NCERT Solutions · Class 12 Mathematics Integrals

The integrand has the form \(\displaystyle \sqrt{a^{2}-x^{2}}\) with \(\displaystyle a=2\); it is real only for \(\displaystyle -2\le x\le 2\).Derive the standard result by integration by parts, taking \(\displaystyle \sqrt{4-x^{2}}\) as the first function and \(\displaystyle 1\) as the second: \[I=\int\sqrt{4-x^{2}}\,dx=x\sqrt{4-x^{2}}-\int x\cdot\frac{-x}{\sqrt{4-x^{2}}}\,dx=x\sqrt{4-x^{2}}+\int\frac{x^{2}}{\sqrt{4-x^{2}}}\,dx.\]The step to get right is rewriting the numerator as \(\displaystyle x^{2}=-\left(4-x^{2}\right)+4\), which sends the last integral back to \(\displaystyle I\): \[\int\frac{x^{2}}{\sqrt{4-x^{2}}}\,dx=-\int\sqrt{4-x^{2}}\,dx+4\int\frac{dx}{\sqrt{4-x^{2}}}=-I+4\sin^{-1}\frac{x}{2}.\]So \(\displaystyle I=x\sqrt{4-x^{2}}-I+4\sin^{-1}\dfrac{x}{2}\), i.e. \(\displaystyle 2I=x\sqrt{4-x^{2}}+4\sin^{-1}\dfrac{x}{2}\).\[\int\sqrt{4-x^{2}}\,dx=\frac{x}{2}\sqrt{4-x^{2}}+2\sin^{-1}\frac{x}{2}+\mathrm{C},\qquad -2\le x\le 2.\]The same argument with \(\displaystyle a\) in place of \(\displaystyle 2\) gives the formula used throughout this exercise: \[\int\sqrt{a^{2}-x^{2}}\,dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}+\mathrm{C}.\]

First make the coefficient of \(\displaystyle x^{2}\) equal to \(\displaystyle 1\) inside the root by taking the \(\displaystyle 4\) out — this is the step most often dropped: \[\sqrt{1-4x^{2}}=\sqrt{4\left(\frac{1}{4}-x^{2}\right)}=2\sqrt{\left(\frac{1}{2}\right)^{2}-x^{2}}.\]Now apply \(\displaystyle \int\sqrt{a^{2}-x^{2}}\,dx=\dfrac{x}{2}\sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2}\sin^{-1}\dfrac{x}{a}+\mathrm{C}\) with \(\displaystyle a=\dfrac{1}{2}\): \[\int\sqrt{1-4x^{2}}\,dx=2\left[\frac{x}{2}\sqrt{\frac{1}{4}-x^{2}}+\frac{1}{8}\sin^{-1}(2x)\right]=x\sqrt{\frac{1}{4}-x^{2}}+\frac{1}{4}\sin^{-1}(2x).\]Since \(\displaystyle \sqrt{\frac{1}{4}-x^{2}}=\frac{1}{2}\sqrt{1-4x^{2}}\), \[\int\sqrt{1-4x^{2}}\,dx=\frac{x}{2}\sqrt{1-4x^{2}}+\frac{1}{4}\sin^{-1}(2x)+\mathrm{C},\qquad -\frac{1}{2}\le x\le\frac{1}{2}.\]

Complete the square: \[x^{2}+4x+6=\left(x^{2}+4x+4\right)+2=(x+2)^{2}+\left(\sqrt{2}\right)^{2}.\]Put \(\displaystyle t=x+2\), so \(\displaystyle dt=dx\); the integral becomes \(\displaystyle \displaystyle\int\sqrt{t^{2}+a^{2}}\,dt\) with \(\displaystyle a=\sqrt{2}\). Using \[\int\sqrt{t^{2}+a^{2}}\,dt=\frac{t}{2}\sqrt{t^{2}+a^{2}}+\frac{a^{2}}{2}\log\left|t+\sqrt{t^{2}+a^{2}}\right|+\mathrm{C},\] we get, with \(\displaystyle \dfrac{a^{2}}{2}=1\), \[\int\sqrt{x^{2}+4x+6}\,dx=\frac{x+2}{2}\sqrt{x^{2}+4x+6}+\log\left|x+2+\sqrt{x^{2}+4x+6}\right|+\mathrm{C}.\]Since \(\displaystyle (x+2)^{2}+2>0\) for every \(\displaystyle x\), this holds for all real \(\displaystyle x\).

Complete the square: \[x^{2}+4x+1=\left(x^{2}+4x+4\right)-3=(x+2)^{2}-\left(\sqrt{3}\right)^{2}.\]This is the \(\displaystyle \sqrt{t^{2}-a^{2}}\) type (note the minus sign before the log, unlike Q3), with \(\displaystyle t=x+2,\ dt=dx,\ a=\sqrt{3}\): \[\int\sqrt{t^{2}-a^{2}}\,dt=\frac{t}{2}\sqrt{t^{2}-a^{2}}-\frac{a^{2}}{2}\log\left|t+\sqrt{t^{2}-a^{2}}\right|+\mathrm{C}.\]Hence \[\int\sqrt{x^{2}+4x+1}\,dx=\frac{x+2}{2}\sqrt{x^{2}+4x+1}-\frac{3}{2}\log\left|x+2+\sqrt{x^{2}+4x+1}\right|+\mathrm{C},\] valid where the radicand is non-negative, i.e. \(\displaystyle |x+2|\ge\sqrt{3}\).

Take out the sign of \(\displaystyle x^{2}\) before completing the square — that is where the sign error usually creeps in: \[1-4x-x^{2}=1-\left(x^{2}+4x\right)=1-\left[(x+2)^{2}-4\right]=5-(x+2)^{2}=\left(\sqrt{5}\right)^{2}-(x+2)^{2}.\]With \(\displaystyle t=x+2,\ dt=dx,\ a=\sqrt{5}\), apply \(\displaystyle \int\sqrt{a^{2}-t^{2}}\,dt=\dfrac{t}{2}\sqrt{a^{2}-t^{2}}+\dfrac{a^{2}}{2}\sin^{-1}\dfrac{t}{a}+\mathrm{C}\): \[\int\sqrt{1-4x-x^{2}}\,dx=\frac{x+2}{2}\sqrt{1-4x-x^{2}}+\frac{5}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+\mathrm{C},\] valid on \(\displaystyle |x+2|\le\sqrt{5}\), i.e. \(\displaystyle -2-\sqrt{5}\le x\le -2+\sqrt{5}\).

Complete the square: \[x^{2}+4x-5=\left(x^{2}+4x+4\right)-9=(x+2)^{2}-3^{2}.\]With \(\displaystyle t=x+2,\ dt=dx,\ a=3\), use \(\displaystyle \int\sqrt{t^{2}-a^{2}}\,dt=\dfrac{t}{2}\sqrt{t^{2}-a^{2}}-\dfrac{a^{2}}{2}\log\left|t+\sqrt{t^{2}-a^{2}}\right|+\mathrm{C}\): \[\int\sqrt{x^{2}+4x-5}\,dx=\frac{x+2}{2}\sqrt{x^{2}+4x-5}-\frac{9}{2}\log\left|x+2+\sqrt{x^{2}+4x-5}\right|+\mathrm{C}.\]The radicand \(\displaystyle (x+5)(x-1)\) is non-negative only for \(\displaystyle x\le -5\) or \(\displaystyle x\ge 1\), so the answer holds on those intervals.

Complete the square after factoring out \(\displaystyle -1\) from the \(\displaystyle x\)-terms: \[1+3x-x^{2}=1-\left(x^{2}-3x\right)=1-\left[\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}\right]=\frac{13}{4}-\left(x-\frac{3}{2}\right)^{2}.\]With \(\displaystyle t=x-\dfrac{3}{2},\ dt=dx,\ a=\dfrac{\sqrt{13}}{2}\), apply \(\displaystyle \int\sqrt{a^{2}-t^{2}}\,dt=\dfrac{t}{2}\sqrt{a^{2}-t^{2}}+\dfrac{a^{2}}{2}\sin^{-1}\dfrac{t}{a}+\mathrm{C}\). Here \(\displaystyle \dfrac{a^{2}}{2}=\dfrac{13}{8}\) and \(\displaystyle \dfrac{t}{a}=\dfrac{2x-3}{\sqrt{13}}\): \[\int\sqrt{1+3x-x^{2}}\,dx=\frac{2x-3}{4}\sqrt{1+3x-x^{2}}+\frac{13}{8}\sin^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+\mathrm{C},\] valid where \(\displaystyle (2x-3)^{2}\le 13\), i.e. \(\displaystyle \dfrac{3-\sqrt{13}}{2}\le x\le\dfrac{3+\sqrt{13}}{2}\).

Complete the square: \[x^{2}+3x=\left(x+\frac{3}{2}\right)^{2}-\frac{9}{4}=\left(x+\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}.\]With \(\displaystyle t=x+\dfrac{3}{2},\ dt=dx,\ a=\dfrac{3}{2}\), use \(\displaystyle \int\sqrt{t^{2}-a^{2}}\,dt=\dfrac{t}{2}\sqrt{t^{2}-a^{2}}-\dfrac{a^{2}}{2}\log\left|t+\sqrt{t^{2}-a^{2}}\right|+\mathrm{C}\), where \(\displaystyle \dfrac{a^{2}}{2}=\dfrac{9}{8}\): \[\int\sqrt{x^{2}+3x}\,dx=\frac{2x+3}{4}\sqrt{x^{2}+3x}-\frac{9}{8}\log\left|x+\frac{3}{2}+\sqrt{x^{2}+3x}\right|+\mathrm{C}.\]The radicand \(\displaystyle x(x+3)\) is non-negative only for \(\displaystyle x\le -3\) or \(\displaystyle x\ge 0\), so the result holds there.

Bring the fraction out of the root first: \[\sqrt{1+\frac{x^{2}}{9}}=\sqrt{\frac{9+x^{2}}{9}}=\frac{1}{3}\sqrt{x^{2}+3^{2}}.\]Apply \(\displaystyle \int\sqrt{x^{2}+a^{2}}\,dx=\dfrac{x}{2}\sqrt{x^{2}+a^{2}}+\dfrac{a^{2}}{2}\log\left|x+\sqrt{x^{2}+a^{2}}\right|+\mathrm{C}\) with \(\displaystyle a=3\): \[\int\sqrt{1+\frac{x^{2}}{9}}\,dx=\frac{1}{3}\left[\frac{x}{2}\sqrt{x^{2}+9}+\frac{9}{2}\log\left|x+\sqrt{x^{2}+9}\right|\right]+\mathrm{C}.\]\[\int\sqrt{1+\frac{x^{2}}{9}}\,dx=\frac{x}{6}\sqrt{x^{2}+9}+\frac{3}{2}\log\left|x+\sqrt{x^{2}+9}\right|+\mathrm{C}\]Equivalently \(\displaystyle \dfrac{x}{2}\sqrt{1+\dfrac{x^{2}}{9}}+\dfrac{3}{2}\log\left(x+\sqrt{x^{2}+9}\right)+\mathrm{C}\); it is valid for all real \(\displaystyle x\), and the modulus may be dropped because \(\displaystyle x+\sqrt{x^{2}+9}>0\) always.

Use \(\displaystyle \int\sqrt{x^{2}+a^{2}}\,dx=\dfrac{x}{2}\sqrt{x^{2}+a^{2}}+\dfrac{a^{2}}{2}\log\left|x+\sqrt{x^{2}+a^{2}}\right|+\mathrm{C}\) with \(\displaystyle a=1\): \[\int\sqrt{1+x^{2}}\,dx=\frac{x}{2}\sqrt{1+x^{2}}+\frac{1}{2}\log\left|x+\sqrt{1+x^{2}}\right|+\mathrm{C}.\]The trap here is option (B): \(\displaystyle \dfrac{2}{3}\left(1+x^{2}\right)^{3/2}\) is what you would write if the chain rule were ignored, but \[\frac{d}{dx}\left[\frac{2}{3}\left(1+x^{2}\right)^{3/2}\right]=2x\sqrt{1+x^{2}}\neq\sqrt{1+x^{2}},\] because \(\displaystyle \sqrt{1+x^{2}}\) is not of the form \(\displaystyle f'(x)\left[f(x)\right]^{n}\). Options (C) and (D) fail the same differentiation test.The correct answer is (A).

Complete the square: \[x^{2}-8x+7=\left(x^{2}-8x+16\right)-9=(x-4)^{2}-3^{2}.\]With \(\displaystyle t=x-4,\ dt=dx,\ a=3\), apply \(\displaystyle \int\sqrt{t^{2}-a^{2}}\,dt=\dfrac{t}{2}\sqrt{t^{2}-a^{2}}-\dfrac{a^{2}}{2}\log\left|t+\sqrt{t^{2}-a^{2}}\right|+\mathrm{C}\), so that \(\displaystyle \dfrac{a^{2}}{2}=\dfrac{9}{2}\) and the log term carries a minus sign: \[\int\sqrt{x^{2}-8x+7}\,dx=\frac{1}{2}(x-4)\sqrt{x^{2}-8x+7}-\frac{9}{2}\log\left|x-4+\sqrt{x^{2}-8x+7}\right|+\mathrm{C}.\]Option (A) has the wrong sign and coefficient on the log, (B) completes the square as \(\displaystyle (x+4)^{2}\), and (C) uses \(\displaystyle 3\sqrt{2}\) instead of \(\displaystyle \dfrac{9}{2}\).The correct answer is (D).