NCERT Solutions · Class 12 Mathematics Integrals

By the Second Fundamental Theorem of Calculus, if \(\displaystyle F\) is an antiderivative of a continuous \(\displaystyle f\) on \(\displaystyle [a,b]\), then \(\displaystyle \int_{a}^{b} f(x)\,dx=F(b)-F(a)\).Here \(\displaystyle f(x)=x+1\) is a polynomial, hence continuous, and by the power rule \[F(x)=\frac{x^{2}}{2}+x.\]Substitute the limits. The step to be careful with is the lower one: squaring makes the first term positive while the second stays negative, \[F(1)=\frac{1}{2}+1=\frac{3}{2},\qquad F(-1)=\frac{(-1)^{2}}{2}+(-1)=\frac{1}{2}-1=-\frac{1}{2}.\]Hence \[\int_{-1}^{1}(x+1)\,dx=\frac{3}{2}-\left(-\frac{1}{2}\right)=2.\]The value of the integral is \(\displaystyle 2\).

The antiderivative of \(\displaystyle \frac{1}{x}\) is \(\displaystyle \log|x|\). On the interval of integration \(\displaystyle [2,3]\) we have \(\displaystyle x>0\), so \(\displaystyle |x|=x\) and the modulus may be dropped; the integrand is also continuous there, so the Second Fundamental Theorem of Calculus applies.\[\int_{2}^{3}\frac{1}{x}\,dx=\bigl[\log x\bigr]_{2}^{3}=\log 3-\log 2.\]Using \(\displaystyle \log a-\log b=\log\dfrac{a}{b}\), \[\int_{2}^{3}\frac{1}{x}\,dx=\log\frac{3}{2}.\]

Integrate term by term using the power rule \(\displaystyle \int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}\ (n\neq-1)\), then apply the Second Fundamental Theorem of Calculus.An antiderivative is \[F(x)=x^{4}-\frac{5x^{3}}{3}+3x^{2}+9x.\]Evaluate at the limits: \[F(2)=16-\frac{40}{3}+12+18=46-\frac{40}{3},\] \[F(1)=1-\frac{5}{3}+3+9=13-\frac{5}{3}.\]Subtracting, and keeping the two fractional parts together so the arithmetic stays clean, \[\int_{1}^{2}\left(4x^{3}-5x^{2}+6x+9\right)dx=(46-13)-\frac{40-5}{3}=33-\frac{35}{3}=\frac{99-35}{3}.\]\[\int_{1}^{2}\left(4x^{3}-5x^{2}+6x+9\right)dx=\frac{64}{3}\]

By the chain rule, \(\displaystyle \dfrac{d}{dx}\left(-\dfrac{\cos 2x}{2}\right)=\sin 2x\), so \[\int \sin 2x\,dx=-\frac{\cos 2x}{2}.\] The factor \(\displaystyle \tfrac12\) coming from the inner derivative \(\displaystyle 2\) is the step most often dropped.By the Second Fundamental Theorem of Calculus, \[\int_{0}^{\pi/4}\sin 2x\,dx=\left[-\frac{\cos 2x}{2}\right]_{0}^{\pi/4}=-\frac{\cos\dfrac{\pi}{2}}{2}+\frac{\cos 0}{2}.\]Since \(\displaystyle \cos\frac{\pi}{2}=0\) and \(\displaystyle \cos 0=1\), \[\int_{0}^{\pi/4}\sin 2x\,dx=0+\frac{1}{2}=\frac{1}{2}.\]

Since \(\displaystyle \dfrac{d}{dx}\left(\dfrac{\sin 2x}{2}\right)=\cos 2x\), \[\int \cos 2x\,dx=\frac{\sin 2x}{2}.\]Applying the Second Fundamental Theorem of Calculus, and remembering that the upper limit \(\displaystyle x=\frac{\pi}{2}\) must be doubled inside the sine (it gives \(\displaystyle \sin\pi\), not \(\displaystyle \sin\frac{\pi}{2}\)), \[\int_{0}^{\pi/2}\cos 2x\,dx=\left[\frac{\sin 2x}{2}\right]_{0}^{\pi/2}=\frac{\sin \pi}{2}-\frac{\sin 0}{2}=\frac{0}{2}-\frac{0}{2}.\]\[\int_{0}^{\pi/2}\cos 2x\,dx=0\]

The exponential function is its own antiderivative: \(\displaystyle \int e^{x}\,dx=e^{x}\). By the Second Fundamental Theorem of Calculus, \[\int_{4}^{5}e^{x}\,dx=\bigl[e^{x}\bigr]_{4}^{5}=e^{5}-e^{4}.\]Taking out the common factor \(\displaystyle e^{4}\), \[\int_{4}^{5}e^{x}\,dx=e^{4}(e-1).\]

The standard result is \(\displaystyle \int \tan x\,dx=\log|\sec x|\) (obtained by putting \(\displaystyle t=\cos x\)). On \(\displaystyle \left[0,\frac{\pi}{4}\right]\) we have \(\displaystyle \cos x>0\), so \(\displaystyle \sec x>0\) and the modulus can be dropped; the integrand is continuous there, so the Second Fundamental Theorem of Calculus applies.\[\int_{0}^{\pi/4}\tan x\,dx=\bigl[\log \sec x\bigr]_{0}^{\pi/4}=\log\sec\frac{\pi}{4}-\log\sec 0.\]Now \(\displaystyle \sec\frac{\pi}{4}=\sqrt{2}\) and \(\displaystyle \sec 0=1\), so \[\int_{0}^{\pi/4}\tan x\,dx=\log\sqrt{2}-\log 1=\frac{1}{2}\log 2.\]

The standard result is \[\int \mathrm{cosec}\,x\,dx=\log\left|\mathrm{cosec}\,x-\cot x\right|.\] On \(\displaystyle \left[\frac{\pi}{6},\frac{\pi}{4}\right]\) the integrand is continuous, so the Second Fundamental Theorem of Calculus applies.At the upper limit \(\displaystyle x=\frac{\pi}{4}\): \(\displaystyle \mathrm{cosec}\,\frac{\pi}{4}=\sqrt{2}\), \(\displaystyle \cot\frac{\pi}{4}=1\), so the bracket is \(\displaystyle \sqrt{2}-1>0\).At the lower limit \(\displaystyle x=\frac{\pi}{6}\): \(\displaystyle \mathrm{cosec}\,\frac{\pi}{6}=2\), \(\displaystyle \cot\frac{\pi}{6}=\sqrt{3}\), so the bracket is \(\displaystyle 2-\sqrt{3}>0\).Both are positive, so the moduli may be dropped: \[\int_{\pi/6}^{\pi/4}\mathrm{cosec}\,x\,dx=\log(\sqrt{2}-1)-\log(2-\sqrt{3}).\]\[\int_{\pi/6}^{\pi/4}\mathrm{cosec}\,x\,dx=\log\left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\](Equivalently \(\displaystyle \log\dfrac{\tan\frac{\pi}{8}}{\tan\frac{\pi}{12}}\), since \(\displaystyle \mathrm{cosec}\,x-\cot x=\tan\frac{x}{2}\).)

By definition of the inverse sine, \(\displaystyle \dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^{2}}}\) for \(\displaystyle x\in(-1,1)\), so \(\displaystyle \sin^{-1}x\) is an antiderivative of the integrand.The integrand itself is unbounded as \(\displaystyle x\to 1^{-}\), but \(\displaystyle \sin^{-1}x\) remains continuous up to \(\displaystyle x=1\), so the value of the integral is obtained by substituting the limits into \(\displaystyle \sin^{-1}x\) as usual.Using principal values, \(\displaystyle \sin^{-1}1=\frac{\pi}{2}\) (the value in \(\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\)) and \(\displaystyle \sin^{-1}0=0\): \[\int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}}=\bigl[\sin^{-1}x\bigr]_{0}^{1}=\frac{\pi}{2}-0.\]\[\int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}}=\frac{\pi}{2}\]

Since \(\displaystyle \dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^{2}}\), the function \(\displaystyle \tan^{-1}x\) is an antiderivative of the integrand, which is continuous on \(\displaystyle [0,1]\). By the Second Fundamental Theorem of Calculus, \[\int_{0}^{1}\frac{dx}{1+x^{2}}=\bigl[\tan^{-1}x\bigr]_{0}^{1}=\tan^{-1}1-\tan^{-1}0.\]Taking principal values (branch \(\displaystyle \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\)), \(\displaystyle \tan^{-1}1=\frac{\pi}{4}\) and \(\displaystyle \tan^{-1}0=0\): \[\int_{0}^{1}\frac{dx}{1+x^{2}}=\frac{\pi}{4}.\]