Class 12: Indefinite Integrals

$\displaystyle \text{Evaluate the following integrals:}$

$\displaystyle \textbf{Question 1: }~\int \sqrt{3+2x-x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \sqrt{3+2x-x^{2}}\,dx.$
$\displaystyle =\int \sqrt{3-(x^{2}-2x)}\,dx.$
$\displaystyle =\int \sqrt{3-(x^{2}-2x+1-1)}\,dx.$
$\displaystyle =\int \sqrt{4-(x-1)^{2}}\,dx.$
$\displaystyle =\int \sqrt{2^{2}-(x-1)^{2}}\,dx.$
$\displaystyle \text{Using } \int \sqrt{a^{2}-x^{2}}\,dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\!\left(\frac{x}{a}\right)+C.$
$\displaystyle =\frac{x-1}{2}\sqrt{2^{2}-(x-1)^{2}}+\frac{2^{2}}{2}\sin^{-1}\!\left(\frac{x-1}{2}\right)+C.$
$\displaystyle =\frac{x-1}{2}\sqrt{3+2x-x^{2}}+\sin^{-1}\!\left(\frac{x-1}{2}\right)+C.$

$\displaystyle \textbf{Question 2: }~\int \sqrt{x^2+x+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \sqrt{x^{2}+x+1}\,dx.$
$\displaystyle =\int \sqrt{x^{2}+x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1}\,dx.$
$\displaystyle =\int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\,dx.$
$\displaystyle =\left(\frac{x+\frac{1}{2}}{2}\right)\sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}+\frac{3}{8}\log\left|x+\frac{1}{2}+\sqrt{x^{2}+x+1}\right|+C.$
$\displaystyle =\frac{2x+1}{4}\sqrt{x^{2}+x+1}+\frac{3}{8}\log\left|2x+1+2\sqrt{x^{2}+x+1}\right|+C.$

$\displaystyle \textbf{Question 3: }~\int \sqrt{x-x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \sqrt{x-x^{2}}\,dx.$
$\displaystyle =\int \sqrt{-(x^{2}-x)}\,dx.$
$\displaystyle =\int \sqrt{-\left(x^{2}-x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}\right)}\,dx.$
$\displaystyle =\int \sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}\,dx.$
$\displaystyle \text{Using } \int \sqrt{a^{2}-u^{2}}\,du=\frac{u}{2}\sqrt{a^{2}-u^{2}}+\frac{a^{2}}{2}\sin^{-1}\!\left(\frac{u}{a}\right)+C.$
$\displaystyle =\frac{x-\frac{1}{2}}{2}\sqrt{x-x^{2}}+\frac{1}{8}\sin^{-1}\!\left(\frac{x-\frac{1}{2}}{\frac{1}{2}}\right)+C.$
$\displaystyle =\frac{2x-1}{4}\sqrt{x-x^{2}}+\frac{1}{8}\sin^{-1}(2x-1)+C.$

$\displaystyle \textbf{Question 4: }~\int \sqrt{1+x-2x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \sqrt{1+x-2x^{2}}\,dx.$
$\displaystyle =\int \sqrt{2\left(\frac{1}{2}+\frac{x}{2}-x^{2}\right)}\,dx.$
$\displaystyle =\sqrt{2}\int \sqrt{\frac{1}{2}-(x^{2}-\frac{x}{2})}\,dx.$
$\displaystyle =\sqrt{2}\int \sqrt{\frac{1}{2}-\left(x^{2}-\frac{x}{2}+\frac{1}{4^{2}}-\frac{1}{4^{2}}\right)}\,dx.$
$\displaystyle =\sqrt{2}\int \sqrt{\frac{1}{2}+\frac{1}{16}-\left(x-\frac{1}{4}\right)^{2}}\,dx.$
$\displaystyle =\sqrt{2}\int \sqrt{\left(\frac{3}{4}\right)^{2}-\left(x-\frac{1}{4}\right)^{2}}\,dx.$
$\displaystyle =\sqrt{2}\left[\frac{x-\frac{1}{4}}{2}\sqrt{\left(\frac{3}{4}\right)^{2}-\left(x-\frac{1}{4}\right)^{2}}+\frac{9}{32}\sin^{-1}\!\left(\frac{x-\frac{1}{4}}{\frac{3}{4}}\right)\right]+C.$
$\displaystyle =\frac{4x-1}{8}\sqrt{1+x-2x^{2}}+\frac{9\sqrt{2}}{32}\sin^{-1}\!\left(\frac{4x-1}{3}\right)+C.$

$\displaystyle \textbf{Question 5: }~\int \cos x\,\sqrt{4-\sin^2 x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \cos x\sqrt{4-\sin^{2}x}\,dx.$
$\displaystyle \text{Putting } \sin x=t.$
$\displaystyle \Rightarrow \cos x\,dx=dt.$
$\displaystyle I=\int \sqrt{2^{2}-t^{2}}\,dt.$
$\displaystyle \text{Using } \int \sqrt{a^{2}-x^{2}}\,dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\!\left(\frac{x}{a}\right)+C.$
$\displaystyle I=\frac{t}{2}\sqrt{2^{2}-t^{2}}+\frac{2^{2}}{2}\sin^{-1}\!\left(\frac{t}{2}\right)+C.$
$\displaystyle I=\frac{\sin x}{2}\sqrt{4-\sin^{2}x}+2\sin^{-1}\!\left(\frac{\sin x}{2}\right)+C.$

$\displaystyle \textbf{Question 6: }~\int e^x\sqrt{e^{2x}+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int e^{x}\sqrt{e^{2x}+1}\,dx.$
$\displaystyle \text{Putting } e^{x}=t.$
$\displaystyle \Rightarrow e^{x}\,dx=dt.$
$\displaystyle \therefore I=\int \sqrt{t^{2}+1}\,dt.$
$\displaystyle \text{Using } \int \sqrt{x^{2}+a^{2}}\,dx=\frac{x}{2}\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2}\log\!\left|x+\sqrt{x^{2}+a^{2}}\right|+C.$
$\displaystyle I=\frac{t}{2}\sqrt{t^{2}+1}+\frac{1}{2}\log\!\left|t+\sqrt{t^{2}+1}\right|+C.$
$\displaystyle I=\frac{e^{x}}{2}\sqrt{e^{2x}+1}+\frac{1}{2}\log\!\left|e^{x}+\sqrt{e^{2x}+1}\right|+C.$

$\displaystyle \textbf{Question 7: }~\int \sqrt{9-x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \sqrt{9-x^{2}}\,dx.$
$\displaystyle =\int \sqrt{3^{2}-x^{2}}\,dx.$
$\displaystyle \text{Using } \int \sqrt{a^{2}-x^{2}}\,dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\!\left(\frac{x}{a}\right)+C.$
$\displaystyle =\frac{x}{2}\sqrt{3^{2}-x^{2}}+\frac{3^{2}}{2}\sin^{-1}\!\left(\frac{x}{3}\right)+C.$
$\displaystyle =\frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\sin^{-1}\!\left(\frac{x}{3}\right)+C.$

$\displaystyle \textbf{Question 8: }~\int \sqrt{16x^2+25}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \sqrt{16x^{2}+25}\,dx.$
$\displaystyle =\int \sqrt{16\left(x^{2}+\frac{25}{16}\right)}\,dx.$
$\displaystyle =4\int \sqrt{x^{2}+\left(\frac{5}{4}\right)^{2}}\,dx.$
$\displaystyle =4\left[\frac{x}{2}\sqrt{x^{2}+\left(\frac{5}{4}\right)^{2}}+\frac{\left(\frac{5}{4}\right)^{2}}{2}\ln\!\left|x+\sqrt{x^{2}+\left(\frac{5}{4}\right)^{2}}\right|\right]+C.$
$\displaystyle =2x\sqrt{x^{2}+\frac{25}{16}}+\frac{25}{8}\ln\!\left|x+\sqrt{x^{2}+\frac{25}{16}}\right|+C.$

$\displaystyle \textbf{Question 9: }~\int \sqrt{4x^2-5}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \sqrt{4x^{2}-5}\,dx.$
$\displaystyle =\int \sqrt{4\left(x^{2}-\frac{5}{4}\right)}\,dx.$
$\displaystyle =2\int \sqrt{x^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}}\,dx.$
$\displaystyle =2\left[\frac{x}{2}\sqrt{x^{2}-\frac{5}{4}}-\frac{\left(\frac{\sqrt{5}}{2}\right)^{2}}{2}\ln\!\left|x+\sqrt{x^{2}-\frac{5}{4}}\right|\right]+C.$
$\displaystyle =x\sqrt{x^{2}-\frac{5}{4}}-\frac{5}{4}\ln\!\left|x+\sqrt{x^{2}-\frac{5}{4}}\right|+C.$

$\displaystyle \textbf{Question 10: }~\int \sqrt{2x^2+3x+4}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \sqrt{2x^{2}+3x+4}\,dx.$
$\displaystyle =\sqrt{2}\int \sqrt{x^{2}+\frac{3}{2}x+2}\,dx.$
$\displaystyle =\sqrt{2}\int \sqrt{x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}+2}\,dx.$
$\displaystyle =\sqrt{2}\int \sqrt{\left(x+\frac{3}{4}\right)^{2}-\frac{9}{16}+2}\,dx.$
$\displaystyle =\sqrt{2}\int \sqrt{\left(x+\frac{3}{4}\right)^{2}+\left(\frac{\sqrt{23}}{4}\right)^{2}}\,dx.$
$\displaystyle =\sqrt{2}\left[\frac{x+\frac{3}{4}}{2}\sqrt{\left(x+\frac{3}{4}\right)^{2}+\left(\frac{\sqrt{23}}{4}\right)^{2}}+\frac{\left(\frac{\sqrt{23}}{4}\right)^{2}}{2}\log\!\left|x+\frac{3}{4}+\sqrt{x^{2}+\frac{3}{2}x+2}\right|\right]+C.$
$\displaystyle =\sqrt{2}\left[\frac{4x+3}{8}\sqrt{x^{2}+\frac{3}{2}x+2}+\frac{23}{32}\log\!\left|x+\frac{3}{4}+\sqrt{x^{2}+\frac{3}{2}x+2}\right|\right]+C.$
$\displaystyle =\frac{4x+3}{8}\sqrt{2x^{2}+3x+4}+\frac{23\sqrt{2}}{32}\log\!\left|x+\frac{3}{4}+\sqrt{x^{2}+\frac{3}{2}x+2}\right|+C.$

$\displaystyle \textbf{Question 11: }~\int \sqrt{3-2x-2x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \sqrt{3-2x-2x^{2}}\,dx.$
$\displaystyle =\int \sqrt{3-(2x^{2}+2x)}\,dx.$
$\displaystyle =\int \sqrt{3-2(x^{2}+x)}\,dx.$
$\displaystyle =\int \sqrt{3-2\left(x^{2}+x+\frac{1}{4}-\frac{1}{4}\right)}\,dx.$
$\displaystyle =\int \sqrt{3-2\left(x+\frac{1}{2}\right)^{2}+\frac{1}{2}}\,dx.$
$\displaystyle =\int \sqrt{\frac{7}{2}-2\left(x+\frac{1}{2}\right)^{2}}\,dx.$
$\displaystyle =\sqrt{2}\int \sqrt{\frac{7}{4}-\left(x+\frac{1}{2}\right)^{2}}\,dx.$
$\displaystyle =\sqrt{2}\int \sqrt{\left(\frac{\sqrt{7}}{2}\right)^{2}-\left(x+\frac{1}{2}\right)^{2}}\,dx.$
$\displaystyle =\sqrt{2}\left[\frac{x+\frac{1}{2}}{2}\sqrt{\left(\frac{\sqrt{7}}{2}\right)^{2}-\left(x+\frac{1}{2}\right)^{2}}+\frac{7}{8}\sin^{-1}\!\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{7}}{2}}\right)\right]+C.$
$\displaystyle =\frac{2x+1}{4}\sqrt{3-2x-2x^{2}}+\frac{7}{4\sqrt{2}}\sin^{-1}\!\left(\frac{2x+1}{\sqrt{7}}\right)+C.$

$\displaystyle \textbf{Question 12: }~\int x\sqrt{x^4+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int x\sqrt{x^{4}+1}\,dx.$
$\displaystyle =\int x\sqrt{(x^{2})^{2}+1}\,dx.$
$\displaystyle \text{Putting } x^{2}=t.$
$\displaystyle \Rightarrow 2x\,dx=dt.$
$\displaystyle \Rightarrow x\,dx=\frac{dt}{2}.$
$\displaystyle \therefore I=\frac{1}{2}\int \sqrt{t^{2}+1}\,dt.$
$\displaystyle =\frac{1}{2}\int \sqrt{t^{2}+1^{2}}\,dt.$
$\displaystyle =\frac{1}{2}\left[\frac{t}{2}\sqrt{t^{2}+1}+\frac{1^{2}}{2}\log\!\left|t+\sqrt{t^{2}+1}\right|\right]+C.$
$\displaystyle =\frac{1}{2}\left[\frac{x^{2}}{2}\sqrt{x^{4}+1}+\frac{1}{2}\log\!\left|x^{2}+\sqrt{x^{4}+1}\right|\right]+C.$
$\displaystyle =\frac{x^{2}}{4}\sqrt{x^{4}+1}+\frac{1}{4}\log\!\left|x^{2}+\sqrt{x^{4}+1}\right|+C.$

$\displaystyle \textbf{Question 13: }~\int x^2\sqrt{a^6-x^6}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int x^{2}\sqrt{a^{6}-x^{6}}\,dx.$
$\displaystyle =\int x^{2}\sqrt{(a^{3})^{2}-(x^{3})^{2}}\,dx.$
$\displaystyle \text{Putting } x^{3}=t.$
$\displaystyle \Rightarrow 3x^{2}\,dx=dt.$
$\displaystyle \Rightarrow x^{2}\,dx=\frac{dt}{3}.$
$\displaystyle \therefore I=\frac{1}{3}\int \sqrt{(a^{3})^{2}-t^{2}}\,dt.$
$\displaystyle \text{Using } \int \sqrt{a^{2}-x^{2}}\,dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\!\left(\frac{x}{a}\right)+C.$
$\displaystyle I=\frac{1}{3}\left[\frac{t}{2}\sqrt{(a^{3})^{2}-t^{2}}+\frac{(a^{3})^{2}}{2}\sin^{-1}\!\left(\frac{t}{a^{3}}\right)\right]+C.$
$\displaystyle I=\frac{x^{3}}{6}\sqrt{a^{6}-x^{6}}+\frac{a^{6}}{6}\sin^{-1}\!\left(\frac{x^{3}}{a^{3}}\right)+C.$

$\displaystyle \textbf{Question 14: }~\int \frac{\sqrt{16+(\log x)^2}}{x}\,dx. \hspace{6.0cm} \text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \sqrt{\frac{16+(\log x)^{2}}{x}}\,dx.$
$\displaystyle \text{Putting } \log x=t.$
$\displaystyle \Rightarrow \frac{1}{x}\,dx=dt.$
$\displaystyle \therefore I=\int \sqrt{16+t^{2}}\,dt.$
$\displaystyle =\int \sqrt{4^{2}+t^{2}}\,dt.$
$\displaystyle =\frac{t}{2}\sqrt{4^{2}+t^{2}}+\frac{4^{2}}{2}\log\!\left|t+\sqrt{4^{2}+t^{2}}\right|+C.$
$\displaystyle =\frac{\log x}{2}\sqrt{16+(\log x)^{2}}+8\log\!\left|\log x+\sqrt{16+(\log x)^{2}}\right|+C.$

$\displaystyle \textbf{Question 15: }~\int \sqrt{2ax-x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \sqrt{2ax-x^{2}}\,dx.$
$\displaystyle =\int \sqrt{a^{2}+2ax-x^{2}-a^{2}}\,dx.$
$\displaystyle =\int \sqrt{a^{2}-(x^{2}-2ax+a^{2})}\,dx.$
$\displaystyle =\int \sqrt{a^{2}-(x-a)^{2}}\,dx.$
$\displaystyle =\frac{x-a}{2}\sqrt{2ax-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\!\left(\frac{x-a}{a}\right)+C.$

$\displaystyle \textbf{Question 16: }~\int \sqrt{3-x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \sqrt{3-x^{2}}\,dx.$
$\displaystyle =\int \sqrt{(\sqrt{3})^{2}-x^{2}}\,dx.$
$\displaystyle =\frac{x}{2}\sqrt{(\sqrt{3})^{2}-x^{2}}+\frac{(\sqrt{3})^{2}}{2}\sin^{-1}\!\left(\frac{x}{\sqrt{3}}\right)+C.$
$\displaystyle =\frac{x}{2}\sqrt{3-x^{2}}+\frac{3}{2}\sin^{-1}\!\left(\frac{x}{\sqrt{3}}\right)+C.$

$\displaystyle \textbf{Question 17: }~\int \sqrt{x^2-2x}\,dx. \hspace{6.0cm} \text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \sqrt{x^{2}-2x}\,dx.$
$\displaystyle \Rightarrow I=\int \sqrt{x^{2}-2x+1-1}\,dx.$
$\displaystyle \Rightarrow I=\int \sqrt{(x-1)^{2}-1^{2}}\,dx.$
$\displaystyle \text{Using } \int \sqrt{x^{2}-a^{2}}\,dx=\frac{x}{2}\sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2}\log\!\left|x+\sqrt{x^{2}-a^{2}}\right|+C.$
$\displaystyle \Rightarrow I=\frac{x-1}{2}\sqrt{(x-1)^{2}-1}-\frac{1}{2}\log\!\left|(x-1)+\sqrt{x^{2}-2x}\right|+C.$

$\displaystyle \textbf{Question 18: }~\int \sqrt{2x-x^2}\,dx. \hspace{6.0cm} \text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \sqrt{2x-x^{2}}\,dx.$
$\displaystyle =\int \sqrt{x(2-x)}\,dx.$
$\displaystyle \text{Let } x=1+\sin u.$
$\displaystyle \text{or, } dx=\cos u\,du.$
$\displaystyle \Rightarrow I=\int \sqrt{(1+\sin u)(1-\sin u)}\cos u\,du.$
$\displaystyle \Rightarrow I=\int \cos^{2}u\,du.$
$\displaystyle \Rightarrow I=\frac{1}{2}\int (\cos 2u+1)\,du.$
$\displaystyle \Rightarrow I=\frac{1}{2}\left(\frac{1}{2}\sin 2u+u\right)+C.$
$\displaystyle \Rightarrow I=\frac{1}{2}(\sin u\cos u+u)+C.$
$\displaystyle \Rightarrow I=\frac{1}{2}\left(\sin u\sqrt{1-\sin^{2}u}+u\right)+C.$
$\displaystyle \therefore I=\frac{1}{2}(x-1)\sqrt{2x-x^{2}}+\frac{1}{2}\sin^{-1}(x-1)+C.$