NCERT Solutions · Class 12 Mathematics Integrals

The limits are symmetric about \(\displaystyle 0\), so test the parity of the integrand. Since \(\displaystyle \sin(-x)=-\sin x\), \[\sin^{2}(-x)=(-\sin x)^{2}=\sin^{2}x,\] so \(\displaystyle \sin^{2}x\) is even, and \(\displaystyle \displaystyle\int_{-a}^{a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx\).\[I=2\int_{0}^{\pi/2}\sin^{2}x\,dx=2\cdot\frac{1}{2}\int_{0}^{\pi/2}(1-\cos 2x)\,dx=\left[x-\frac{\sin 2x}{2}\right]_{0}^{\pi/2}.\]\[I=\left(\frac{\pi}{2}-\frac{\sin\pi}{2}\right)-(0-0)=\frac{\pi}{2}.\]Hence \(\displaystyle \displaystyle\int_{-\pi/2}^{\pi/2}\sin^{2}x\,dx=\frac{\pi}{2}\).

Let \(\displaystyle I=\displaystyle\int_{0}^{\pi}\frac{x}{1+\sin x}\,dx\) and use \(\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\) with \(\displaystyle a=\pi\). Since \(\displaystyle \sin(\pi-x)=\sin x\), only the numerator changes: \[I=\int_{0}^{\pi}\frac{\pi-x}{1+\sin x}\,dx.\]Adding the two expressions kills the \(\displaystyle x\): \[2I=\int_{0}^{\pi}\frac{x+(\pi-x)}{1+\sin x}\,dx=\pi\int_{0}^{\pi}\frac{dx}{1+\sin x}.\]Rationalise the denominator by multiplying above and below by \(\displaystyle 1-\sin x\): \[\frac{1}{1+\sin x}=\frac{1-\sin x}{1-\sin^{2}x}=\frac{1-\sin x}{\cos^{2}x}=\sec^{2}x-\sec x\tan x.\]An antiderivative is \(\displaystyle \tan x-\sec x\). This is the step to be careful with: \(\displaystyle \tan x\) and \(\displaystyle \sec x\) each blow up at \(\displaystyle x=\frac{\pi}{2}\), which is inside \(\displaystyle [0,\pi]\), so rewrite the antiderivative in a form that is continuous there: \[\tan x-\sec x=\frac{\sin x-1}{\cos x}=\frac{(\sin x-1)(\sin x+1)}{\cos x(1+\sin x)}=\frac{-\cos^{2}x}{\cos x(1+\sin x)}=\frac{-\cos x}{1+\sin x}.\]This last expression is continuous on all of \(\displaystyle [0,\pi]\), so it may be used at the limits: \[\int_{0}^{\pi}\frac{dx}{1+\sin x}=\left[\frac{-\cos x}{1+\sin x}\right]_{0}^{\pi}=\frac{-\cos\pi}{1+0}-\frac{-\cos 0}{1+0}=1+1=2.\]Hence \(\displaystyle 2I=\pi\cdot 2\), so \[I=\pi.\]

The limits are symmetric about \(\displaystyle 0\), so check parity. Since \(\displaystyle \sin(-x)=-\sin x\), \[\sin^{7}(-x)=(-\sin x)^{7}=-\sin^{7}x,\] an odd power of an odd function, so \(\displaystyle f(x)=\sin^{7}x\) is odd.By the property \(\displaystyle \displaystyle\int_{-a}^{a}f(x)\,dx=0\) when \(\displaystyle f\) is odd, \[\int_{-\pi/2}^{\pi/2}\sin^{7}x\,dx=0.\]Hence the value is \(\displaystyle 0\).

Use \(\displaystyle \displaystyle\int_{0}^{2a}f(x)\,dx=\int_{0}^{a}f(x)\,dx+\int_{0}^{a}f(2a-x)\,dx\) with \(\displaystyle a=\pi\).Since \(\displaystyle \cos(2\pi-x)=\cos x\), we get \(\displaystyle f(2\pi-x)=\cos^{5}x=f(x)\), so \[\int_{0}^{2\pi}\cos^{5}x\,dx=2\int_{0}^{\pi}\cos^{5}x\,dx.\]Now apply the same idea on \(\displaystyle [0,\pi]\) with \(\displaystyle a=\frac{\pi}{2}\). Here \(\displaystyle \cos(\pi-x)=-\cos x\), so \(\displaystyle \cos^{5}(\pi-x)=-\cos^{5}x\): \[\int_{0}^{\pi}\cos^{5}x\,dx=\int_{0}^{\pi/2}\left[\cos^{5}x+\cos^{5}(\pi-x)\right]dx=\int_{0}^{\pi/2}\left[\cos^{5}x-\cos^{5}x\right]dx=0.\]Hence \[\int_{0}^{2\pi}\cos^{5}x\,dx=2\cdot 0=0.\]

Let \[I=\int_{0}^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx\] and apply \(\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\) with \(\displaystyle a=\frac{\pi}{2}\).Under \(\displaystyle x\mapsto\frac{\pi}{2}-x\), \(\displaystyle \sin x\) and \(\displaystyle \cos x\) swap. The denominator \(\displaystyle 1+\sin x\cos x\) is symmetric in them, so it is unchanged; the numerator reverses sign: \[I=\int_{0}^{\pi/2}\frac{\cos x-\sin x}{1+\sin x\cos x}\,dx=-\int_{0}^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx=-I.\]Hence \(\displaystyle 2I=0\), so \[I=0.\]

Let \(\displaystyle I=\displaystyle\int_{0}^{\pi}\log(1+\cos x)\,dx\) and apply \(\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\) with \(\displaystyle a=\pi\). Since \(\displaystyle \cos(\pi-x)=-\cos x\), \[I=\int_{0}^{\pi}\log(1-\cos x)\,dx.\]Add the two forms and use \(\displaystyle \log A+\log B=\log AB\): \[2I=\int_{0}^{\pi}\log\left[(1+\cos x)(1-\cos x)\right]dx=\int_{0}^{\pi}\log\left(\sin^{2}x\right)dx=2\int_{0}^{\pi}\log\sin x\,dx.\]So everything reduces to the standard value of \(\displaystyle \displaystyle\int_{0}^{\pi/2}\log\sin x\,dx\). Call it \(\displaystyle A\). By the same reflection property, \(\displaystyle A=\displaystyle\int_{0}^{\pi/2}\log\cos x\,dx\), so \[2A=\int_{0}^{\pi/2}\log(\sin x\cos x)\,dx=\int_{0}^{\pi/2}\log\left(\frac{\sin 2x}{2}\right)dx=\int_{0}^{\pi/2}\log\sin 2x\,dx-\frac{\pi}{2}\log 2.\]In the remaining integral put \(\displaystyle t=2x\), \(\displaystyle dt=2\,dx\); the limits become \(\displaystyle 0\) to \(\displaystyle \pi\): \[\int_{0}^{\pi/2}\log\sin 2x\,dx=\frac{1}{2}\int_{0}^{\pi}\log\sin t\,dt=\frac{1}{2}\left(2A\right)=A,\] where \(\displaystyle \displaystyle\int_{0}^{\pi}\log\sin t\,dt=2A\) because \(\displaystyle \sin(\pi-t)=\sin t\) makes the graph symmetric about \(\displaystyle t=\frac{\pi}{2}\).Therefore \(\displaystyle 2A=A-\frac{\pi}{2}\log 2\), giving \(\displaystyle A=-\frac{\pi}{2}\log 2\) and \(\displaystyle \displaystyle\int_{0}^{\pi}\log\sin x\,dx=2A=-\pi\log 2\).Substituting back, \[2I=2(-\pi\log 2)\quad\Longrightarrow\quad I=-\pi\log 2.\]Hence \(\displaystyle \displaystyle\int_{0}^{\pi}\log(1+\cos x)\,dx=-\pi\log 2\).

Apply \(\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\). Let \[I=\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx.\]Replacing \(\displaystyle x\) by \(\displaystyle a-x\) interchanges \(\displaystyle \sqrt{x}\) and \(\displaystyle \sqrt{a-x}\): \[I=\int_{0}^{a}\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\,dx.\]The denominators agree, so adding the two expressions gives \[2I=\int_{0}^{a}\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\,dx=\int_{0}^{a}1\,dx=a.\]Hence \(\displaystyle I=\dfrac{a}{2}\) (for \(\displaystyle a>0\)).

\(\displaystyle |x-1|\) changes rule at \(\displaystyle x=1\), which lies inside \(\displaystyle [0,4]\), so split the integral there.On \(\displaystyle 0\le x\le 1\), \(\displaystyle x-1\le 0\) so \(\displaystyle |x-1|=1-x\); on \(\displaystyle 1\le x\le 4\), \(\displaystyle |x-1|=x-1\).\[\int_{0}^{4}|x-1|\,dx=\int_{0}^{1}(1-x)\,dx+\int_{1}^{4}(x-1)\,dx.\]First piece: \[\left[x-\frac{x^{2}}{2}\right]_{0}^{1}=\left(1-\frac{1}{2}\right)-0=\frac{1}{2}.\]Second piece: \[\left[\frac{x^{2}}{2}-x\right]_{1}^{4}=(8-4)-\left(\frac{1}{2}-1\right)=4+\frac{1}{2}=\frac{9}{2}.\]\[\int_{0}^{4}|x-1|\,dx=\frac{1}{2}+\frac{9}{2}=5.\]

Write \(\displaystyle I=\displaystyle\int_{0}^{a}f(x)g(x)\,dx\) and apply the property \(\displaystyle \displaystyle\int_{0}^{a}h(x)\,dx=\int_{0}^{a}h(a-x)\,dx\) to the whole integrand \(\displaystyle h=fg\): \[I=\int_{0}^{a}f(a-x)\,g(a-x)\,dx.\]Now use the two given conditions. First \(\displaystyle f(a-x)=f(x)\). Second, \(\displaystyle g(x)+g(a-x)=4\) gives \(\displaystyle g(a-x)=4-g(x)\). Substituting both: \[I=\int_{0}^{a}f(x)\left[4-g(x)\right]dx=4\int_{0}^{a}f(x)\,dx-\int_{0}^{a}f(x)g(x)\,dx=4\int_{0}^{a}f(x)\,dx-I.\]The integral \(\displaystyle I\) has reappeared, so bring it across: \[2I=4\int_{0}^{a}f(x)\,dx\quad\Longrightarrow\quad I=2\int_{0}^{a}f(x)\,dx.\]That is, \(\displaystyle \displaystyle\int_{0}^{a}f(x)g(x)\,dx=2\int_{0}^{a}f(x)\,dx\), as required.

The limits \(\displaystyle -\frac{\pi}{2}\) and \(\displaystyle \frac{\pi}{2}\) are symmetric about \(\displaystyle 0\), so split the integrand by parity and use \[\int_{-a}^{a}f(x)\,dx=0\ (f\ \text{odd}),\qquad \int_{-a}^{a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx\ (f\ \text{even}).\]Check each term: All three odd terms integrate to \(\displaystyle 0\) over the symmetric interval, leaving only the constant: \[\int_{-\pi/2}^{\pi/2}1\,dx=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi.\]Hence the value is \(\displaystyle \pi\) — option (C).

Let \[I=\int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right)dx\] and apply \(\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\) with \(\displaystyle a=\frac{\pi}{2}\), which swaps \(\displaystyle \sin x\) and \(\displaystyle \cos x\): \[I=\int_{0}^{\pi/2}\log\left(\frac{4+3\cos x}{4+3\sin x}\right)dx.\]The new integrand is the logarithm of the reciprocal of the old one, and \(\displaystyle \log\frac{1}{t}=-\log t\), so \[I=-\int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right)dx=-I.\]Hence \(\displaystyle 2I=0\), so \(\displaystyle I=0\) — option (C).