NCERT Solutions · Class 12 Mathematics Integrals

Factorise the denominator and split by partial fractions: \[\frac{1}{x^{2}-1}=\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}.\] Then \(\displaystyle 1=A(x+1)+B(x-1)\); putting \(\displaystyle x=1\) gives \(\displaystyle A=\frac{1}{2}\) and \(\displaystyle x=-1\) gives \(\displaystyle B=-\frac{1}{2}\). So \[\frac{1}{x^{2}-1}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right).\]On \(\displaystyle [2,3]\) both \(\displaystyle x-1\) and \(\displaystyle x+1\) are positive, so the moduli drop, and the integrand is continuous there: \[\int_{2}^{3}\frac{dx}{x^{2}-1}=\frac{1}{2}\left[\log\frac{x-1}{x+1}\right]_{2}^{3}=\frac{1}{2}\left(\log\frac{2}{4}-\log\frac{1}{3}\right).\]Since \(\displaystyle \log\frac{1}{2}-\log\frac{1}{3}=\log\frac{3}{2}\), \[\int_{2}^{3}\frac{dx}{x^{2}-1}=\frac{1}{2}\log\frac{3}{2}.\]

A power of cosine cannot be integrated directly; use the double-angle identity \(\displaystyle \cos 2x=2\cos^{2}x-1\), i.e. \[\cos^{2}x=\frac{1+\cos 2x}{2}.\]Hence \[\int_{0}^{\pi/2}\cos^{2}x\,dx=\frac{1}{2}\int_{0}^{\pi/2}(1+\cos 2x)\,dx=\frac{1}{2}\left[x+\frac{\sin 2x}{2}\right]_{0}^{\pi/2}.\]At \(\displaystyle x=\frac{\pi}{2}\) the sine term is \(\displaystyle \frac{\sin\pi}{2}=0\), and at \(\displaystyle x=0\) everything vanishes, so \[\int_{0}^{\pi/2}\cos^{2}x\,dx=\frac{1}{2}\left(\frac{\pi}{2}+0\right)=\frac{\pi}{4}.\]

The numerator is (a constant times) the derivative of the denominator, so use the substitution \[t=x^{2}+1,\qquad dt=2x\,dx\ \Rightarrow\ x\,dx=\frac{dt}{2}.\]Change the limits with the variable (this is the step usually forgotten): \(\displaystyle x=2\Rightarrow t=5\) and \(\displaystyle x=3\Rightarrow t=10\). Then \[\int_{2}^{3}\frac{x\,dx}{x^{2}+1}=\frac{1}{2}\int_{5}^{10}\frac{dt}{t}=\frac{1}{2}\bigl[\log t\bigr]_{5}^{10}=\frac{1}{2}\left(\log 10-\log 5\right).\]\[\int_{2}^{3}\frac{x\,dx}{x^{2}+1}=\frac{1}{2}\log 2\]

Split the numerator into the part that is proportional to the derivative of the denominator and a constant part: \[\int_{0}^{1}\frac{2x+3}{5x^{2}+1}\,dx=\int_{0}^{1}\frac{2x}{5x^{2}+1}\,dx+3\int_{0}^{1}\frac{dx}{5x^{2}+1}=I_{1}+I_{2}.\]\(\displaystyle I_{1}\): put \(\displaystyle t=5x^{2}+1\), so \(\displaystyle dt=10x\,dx\), i.e. \(\displaystyle 2x\,dx=\frac{dt}{5}\). The limits become \(\displaystyle x=0\Rightarrow t=1\), \(\displaystyle x=1\Rightarrow t=6\): \[I_{1}=\frac{1}{5}\int_{1}^{6}\frac{dt}{t}=\frac{1}{5}\bigl[\log t\bigr]_{1}^{6}=\frac{1}{5}\log 6.\]\(\displaystyle I_{2}\): take the coefficient of \(\displaystyle x^{2}\) out first, then use \(\displaystyle \int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a}\) with \(\displaystyle a=\frac{1}{\sqrt5}\): \[I_{2}=\frac{3}{5}\int_{0}^{1}\frac{dx}{x^{2}+\left(\dfrac{1}{\sqrt5}\right)^{2}}=\frac{3}{5}\cdot\sqrt5\left[\tan^{-1}\left(\sqrt5\,x\right)\right]_{0}^{1}=\frac{3}{\sqrt5}\tan^{-1}\sqrt5 .\]Adding, \[\int_{0}^{1}\frac{2x+3}{5x^{2}+1}\,dx=\frac{1}{5}\log 6+\frac{3}{\sqrt{5}}\tan^{-1}\sqrt{5}.\]

The factor \(\displaystyle x\) outside is, up to a constant, the derivative of the exponent \(\displaystyle x^{2}\), so substitute \[t=x^{2},\qquad dt=2x\,dx\ \Rightarrow\ x\,dx=\frac{dt}{2}.\]Change the limits along with the variable: \(\displaystyle x=0\Rightarrow t=0\) and \(\displaystyle x=1\Rightarrow t=1\). Then \[\int_{0}^{1}x\,e^{x^{2}}dx=\frac{1}{2}\int_{0}^{1}e^{t}\,dt=\frac{1}{2}\bigl[e^{t}\bigr]_{0}^{1}=\frac{1}{2}\left(e^{1}-e^{0}\right).\]\[\int_{0}^{1}x\,e^{x^{2}}dx=\frac{e-1}{2}\]

The integrand \(\displaystyle \dfrac{5x^{2}}{x^{2}+4x+3}\) is an improper rational function (degree of numerator = degree of denominator), so divide first: \[5x^{2}=5\left(x^{2}+4x+3\right)-(20x+15)\ \Longrightarrow\ \frac{5x^{2}}{x^{2}+4x+3}=5-\frac{20x+15}{(x+1)(x+3)}.\]Now split the proper part by partial fractions: \[\frac{20x+15}{(x+1)(x+3)}=\frac{A}{x+1}+\frac{B}{x+3},\qquad 20x+15=A(x+3)+B(x+1).\] Putting \(\displaystyle x=-1\): \(\displaystyle -5=2A\Rightarrow A=-\frac{5}{2}\). Putting \(\displaystyle x=-3\): \(\displaystyle -45=-2B\Rightarrow B=\frac{45}{2}\).So the integrand becomes \[5+\frac{5}{2}\cdot\frac{1}{x+1}-\frac{45}{2}\cdot\frac{1}{x+3}\] (note the two sign changes: the partial fractions are subtracted from \(\displaystyle 5\)).On \(\displaystyle [1,2]\) both \(\displaystyle x+1\) and \(\displaystyle x+3\) are positive, so the moduli drop: \[\int_{1}^{2}\frac{5x^{2}}{x^{2}+4x+3}\,dx=\left[5x+\frac{5}{2}\log (x+1)-\frac{45}{2}\log (x+3)\right]_{1}^{2}\] \[=\left(10+\frac{5}{2}\log 3-\frac{45}{2}\log 5\right)-\left(5+\frac{5}{2}\log 2-\frac{45}{2}\log 4\right).\]Collecting the logarithms, \[\int_{1}^{2}\frac{5x^{2}}{x^{2}+4x+3}\,dx=5+\frac{5}{2}\log\frac{3}{2}-\frac{45}{2}\log\frac{5}{4}.\]

Integrate term by term, using \(\displaystyle \int \sec^{2}x\,dx=\tan x\) and the power rule.An antiderivative is \[F(x)=2\tan x+\frac{x^{4}}{4}+2x.\]By the Second Fundamental Theorem of Calculus, with \(\displaystyle F(0)=0\) and \(\displaystyle \tan\frac{\pi}{4}=1\), \[\int_{0}^{\pi/4}\left(2\sec^{2}x+x^{3}+2\right)dx=2(1)+\frac{1}{4}\left(\frac{\pi}{4}\right)^{4}+2\cdot\frac{\pi}{4}.\]Since \(\displaystyle \dfrac{1}{4}\cdot\dfrac{\pi^{4}}{256}=\dfrac{\pi^{4}}{1024}\), \[\int_{0}^{\pi/4}\left(2\sec^{2}x+x^{3}+2\right)dx=2+\frac{\pi}{2}+\frac{\pi^{4}}{1024}.\]

Do not integrate the two squares separately; recognise the double-angle identity \[\cos\theta=\cos^{2}\frac{\theta}{2}-\sin^{2}\frac{\theta}{2}\ \Longrightarrow\ \sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}=-\cos x.\]Hence \[\int_{0}^{\pi}\left(\sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}\right)dx=-\int_{0}^{\pi}\cos x\,dx=-\bigl[\sin x\bigr]_{0}^{\pi}=-\left(\sin\pi-\sin 0\right).\]Both sines vanish, so \[\int_{0}^{\pi}\left(\sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}\right)dx=0.\]

Split the numerator so that one part is a multiple of \(\displaystyle \frac{d}{dx}\left(x^{2}+4\right)=2x\): \[\int_{0}^{2}\frac{6x+3}{x^{2}+4}\,dx=3\int_{0}^{2}\frac{2x}{x^{2}+4}\,dx+3\int_{0}^{2}\frac{dx}{x^{2}+4}.\]For the first integral put \(\displaystyle t=x^{2}+4\) (so \(\displaystyle dt=2x\,dx\), limits \(\displaystyle t=4\) to \(\displaystyle t=8\)): \[3\int_{4}^{8}\frac{dt}{t}=3\bigl[\log t\bigr]_{4}^{8}=3\left(\log 8-\log 4\right)=3\log 2.\]For the second use \(\displaystyle \int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a}\) with \(\displaystyle a=2\): \[3\cdot\frac{1}{2}\left[\tan^{-1}\frac{x}{2}\right]_{0}^{2}=\frac{3}{2}\left(\tan^{-1}1-\tan^{-1}0\right)=\frac{3}{2}\cdot\frac{\pi}{4}=\frac{3\pi}{8}.\]Adding, \[\int_{0}^{2}\frac{6x+3}{x^{2}+4}\,dx=3\log 2+\frac{3\pi}{8}.\]

Split the integral into two parts.Part $\displaystyle 1$: \(\displaystyle \displaystyle\int_{0}^{1}x\,e^{x}dx\) needs integration by parts, \(\displaystyle \int u\,v'\,dx=uv-\int u'v\,dx\). By ILATE take \(\displaystyle u=x\) (algebraic) and \(\displaystyle v'=e^{x}\): \[\int x\,e^{x}dx=x e^{x}-\int e^{x}dx=e^{x}(x-1).\] Hence \[\int_{0}^{1}x\,e^{x}dx=\bigl[e^{x}(x-1)\bigr]_{0}^{1}=e^{1}(0)-e^{0}(-1)=1.\]Part $\displaystyle 2$: \(\displaystyle \displaystyle\int_{0}^{1}\sin\frac{\pi x}{4}\,dx\); the chain-rule factor is \(\displaystyle \frac{4}{\pi}\): \[\int_{0}^{1}\sin\frac{\pi x}{4}\,dx=\left[-\frac{4}{\pi}\cos\frac{\pi x}{4}\right]_{0}^{1}=-\frac{4}{\pi}\left(\cos\frac{\pi}{4}-\cos 0\right)=-\frac{4}{\pi}\left(\frac{1}{\sqrt{2}}-1\right).\] That is \(\displaystyle \dfrac{4}{\pi}-\dfrac{4}{\pi\sqrt{2}}=\dfrac{4}{\pi}-\dfrac{2\sqrt{2}}{\pi}\).Adding the two parts, \[\int_{0}^{1}\left(x e^{x}+\sin\frac{\pi x}{4}\right)dx=1+\frac{4}{\pi}-\frac{2\sqrt{2}}{\pi}.\]

The antiderivative of \(\displaystyle \dfrac{1}{1+x^{2}}\) is \(\displaystyle \tan^{-1}x\), and the integrand is continuous on \(\displaystyle \left[1,\sqrt{3}\right]\), so by the Second Fundamental Theorem of Calculus \[\int_{1}^{\sqrt{3}}\frac{dx}{1+x^{2}}=\bigl[\tan^{-1}x\bigr]_{1}^{\sqrt{3}}=\tan^{-1}\sqrt{3}-\tan^{-1}1.\]Using principal values, \(\displaystyle \tan^{-1}\sqrt{3}=\frac{\pi}{3}\) and \(\displaystyle \tan^{-1}1=\frac{\pi}{4}\). The trap here is stopping at \(\displaystyle \frac{\pi}{3}\) and forgetting the lower limit: \[\frac{\pi}{3}-\frac{\pi}{4}=\frac{4\pi-3\pi}{12}=\frac{\pi}{12}.\]The value is \(\displaystyle \dfrac{\pi}{12}\), so the correct option is (D).

Take the coefficient of \(\displaystyle x^{2}\) outside so the standard form \(\displaystyle \int\dfrac{dx}{x^{2}+a^{2}}=\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}\) can be used: \[\frac{1}{4+9x^{2}}=\frac{1}{9}\cdot\frac{1}{x^{2}+\left(\dfrac{2}{3}\right)^{2}}.\]With \(\displaystyle a=\frac{2}{3}\), \[\int\frac{dx}{4+9x^{2}}=\frac{1}{9}\cdot\frac{3}{2}\tan^{-1}\frac{3x}{2}=\frac{1}{6}\tan^{-1}\frac{3x}{2}.\]Now apply the limits \(\displaystyle x=0\) to \(\displaystyle x=\frac{2}{3}\); at the upper limit \(\displaystyle \frac{3x}{2}=1\): \[\int_{0}^{2/3}\frac{dx}{4+9x^{2}}=\frac{1}{6}\left(\tan^{-1}1-\tan^{-1}0\right)=\frac{1}{6}\cdot\frac{\pi}{4}=\frac{\pi}{24}.\]The value is \(\displaystyle \dfrac{\pi}{24}\), so the correct option is (C).