\(=x\cot^{-1}{x}+\frac{1}{2}\ln{|1+x^2|}+c\) ➜Note \(\because t=1+x^2\)

ধরি,
\(I=\int{e^x\sin{x}dx}\)
\(\Rightarrow I=\int{\sin{x}.e^xdx}\)
\(\Rightarrow I=\sin{x}\int{e^xdx}-\int{\left\{\frac{d}{dx}(\sin{x})\int{e^xdx}\right\}dx}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\sin{x}.e^x-\int{\left\{\cos{x}.e^x\right\}dx}\) ➜Note \(\because \int{e^xdx}=e^x, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow I=e^x\sin{x}-\int{\cos{x}.e^xdx}\)
\(\Rightarrow I=e^x\sin{x}-\left[\cos{x}\int{e^xdx}-\int{\left\{\frac{d}{dx}(\cos{x})\int{e^xdx}\right\}dx}\right]\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=e^x\sin{x}-\left[\cos{x}.e^x-\int{(-\sin{x}).e^xdx}\right]\) ➜Note \(\because \int{e^xdx}=e^x, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow I=e^x\sin{x}-\left[e^x\cos{x}+\int{e^x\sin{x}dx}\right]\)
\(\Rightarrow I=e^x\sin{x}-\left[e^x\cos{x}+I\right]\) ➜Note \(\because \int{e^x\sin{x}dx}=I\)
\(\Rightarrow I=e^x\sin{x}-e^x\cos{x}-I\)
\(\Rightarrow I+I=e^x\sin{x}-e^x\cos{x}\)
\(\Rightarrow 2I=e^x\sin{x}-e^x\cos{x}\)
\(\Rightarrow I=\frac{1}{2}(e^x\sin{x}-e^x\cos{x})+c\) ➜Note\(c\) যোগজীকরণ ধ্রুবক।
\(\therefore \int{e^x\sin{x}dx}=\frac{1}{2}(e^x\sin{x}-e^x\cos{x})+c\)

\(L.S=I=\int{e^{ax}\sin{bx}dx}\)
\(\Rightarrow I=\int{\sin{bx}.e^{ax}dx}\)
\(\Rightarrow I=\sin{bx}\int{e^{ax}dx}-\int{\left\{\frac{d}{dx}(\sin{bx})\int{e^{ax}dx}\right\}dx}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\sin{bx}.\frac{e^{ax}}{a}-\int{\left\{\cos{bx}.\frac{d}{dx}(bx).\frac{e^{ax}}{a}\right\}dx}\) ➜Note \(\because \int{e^{ax}dx}=\frac{e^{ax}}{a}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\int{\left\{\cos{bx}.b.\frac{e^{ax}}{a}\right\}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a}\int{\cos{bx}.e^{ax}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a}\left[\cos{bx}\int{e^{ax}dx}-\int{\left\{\frac{d}{dx}(\cos{bx})\int{e^{ax}dx}\right\}dx}\right]\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a}\left[\cos{bx}.\frac{e^{ax}}{a}-\int{(-\sin{bx}).\frac{d}{dx}(bx).\frac{e^{ax}}{a}dx}\right]\) ➜Note \(\because \int{e^{ax}dx}=\frac{e^{ax}}{a}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a}\left[\frac{1}{a}e^{ax}\cos{bx}+\int{\sin{bx}.b.\frac{e^{ax}}{a}dx}\right]\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a}\left[\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a}\int{\sin{bx}.e^{ax}dx}\right]\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a^2}e^{ax}\cos{bx}-\frac{b^2}{a^2}\int{\sin{bx}.e^{ax}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a^2}e^{ax}\cos{bx}-\frac{b^2}{a^2}I\) ➜Note \(\because \int{\sin{bx}.e^{ax}dx}=I\)
\(\Rightarrow I+\frac{b^2}{a^2}I=\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a^2}e^{ax}\cos{bx}\)
\(\Rightarrow I\left(1+\frac{b^2}{a^2}\right)=\frac{1}{a^2}e^{ax}(a\sin{bx}-b\cos{bx})\)
\(\Rightarrow I\left(\frac{a^2+b^2}{a^2}\right)=\frac{1}{a^2}e^{ax}(a\sin{bx}-b\cos{bx})\)
\(\Rightarrow I=\frac{a^2}{a^2+b^2}\times{\frac{1}{a^2}}e^{ax}(a\sin{bx}-b\cos{bx})+c\) ➜Note\(c\) যোগজীকরণ ধ্রুবক।
\(\Rightarrow I=\frac{e^{ax}}{a^2+b^2}(a\sin{bx}-b\cos{bx})+c\)
\(\Rightarrow I=R.S\)
\(\therefore L.S=R.S\)
(Proved)

\(L.S=I=\int{e^{ax}\cos{bx}dx}\)
\(\Rightarrow I=\int{\cos{bx}.e^{ax}dx}\)
\(\Rightarrow I=\cos{bx}\int{e^{ax}dx}-\int{\left\{\frac{d}{dx}(\cos{bx})\int{e^{ax}dx}\right\}dx}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\cos{bx}.\frac{e^{ax}}{a}-\int{\left\{(-\sin{bx}).\frac{d}{dx}(bx).\frac{e^{ax}}{a}\right\}dx}\) ➜Note \(\because \int{e^{ax}dx}=\frac{e^{ax}}{a}, \frac{d}{dx}(\cos{x})=-\sin{x}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\int{\left\{\sin{bx}.b.\frac{e^{ax}}{a}\right\}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a}\int{\sin{bx}.e^{ax}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a}\left[\sin{bx}\int{e^{ax}dx}-\int{\left\{\frac{d}{dx}(\sin{bx})\int{e^{ax}dx}\right\}dx}\right]\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a}\left[\sin{bx}.\frac{e^{ax}}{a}-\int{\cos{bx}.\frac{d}{dx}(bx).\frac{e^{ax}}{a}dx}\right]\) ➜Note \(\because \int{e^{ax}dx}=\frac{e^{ax}}{a}, \frac{d}{dx}(\sin{x})=\cos{x}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a}\left[\frac{1}{a}e^{ax}\sin{bx}-\int{\cos{bx}.b.\frac{e^{ax}}{a}dx}\right]\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a}\left[\frac{1}{a}e^{ax}\sin{bx}-\frac{b}{a}\int{\cos{bx}.e^{ax}dx}\right]\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a^2}e^{ax}\sin{bx}-\frac{b^2}{a^2}\int{\cos{bx}.e^{ax}dx}\)
\(\Rightarrow I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a^2}e^{ax}\sin{bx}-\frac{b^2}{a^2}I\) ➜Note \(\because \int{\cos{bx}.e^{ax}dx}=I\)
\(\Rightarrow I+\frac{b^2}{a^2}I=\frac{1}{a}e^{ax}\cos{bx}+\frac{b}{a^2}e^{ax}\sin{bx}\)
\(\Rightarrow I\left(1+\frac{b^2}{a^2}\right)=\frac{1}{a^2}e^{ax}(a\cos{bx}+b\sin{bx})\)
\(\Rightarrow I\left(\frac{a^2+b^2}{a^2}\right)=\frac{1}{a^2}e^{ax}(a\cos{bx}+b\sin{bx})\)
\(\Rightarrow I=\frac{a^2}{a^2+b^2}\times{\frac{1}{a^2}}e^{ax}(a\cos{bx}+b\sin{bx})+c\) ➜Note\(c\) যোগজীকরণ ধ্রুবক।
\(\Rightarrow I=\frac{e^{ax}}{a^2+b^2}(a\cos{bx}+b\sin{bx})+c\)
\(\Rightarrow I=R.S\)
\(\therefore L.S=R.S\)
(Proved)

\(L.S=\int{\sqrt{x^2-a^2}dx}\)
\(=\int{\sqrt{a^2\sec^2{\theta}-a^2}.a\sec{\theta}\tan{\theta}d\theta}\)
\(=\int{\sqrt{a^2(\sec^2{\theta}-1)}.a\sec{\theta}\tan{\theta}d\theta}\)
\(=\int{a\sqrt{\tan^2{\theta}}.a\sec{\theta}\tan{\theta}d\theta}\)
\(=a^2\int{\tan{\theta}.\sec{\theta}\tan{\theta}d\theta}\)
\(=a^2\left[\tan{\theta}\int{\sec{\theta}\tan{\theta}d\theta}-\int{\left\{\frac{d}{d\theta}(\tan{\theta})\int{\sec{\theta}\tan{\theta}d\theta}\right\}d\theta}\right]\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=a^2\left[\tan{\theta}\sec{\theta}-\int{\sec^2{\theta}.\sec{\theta}d\theta}\right]\)
\(=a^2\tan{\theta}\sec{\theta}-a^2\int{\sec^3{\theta}d\theta}\)
\(=a^2\tan{\theta}\sec{\theta}-a^2I\)
ধরি,
\(I=\int{\sec^3{\theta}d\theta}\)
\(\Rightarrow I=\int{\sec^2{\theta}.\sec{\theta}d\theta}\)
\(\Rightarrow I=\sec{\theta}\int{\sec^2{\theta}d\theta}-\int{\left\{\frac{d}{d\theta}(\sec{\theta})\int{\sec^2{\theta}d\theta}\right\}d\theta}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec{\theta}\tan{\theta}.\tan{\theta}d\theta}\) ➜Note \(\because \int{\sec^2{x}dx}=\tan{x}, \frac{d}{dx}(sec{x})=sec{x}\tan{x}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec{\theta}\tan^2{\theta}d\theta}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec{\theta}(\sec^2{\theta}-1)d\theta}\) ➜Note \(\because tan^2{x}=\sec^2{x}-1\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec^3{\theta}d\theta}+\int{\sec{\theta}d\theta}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-I+\int{\sec{\theta}d\theta}\) ➜Note \(\because \int{\sec^3{\theta}d\theta}=I\)
\(\Rightarrow I+I=\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\)
\(\Rightarrow 2I=\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\)
\(\Rightarrow I=\frac{1}{2}\left(\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\right)+c_{1}\) ➜Note \(\because \int{\sec{\theta}d\theta}=\ln{\left|\sec{x}+\tan{x}\right|}\) এবং \(c_{1}\) যোগজীকরণ ধ্রুবক।
এখন,
\(L.S=a^2\tan{\theta}\sec{\theta}-a^2I\)
\(=a^2\tan{\theta}\sec{\theta}-a^2\left\{\frac{1}{2}\left(\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\right)+c_{1}\right\}\)
\(=a^2\tan{\theta}\sec{\theta}-\frac{a^2}{2}\left(\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\right)-a^2c_{1}\)
\(=a^2\tan{\theta}\sec{\theta}-\frac{a^2}{2}\sec{\theta}\tan{\theta}-\frac{a^2}{2}\ln{\left|\sec{\theta}+\tan{\theta}\right|}-a^2c_{1}\)
\(=\frac{2a^2-a^2}{2}\sec{\theta}\tan{\theta}-\frac{a^2}{2}\ln{\left|\sec{\theta}+\tan{\theta}\right|}-a^2c_{1}\)
\(=\frac{a^2}{2}\sec{\theta}\sqrt{\sec^2{\theta}-1}-\frac{a^2}{2}\ln{\left|\sec{\theta}+\sqrt{\sec^2{\theta}-1}\right|}-a^2c_{1}\) ➜Note \(\because \tan{x}=\sqrt{\sec^2{x}-1}\)
\(=\frac{a^2}{2}\times{\frac{x}{a}}\sqrt{\left(\frac{x}{a}\right)^2-1}-\frac{a^2}{2}\ln{\left|\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^2-1}\right|}-a^2c_{1}\) ➜Note \(\because \sec{\theta}=\frac{x}{a}\)
\(=\frac{ax}{2}.\sqrt{\frac{x^2}{a^2}-1}-\frac{a^2}{2}\ln{\left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right|}-a^2c_{1}\)
\(=\frac{ax}{2}.\sqrt{\frac{x^2-a^2}{a^2}}-\frac{a^2}{2}\ln{\left|\frac{x}{a}+\sqrt{\frac{x^2-a^2}{a^2}}\right|}-a^2c_{1}\)
\(=\frac{ax}{2}.\frac{\sqrt{x^2-a^2}}{a}-\frac{a^2}{2}\ln{\left|\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a}\right|}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\ln{\left|\frac{x+\sqrt{x^2-a^2}}{a}\right|}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\ln{\frac{\left|x+\sqrt{x^2-a^2}\right|}{|a|}}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\left(\ln{\left|x+\sqrt{x^2-a^2}\right|}-\ln{|a|}\right)-a^2c_{1}\) ➜Note \(\because \ln{\frac{M}{N}}=\ln{M}-\ln{N}\)
\(=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\ln{\left|x+\sqrt{x^2-a^2}\right|}+\frac{a^2}{2}\ln{|a|}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\ln{\left|x+\sqrt{x^2-a^2}\right|}+c\) ➜Note \(\because c=\frac{a^2}{2}\ln{|a|}-a^2c_{1}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(Proved)

\(L.S=\int{\sqrt{x^2+a^2}dx}\)
\(=\int{\sqrt{a^2\tan^2{\theta}+a^2}.a\sec^2{\theta}d\theta}\)
\(=\int{\sqrt{a^2(\tan^2{\theta}+1)}.a\sec^2{\theta}d\theta}\)
\(=\int{a\sqrt{\sec^2{\theta}}.a\sec^2{\theta}d\theta}\)
\(=a^2\int{\sec{\theta}.\sec^2{\theta}d\theta}\)
\(=a^2\int{\sec^3{\theta}d\theta}\)
\(=a^2I\)
ধরি,
\(I=\int{\sec^3{\theta}d\theta}\)
\(\Rightarrow I=\int{\sec^2{\theta}.\sec{\theta}d\theta}\)
\(\Rightarrow I=\sec{\theta}\int{\sec^2{\theta}d\theta}-\int{\left\{\frac{d}{d\theta}(\sec{\theta})\int{\sec^2{\theta}d\theta}\right\}d\theta}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec{\theta}\tan{\theta}.\tan{\theta}d\theta}\) ➜Note \(\because \int{\sec^2{x}dx}=\tan{x}, \frac{d}{dx}(sec{x})=sec{x}\tan{x}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec{\theta}\tan^2{\theta}d\theta}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec{\theta}(\sec^2{\theta}-1)d\theta}\) ➜Note \(\because tan^2{x}=\sec^2{x}-1\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-\int{\sec^3{\theta}d\theta}+\int{\sec{\theta}d\theta}\)
\(\Rightarrow I=\sec{\theta}\tan{\theta}-I+\int{\sec{\theta}d\theta}\) ➜Note \(\because \int{\sec^3{\theta}d\theta}=I\)
\(\Rightarrow I+I=\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\)
\(\Rightarrow 2I=\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\)
\(\Rightarrow I=\frac{1}{2}\left(\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\right)+c_{1}\) ➜Note \(\because \int{\sec{\theta}d\theta}=\ln{\left|\sec{x}+\tan{x}\right|}\) এবং \(c_{1}\) যোগজীকরণ ধ্রুবক।
এখন,
\(L.S=a^2I\)
\(=a^2\left\{\frac{1}{2}\left(\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\right)+c_{1}\right\}\)
\(=\frac{a^2}{2}\left(\sec{\theta}\tan{\theta}+\ln{\left|\sec{\theta}+\tan{\theta}\right|}\right)-a^2c_{1}\)
\(=\frac{a^2}{2}\sec{\theta}\tan{\theta}+\frac{a^2}{2}\ln{\left|\tan{\theta}+\sec{\theta}\right|}-a^2c_{1}\)
\(=\frac{a^2}{2}\tan{\theta}\sqrt{\tan^2{\theta}+1}+\frac{a^2}{2}\ln{\left|\tan{\theta}+\sqrt{\tan^2{\theta}+1}\right|}-a^2c_{1}\) ➜Note \(\because \sec{x}=\sqrt{\tan^2{x}+1}\)
\(=\frac{a^2}{2}\times{\frac{x}{a}}\sqrt{\left(\frac{x}{a}\right)^2+1}+\frac{a^2}{2}\ln{\left|\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^2+1}\right|}-a^2c_{1}\) ➜Note \(\because \tan{\theta}=\frac{x}{a}\)
\(=\frac{ax}{2}.\sqrt{\frac{x^2}{a^2}+1}+\frac{a^2}{2}\ln{\left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}+1}\right|}-a^2c_{1}\)
\(=\frac{ax}{2}.\sqrt{\frac{x^2+a^2}{a^2}}+\frac{a^2}{2}\ln{\left|\frac{x}{a}+\sqrt{\frac{x^2+a^2}{a^2}}\right|}-a^2c_{1}\)
\(=\frac{ax}{2}.\frac{\sqrt{x^2+a^2}}{a}+\frac{a^2}{2}\ln{\left|\frac{x}{a}+\frac{\sqrt{x^2+a^2}}{a}\right|}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln{\left|\frac{x+\sqrt{x^2+a^2}}{a}\right|}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln{\frac{\left|x+\sqrt{x^2+a^2}\right|}{|a|}}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\left(\ln{\left|x+\sqrt{x^2+a^2}\right|}-\ln{|a|}\right)-a^2c_{1}\) ➜Note \(\because \ln{\frac{M}{N}}=\ln{M}-\ln{N}\)
\(=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln{\left|x+\sqrt{x^2+a^2}\right|}-\frac{a^2}{2}\ln{|a|}-a^2c_{1}\)
\(=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln{\left|x+\sqrt{x^2+a^2}\right|}+c\) ➜Note \(\because c=-\frac{a^2}{2}\ln{|a|}-a^2c_{1}\)
\(=R.S\)
\(\therefore L.S=R.S\)
(Proved)

\(L.S=\int{e^x\{f(x)+f^{\prime}(x)\}dx}\)
\(=\int{\{e^xf(x)+e^xf^{\prime}(x)\}dx}\)
\(=\int{e^xf(x)dx}+\int{e^xf^{\prime}(x)dx}\)
\(=\int{e^xf(x)dx}+e^x\int{f^{\prime}(x)dx}-\int{\left\{\frac{d}{dx}(e^x)\int{f^{\prime}(x)dx}\right\}dx}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\int{e^xf(x)dx}+e^xf(x)-\int{e^xf(x)dx}\) ➜Note \(\because \int{f^{\prime}(x)dx}=f(x), \frac{d}{dx}(e^x)=e^x\)
\(=e^xf(x)+c\) ➜Note \(c\) যোগজীকরণ ধ্রুবক।
\(=R.S\)
\(\therefore L.S=R.S\)
(Proved)

\(L.S=\int{e^{ax}\{af(x)+f^{\prime}(x)\}dx}\)
\(=\int{\{ae^{ax}f(x)+e^{ax}f^{\prime}(x)\}dx}\)
\(=\int{ae^{ax}f(x)dx}+\int{e^{ax}f^{\prime}(x)dx}\)
\(=\int{ae^{ax}f(x)dx}+e^{ax}\int{f^{\prime}(x)dx}-\int{\left\{\frac{d}{dx}(e^{ax})\int{f^{\prime}(x)dx}\right\}dx}\) ➜Note \(\because \int{(uv)dx}=u\int{vdx}-\int{\left\{\frac{d}{dx}(u)\int{vdx}\right\}dx}\)
\(=\int{ae^{ax}f(x)dx}+e^{ax}f(x)-\int{e^{ax}.\frac{d}{dx}(ax).f(x)dx}\) ➜Note \(\because \int{f^{\prime}(x)dx}=f(x), \frac{d}{dx}(e^x)=e^x\)
\(=\int{ae^{ax}f(x)dx}+e^{ax}f(x)-\int{e^{ax}.a.f(x)dx}\)
\(=\int{ae^{ax}f(x)dx}+e^{ax}f(x)-\int{ae^{ax}f(x)dx}\)
\(=e^{ax}f(x)+c\) ➜Note \(c\) যোগজীকরণ ধ্রুবক।
\(=R.S\)
\(\therefore L.S=R.S\)
(Proved)