Class 12: Indefinite Integrals

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

$\displaystyle \textbf{Question 1: }~\int \frac{1}{\sqrt{2x-x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{dx}{\sqrt{2x-x^{2}}}$
$\displaystyle =\int \frac{dx}{\sqrt{2x-x^{2}-1+1}}$
$\displaystyle =\int \frac{dx}{\sqrt{1-(x^{2}-2x+1)}}$
$\displaystyle =\int \frac{dx}{\sqrt{1-(x-1)^{2}}}$
$\displaystyle =\sin^{-1}(x-1)+C \;\; \left[\because \int \frac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\left(\frac{x}{a}\right)+C\right]$
$\\$
$\displaystyle \textbf{Question 2: }~\int \frac{1}{\sqrt{8+3x-x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{dx}{\sqrt{8+3x-x^{2}}}$
$\displaystyle \Rightarrow \int \frac{dx}{\sqrt{8-(x^{2}-3x)}}$
$\displaystyle \Rightarrow \int \frac{dx}{\sqrt{8-\left(x^{2}-3x+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}\right)}}$
$\displaystyle \Rightarrow \int \frac{dx}{\sqrt{8-\left(x-\frac{3}{2}\right)^{2}+\frac{9}{4}}}$
$\displaystyle \Rightarrow \int \frac{dx}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}}$
$\displaystyle \Rightarrow \sin^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)+C$
$\displaystyle \Rightarrow \sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+C$
$\\$
$\displaystyle \textbf{Question 3: }~\int \frac{1}{\sqrt{5-4x-2x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{dx}{\sqrt{5-4x-2x^{2}}}$
$\displaystyle =\int \frac{dx}{\sqrt{2\left[\frac{5}{2}-2x-x^{2}\right]}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\frac{5}{2}-2x-x^{2}}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\frac{5}{2}-(x^{2}+2x)}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\frac{5}{2}-(x^{2}+2x+1-1)}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\frac{5}{2}-(x+1)^{2}+1}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\frac{7}{2}-(x+1)^{2}}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\left(\frac{\sqrt{7}}{\sqrt{2}}\right)^{2}-(x+1)^{2}}}$
$\displaystyle =\frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{(x+1)\sqrt{2}}{\sqrt{7}}\right)+C$
$\displaystyle =\frac{1}{\sqrt{2}}\sin^{-1}\left(\sqrt{\frac{2}{7}}(x+1)\right)+C$
$\\$
$\displaystyle \textbf{Question 4: }~\int \frac{1}{\sqrt{3x^2+5x+7}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{dx}{\sqrt{3x^{2}+5x+7}}$
$\displaystyle =\int \frac{dx}{\sqrt{3\left(x^{2}+\frac{5}{3}x+\frac{7}{3}\right)}}$
$\displaystyle =\frac{1}{\sqrt{3}}\int \frac{dx}{\sqrt{x^{2}+\frac{5}{3}x+\left(\frac{5}{6}\right)^{2}-\left(\frac{5}{6}\right)^{2}+\frac{7}{3}}}$
$\displaystyle =\frac{1}{\sqrt{3}}\int \frac{dx}{\sqrt{\left(x+\frac{5}{6}\right)^{2}-\frac{25}{36}+\frac{7}{3}}}$
$\displaystyle =\frac{1}{\sqrt{3}}\int \frac{dx}{\sqrt{\left(x+\frac{5}{6}\right)^{2}+\frac{-25+84}{36}}}$
$\displaystyle =\frac{1}{\sqrt{3}}\int \frac{dx}{\sqrt{\left(x+\frac{5}{6}\right)^{2}+\frac{59}{36}}}$
$\displaystyle =\frac{1}{\sqrt{3}}\int \frac{dx}{\sqrt{\left(x+\frac{5}{6}\right)^{2}+\left(\frac{\sqrt{59}}{6}\right)^{2}}}$
$\displaystyle =\frac{1}{\sqrt{3}}\log\left|x+\frac{5}{6}+\sqrt{\left(x+\frac{5}{6}\right)^{2}+\frac{59}{36}}\right|+C$
$\displaystyle =\frac{1}{\sqrt{3}}\log\left|x+\frac{5}{6}+\sqrt{x^{2}+\frac{5}{3}x+\frac{7}{3}}\right|+C$
$\\$
$\displaystyle \textbf{Question 5: }~\int \frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,dx,\ (\beta>\alpha).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int \frac{dx}{\sqrt{(x-\alpha)(\beta-x)}}$
$\displaystyle =\int \frac{dx}{\sqrt{\beta x-x^{2}-\alpha\beta+\alpha x}}$
$\displaystyle =\int \frac{dx}{\sqrt{-x^{2}+(\alpha+\beta)x-\alpha\beta}}$
$\displaystyle =\int \frac{dx}{\sqrt{-\left[x^{2}-(\alpha+\beta)x+\alpha\beta\right]}}$
$\displaystyle =\int \frac{dx}{\sqrt{-\left[x^{2}-(\alpha+\beta)x+\left(\frac{\alpha+\beta}{2}\right)^{2}-\left(\frac{\alpha+\beta}{2}\right)^{2}+\alpha\beta\right]}}$
$\displaystyle =\int \frac{dx}{\sqrt{-\left\{x-\left(\frac{\alpha+\beta}{2}\right)\right\}^{2}+\left(\frac{\alpha+\beta}{2}\right)^{2}-\alpha\beta}}$
$\displaystyle =\int \frac{dx}{\sqrt{-\left[x-\left(\frac{\alpha+\beta}{2}\right)\right]^{2}+\frac{(\alpha+\beta)^{2}-4\alpha\beta}{4}}}$
$\displaystyle =\int \frac{dx}{\sqrt{-\left[x-\left(\frac{\alpha+\beta}{2}\right)\right]^{2}+\left(\frac{\alpha-\beta}{2}\right)^{2}}}$
$\displaystyle =\int \frac{dx}{\sqrt{\left(\frac{\alpha-\beta}{2}\right)^{2}-\left(x-\frac{\alpha+\beta}{2}\right)^{2}}}$
$\displaystyle =\sin^{-1}\left[\frac{x-\left(\frac{\alpha+\beta}{2}\right)}{\frac{\alpha-\beta}{2}}\right]+C$
$\displaystyle =\sin^{-1}\left(\frac{2x-\alpha-\beta}{\alpha-\beta}\right)+C$
$\\$
$\displaystyle \textbf{Question 6: }~\int \frac{1}{\sqrt{7-3x-2x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{dx}{\sqrt{7-3x-2x^{2}}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\frac{7}{2}-\frac{3}{2}x-x^{2}}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\frac{7}{2}-(x^{2}-\frac{3}{2}x)}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\left(\frac{\sqrt{7}}{\sqrt{2}}\right)^{2}-\left(x^{2}-\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}\right)}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\left(\frac{\sqrt{7}}{\sqrt{2}}\right)^{2}-\left(x+\frac{3}{4}\right)^{2}+\frac{9}{16}}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\frac{7}{2}+\frac{9}{16}-\left(x+\frac{3}{4}\right)^{2}}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\frac{56+9}{16}-\left(x+\frac{3}{4}\right)^{2}}}$
$\displaystyle =\frac{1}{\sqrt{2}}\int \frac{dx}{\sqrt{\left(\frac{\sqrt{65}}{4}\right)^{2}-\left(x+\frac{3}{4}\right)^{2}}}$
$\displaystyle =\frac{1}{\sqrt{2}}\sin^{-1}\left[\frac{x+\frac{3}{4}}{\frac{\sqrt{65}}{4}}\right]+C$
$\displaystyle =\frac{1}{\sqrt{2}}\sin^{-1}\left(\frac{4x+3}{\sqrt{65}}\right)+C$
$\\$
$\displaystyle \textbf{Question 7: }~\int \frac{1}{\sqrt{16-6x-x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{dx}{\sqrt{16-6x-x^{2}}}$
$\displaystyle =\int \frac{dx}{\sqrt{16-(x^{2}+6x)}}$
$\displaystyle =\int \frac{dx}{\sqrt{16-(x^{2}+6x+3^{2}-3^{2})}}$
$\displaystyle =\int \frac{dx}{\sqrt{16+9-(x+3)^{2}}}$
$\displaystyle =\int \frac{dx}{\sqrt{5^{2}-(x+3)^{2}}}$
$\displaystyle =\sin^{-1}\left(\frac{x+3}{5}\right)+C$
$\\$
$\displaystyle \textbf{Question 8: }~\int \frac{1}{\sqrt{7-6x-x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{dx}{\sqrt{7-6x-x^{2}}}$
$\displaystyle =\int \frac{dx}{\sqrt{7-(x^{2}+6x)}}$
$\displaystyle =\int \frac{dx}{\sqrt{7-\left[x^{2}+6x+3^{2}-3^{2}\right]}}$
$\displaystyle =\int \frac{dx}{\sqrt{7+9-(x+3)^{2}}}$
$\displaystyle =\int \frac{dx}{\sqrt{4^{2}-(x+3)^{2}}}$
$\displaystyle =\sin^{-1}\left(\frac{x+3}{4}\right)+C$
$\\$
$\displaystyle \textbf{Question 9: }~\int \frac{1}{\sqrt{5x^2-2x}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int \frac{dx}{\sqrt{5x^{2}-2x}}$
$\displaystyle =\int \frac{dx}{\sqrt{5\left(x^{2}-\frac{2}{5}x\right)}}$
$\displaystyle =\frac{1}{\sqrt{5}}\int \frac{dx}{\sqrt{x^{2}-\frac{2}{5}x+\left(\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}}$
$\displaystyle =\frac{1}{\sqrt{5}}\int \frac{dx}{\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}}$
$\displaystyle =\frac{1}{\sqrt{5}}\log\left|x-\frac{1}{5}+\sqrt{\left(x-\frac{1}{5}\right)^{2}+\left(\frac{1}{5}\right)^{2}}\right|+C$
$\displaystyle =\frac{1}{\sqrt{5}}\log\left|\frac{5x-1}{5}+\frac{\sqrt{5x^{2}-2x}}{\sqrt{5}}\right|+C$