\displaystyle \textbf{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~\int \frac{x^2}{(a^2-x^2)^{3/2}}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{x^{2}}{(a^{2}-x^{2})^{3/2}}\,dx

\displaystyle \text{Let } x=a\cos\theta

\displaystyle \text{Differentiating both sides, we get } dx=-a\sin\theta\,d\theta

\displaystyle I=\int \frac{a^{2}\cos^{2}\theta}{\left(a^{2}-a^{2}\cos^{2}\theta\right)^{3/2}}\cdot(-a\sin\theta)\,d\theta

\displaystyle I=-\int \frac{a^{3}\cos^{2}\theta\sin\theta}{a^{3}(1-\cos^{2}\theta)^{3/2}}\,d\theta

\displaystyle I=-\int \frac{\cos^{2}\theta\sin\theta}{\sin^{3}\theta}\,d\theta

\displaystyle I=-\int \cot^{2}\theta\,d\theta

\displaystyle I=-\int (\mathrm{cosec}^{2}\theta-1)\,d\theta

\displaystyle I=-(-\cot\theta-\theta)+C

\displaystyle I=\cot\theta+\theta+C

\displaystyle I=\frac{x}{\sqrt{a^{2}-x^{2}}}+\cos^{-1}\frac{x}{a}+C

\displaystyle \textbf{Question 2: }~\int \frac{x^7}{(a^2-x^2)^5}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{x^{7}}{(a^{2}-x^{2})^{5}}\,dx

\displaystyle \text{Let } x=a\sin\theta

\displaystyle \text{Differentiating both sides, we get } dx=a\cos\theta\,d\theta

\displaystyle I=\int \frac{a^{7}\sin^{7}\theta\cdot a\cos\theta}{(a^{2}-a^{2}\sin^{2}\theta)^{5}}\,d\theta

\displaystyle I=\int \frac{a^{8}\sin^{7}\theta\cos\theta}{a^{10}(1-\sin^{2}\theta)^{5}}\,d\theta

\displaystyle I=\int \frac{\sin^{7}\theta}{a^{2}\cos^{9}\theta}\,d\theta

\displaystyle I=\frac{1}{a^{2}}\int \tan^{7}\theta\sec^{2}\theta\,d\theta

\displaystyle \text{Let } \tan\theta=t

\displaystyle \text{Then } \sec^{2}\theta\,d\theta=dt

\displaystyle I=\frac{1}{a^{2}}\int t^{7}\,dt

\displaystyle I=\frac{1}{a^{2}}\cdot\frac{t^{8}}{8}+C

\displaystyle I=\frac{1}{8a^{2}}\tan^{8}\theta+C

\displaystyle I=\frac{1}{8a^{2}}\left(\frac{x}{\sqrt{a^{2}-x^{2}}}\right)^{8}+C

\displaystyle I=\frac{x^{8}}{8a^{2}(a^{2}-x^{2})^{4}}+C

\displaystyle \textbf{Question 3: }~\int \cos\!\left(2\cot^{-1}\sqrt{\frac{1+x}{1-x}}\right)\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \cos\!\left\{2\cot^{-1}\sqrt{\frac{1+x}{1-x}}\right\}\,dx

\displaystyle \text{Let } x=\cos 2\theta

\displaystyle \text{On differentiating both sides, we get } dx=-2\sin 2\theta\,d\theta

\displaystyle I=-2\int \cos\!\left\{2\cot^{-1}\sqrt{\frac{1+\cos 2\theta}{1-\cos 2\theta}}\right\}\sin 2\theta\,d\theta

\displaystyle I=-2\int \cos\!\left\{2\cot^{-1}\sqrt{\frac{2\cos^{2}\theta}{2\sin^{2}\theta}}\right\}\sin 2\theta\,d\theta

\displaystyle I=-2\int \cos\!\left\{2\cot^{-1}(\cot\theta)\right\}\sin 2\theta\,d\theta

\displaystyle I=-2\int \cos 2\theta\sin 2\theta\,d\theta

\displaystyle I=-\int \sin 4\theta\,d\theta

\displaystyle I=\frac{\cos 4\theta}{4}+C_{1}

\displaystyle I=\frac{1}{4}(2\cos^{2}2\theta-1)+C_{1}

\displaystyle I=\frac{1}{2}\cos^{2}2\theta-\frac{1}{4}+C_{1}

\displaystyle I=\frac{1}{2}x^{2}+C

\displaystyle \text{Hence, } \int \cos\!\left\{2\cot^{-1}\sqrt{\frac{1+x}{1-x}}\right\}\,dx=\frac{1}{2}x^{2}+C

\displaystyle \textbf{Question 4: }~\int \frac{\sqrt{1+x^2}}{x^4}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{\sqrt{1+x^{2}}}{x^{4}}\,dx

\displaystyle \text{Let } x=\tan\theta

\displaystyle \text{On differentiating both sides, we get } dx=\sec^{2}\theta\,d\theta

\displaystyle I=\int \frac{\sqrt{1+\tan^{2}\theta}}{\tan^{4}\theta}\sec^{2}\theta\,d\theta

\displaystyle I=\int \frac{\sec^{3}\theta}{\tan^{4}\theta}\,d\theta

\displaystyle I=\int \frac{\cos\theta}{\sin^{4}\theta}\,d\theta

\displaystyle I=\int \cot\theta\,\mathrm{cosec}^{3}\theta\,d\theta

\displaystyle \text{Let } \mathrm{cosec}^{3}\theta=t

\displaystyle \text{Then } dt=-3\,\mathrm{cosec}^{3}\theta\cot\theta\,d\theta

\displaystyle I=-\frac{1}{3}\int dt

\displaystyle I=-\frac{t}{3}+C

\displaystyle I=-\frac{1}{3}\mathrm{cosec}^{3}\theta+C

\displaystyle I=-\frac{1}{3}\left(\frac{\sqrt{1+x^{2}}}{x}\right)^{3}+C

\displaystyle \text{Hence, } \int \frac{\sqrt{1+x^{2}}}{x^{4}}\,dx=-\frac{1}{3}\left(\frac{\sqrt{1+x^{2}}}{x}\right)^{3}+C

\displaystyle \textbf{Question 5: }~\int \frac{1}{(x^2+2x+10)^2}\,dx.
\displaystyle \text{Answer:}

\displaystyle I=\int \frac{1}{(x^{2}+2x+10)^{2}}\,dx

\displaystyle I=\int \frac{1}{\left[(x+1)^{2}+3^{2}\right]^{2}}\,dx

\displaystyle \text{Let } x+1=3\tan\theta

\displaystyle \text{On differentiating both sides, we get } dx=3\sec^{2}\theta\,d\theta

\displaystyle I=\int \frac{3\sec^{2}\theta}{\left[9\tan^{2}\theta+9\right]^{2}}\,d\theta

\displaystyle I=\frac{1}{27}\int \frac{\sec^{2}\theta}{\sec^{4}\theta}\,d\theta

\displaystyle I=\frac{1}{27}\int \frac{1}{\sec^{2}\theta}\,d\theta

\displaystyle I=\frac{1}{27}\int \cos^{2}\theta\,d\theta

\displaystyle I=\frac{1}{54}\int (1+\cos 2\theta)\,d\theta

\displaystyle I=\frac{1}{54}\left(\theta+\frac{\sin 2\theta}{2}\right)+C

\displaystyle I=\frac{1}{54}\left(\theta+\frac{\tan\theta}{1+\tan^{2}\theta}\right)+C

\displaystyle I=\frac{1}{54}\left(\tan^{-1}\frac{x+1}{3}+\frac{\frac{x+1}{3}}{1+\left(\frac{x+1}{3}\right)^{2}}\right)+C

\displaystyle I=\frac{1}{54}\left(\tan^{-1}\frac{x+1}{3}+\frac{3(x+1)}{x^{2}+2x+10}\right)+C

\displaystyle \text{Hence, } \int \frac{1}{(x^{2}+2x+10)^{2}}\,dx=\frac{1}{54}\left(\tan^{-1}\frac{x+1}{3}+\frac{3(x+1)}{x^{2}+2x+10}\right)+C

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.