Exercise 7.9(Integration)

Evaluate the definite integrals in Exercises 1 to 20.(Ex 7.9 integration ncert maths solution class 12)

Question 1: $\int_{-1}^1(x+1) d x$

Solution: Let $I=\int_{-1}^1(x+1) d x$

$=\left[\frac{x^2}{2}+x\right]_{-1}^1$

$=\left(\frac{1}{2}+1\right)-\left(\frac{1}{2}-1\right)$

$=\frac{1}{2}+1-\frac{1}{2}+1=2$

Question 2: $\int_2^3 \frac{1}{x} d x$

Solution: $I=\int \frac{1}{x} d x$

$=\left.\log x\right|_2 ^3$

$=\log |3|-\log |2|=\log \left|\frac{3}{2}\right|$

Question 3: $\int_1^2\left(4 x^3-5 x^2+6 x+9\right) d x$

Solution: $I=\int_1^2\left(4 x^3-5 x^2+6 x+9\right) d x$

$=\left[4\left(\frac{x^4}{4}\right)-5\left(\frac{x^3}{3}\right)+6\left(\frac{x^2}{2}\right)+9(x)\right]_1^2$

$=\left[x^4-\frac{5 x^3}{3}+3 x^2+9 x\right]_1^2$

$=\left[2^4-\frac{5(2)^3}{3}+3(2)^2+9(2)\right]-\left[(1)^4-\frac{5(1)^3}{3}+3(1)^2+9(1)\right]$

$=\left\{16-\frac{40}{3}+12+18\right\}-\left\{1-\frac{5}{3}+3+9\right\}$

$=16-\frac{40}{3}+12+18-1+\frac{5}{3}-3-9$

$=33-\frac{35}{3}=\frac{64}{3}$

Question 4: $\int_0^{\frac{\pi}{4}} \sin 2 x d x$

Solution: $I=\int_0^{\frac{\pi}{4}} \sin 2 x d x$

$=\left[\frac{-\cos 2 x}{2}\right]_0^{\frac{\pi}{4}}$

$=-\frac{1}{2}\left[\cos 2\left(\frac{\pi}{4}\right)-\cos 0\right]$

$=-\frac{1}{2}\left[\cos \left(\frac{\pi}{2}\right)-\cos 0\right]$

$=-\frac{1}{2}(0-1)=\frac{1}{2}$

Question 5: $\int_0^{\frac{\pi}{4}} \cos 2 x d x$

Solution: $I=\int_0^{\frac{\pi}{4}} \cos 2 x d x=\left[\frac{\sin 2 x}{2}\right]_0^\frac{\pi}{4}$

$=\frac{1}{2}\left[\sin 2\left(\frac{\pi}{2}\right)-\sin 0\right]$

$=\frac{1}{2}[\sin \pi-\sin 0]$

$=\frac{1}{2}(0-0)=0$

Question 6: $\int_4^5 e^x d x$

Solution: $I=\int_4^5 e^x d x$

$=\left[e^x\right]_4^5$

$=e^5-e^4=e^4(e-1)$

Question 7: $\int_0^{\frac{\pi}{4}} \tan x d x$

Solution : $I=\int_0^{\frac{\pi}{4}} \tan x d x=[-\log |\cos x|]_0^{\frac{\pi}{4}}$

$=-\log \left|\cos \frac{\pi}{4}\right|+\log |\cos 0|$

$=-\log \left|\frac{1}{\sqrt{2}}\right|+\log |1|$

$=-\log (2)^{-\frac{1}{2}}=\frac{1}{2} \log 2$

Question 8: $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x$

Solution : $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x$

$=[\log |\operatorname{cosec} x-\cot x|]_{\frac{\pi}{6}}^{\frac{\pi}{4}}$

$=\log \left|\operatorname{cosec} \frac{\pi}{4}-\cot \frac{\pi}{4}\right|-\log \left|\operatorname{cosec} \frac{\pi}{6}-\cot \frac{\pi}{6}\right|$

$=\log |\sqrt{2}-1|-\log |2-\sqrt{3}|$

$=\log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)$

Question 9: $\int_0^1 \frac{d x}{\sqrt{1-x^2}}$

Solution : $I=\int_0^1 \frac{d x}{\sqrt{1-x^2}}$

$=\left[\sin ^{-1} x\right]_0^1$

$=\sin ^{-1}(1)-\sin ^{-1}(0)$

$=\frac{\pi}{2}-0=\frac{\pi}{2}$

Question 10: $\int_0^1 \frac{d x}{1+x^2}$

Solution: Let $I=\int_0^1 \frac{d x}{1+x^2}$

$=\left[\tan ^{-1} x\right]_0^1$

$=\tan ^{-1}(1)-\tan ^{-1}(0)=\frac{\pi}{4}$

Question 11: $\int_0^1 \frac{d x}{x^2-1}$

Solution: Let $I=\int_0^1 \frac{d x}{x^2-1}$

$=\left[\frac{1}{2} \log \left|\frac{x-1}{x+1}\right|\right]_0^1$

$=\frac{1}{2}\left[\log \left|\frac{3-1}{3+1}\right|-\log \left|\frac{3-1}{3+1}\right|\right]$

$=\frac{1}{2}\left[\log \left|\frac{2}{4}\right|-\log \left|\frac{1}{3}\right|\right]$

$=\frac{1}{2}\left[\log \frac{1}{2}-\log \frac{1}{3}\right]$

$=\frac{1}{2}\left[\log \frac{3}{2}\right]$

Question 12: $\int_0^{\frac{\pi}{2}} \cos ^2 x d x$

Solution: $I=\int_0^{\frac{\pi}{2}} \cos ^2 x d x$

$=\int_0^{\frac{\pi}{2}}\left(\frac{1+\cos 2 x}{2}\right) d x$

$=\left[\frac{x}{2}+\frac{\sin 2 x}{4}\right] \frac{\pi}{2}$

$=\left[\frac{1}{2}\left(x+\frac{\sin 2 x}{2}\right)\right] \frac{\pi}{2}$

$=\frac{1}{2}\left[\left(\frac{\pi}{2}-\frac{\sin \pi}{2}\right)-\left(0+\frac{\sin \pi}{2}\right)\right]$

$=\frac{1}{2}\left[\frac{\pi}{2}+0-0-0\right]=\frac{\pi}{4}$

Question 13: $\int_2^3 \frac{x d x}{x^2+1}$

Solution : $I=\int_2^3 \frac{x}{x^2+1} d x$

$=\frac{1}{2} \int_2^3 \frac{2 x}{x^2+1} d x$

$=\left[\frac{1}{2} \log \left(1+x^2\right)\right]_2^3$

$=\frac{1}{2}\left[\log \left\{1+(3)^2\right\}-\log \left\{1+(2)^2\right\}\right]$

$=\frac{1}{2}[\log (10)-\log (5)]$

$=\frac{1}{2}\left[\log \frac{10}{5}\right]=\frac{1}{2} \log 2$

Question 14: $\int_0^1 \frac{2 x+3}{5 x^2+1} d x$

Solution: $I=\int_0^1 \frac{2 x+3}{5 x^2+1} d x$

$=\frac{1}{5} \int_0^1 \frac{5(2 x+3)}{5 x^2+1} d x$

$=\frac{1}{5} \int_0^1 \frac{10 x+15}{5 x^2+1} d x$

$=\frac{1}{5} \int_0^1 \frac{10 x}{5 x^2+1} d x+3 \int_0^1 \frac{1}{5 x^2+1} d x$

$=\frac{1}{5} \int_0^1 \frac{10 x}{5 x^2+1} d x+3 \int_0^1 \frac{1}{5\left(x^2+\frac{1}{5}\right)} d x$

$=\left[\frac{1}{5} \log \left(5 x^2+1\right)+\frac{3}{5} \cdot \frac{1}{\frac{1}{\sqrt{5}}} \tan ^{-1} \frac{x}{\frac{1}{\sqrt{5}}}\right]_0^1$

$=\left[\frac{1}{5} \log \left(5 x^2+1\right)+\frac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5} x)\right]_0^1$

$I=\left\{\frac{1}{5} \log (5+1)+\frac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5})\right\}-\left\{\frac{1}{5} \log (1)+\frac{3}{\sqrt{5}} \tan ^{-1}(0)\right\}$

$=\frac{1}{5} \log (6)+\frac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5})$

Question 15: $\int_0^1 x e^{x^2} d x$

Solution: $I=\int_0^1 x e^{x^2} d x$

Put $x^2=t \quad \Rightarrow 2 x d x=d t$

If $x \rightarrow 0, t \rightarrow 0 \quad$ and if $\quad x \rightarrow 1, t \rightarrow 1$

$\quad I=\frac{1}{2} \int_0^1 e^t d t=\frac{1}{2}\left[e^t\right]_0^1=\frac{1}{2} e-\frac{1}{2} e^0=\frac{1}{2}(e-1)$

Question 16: $\int_1^2 \dfrac{5x^2}{x^2+4x+3}dx$

Solution: Let $I=\int_1^2 \dfrac{5x^2}{x^2+4x+3}d$

$=\int_1^2 \dfrac{5x^2}{(x+1)(x+3)}d$

Since, $\dfrac{5x^2}{(x+1)(x+3)}=5+\dfrac{A}{(x+1)}+\dfrac{B}{x+3}--(i)$

$\Rightarrow \dfrac{5x^2}{(x+1)(x+3)}=\dfrac{5(x+1(x+3)+A(x+3)+B(x+1)}{(x+1)(x+3)}$

$\Rightarrow 5x^2=5(x+1)(x+3)+A(x+3)+B(x+1)$

Let $x =-1$

$5\times (-1)^2 = 0+A(-1+3)+0$

$\Rightarrow 5 = 2A$

$\Rightarrow A =5/2$

Let $x =-3$

$5\times (-3)^2 = 0+A(-3+3)+B(-3+1)$

$\Rightarrow 45 = -2B$

$\Rightarrow B =-45/2$

Putting in (i) and integrate

$\displaystyle\int_1^2\dfrac{5x^2}{(x+1)(x+3)}dx=\int_1^2\left[5+\dfrac{5/2}{(x+1)}+\dfrac{-45/2}{x+3}\right]dx$

$=\left[5x+\frac{5}{2}\log|x+1|-\frac{45}{2}\log|x+3|\right]_1^2$

$=\left[(5\times 2+\frac{5}{2}\log|2+1|-\frac{45}{2}\log|2+3|)-(5\times 1+\frac{5}{2}\log|1+1|-\frac{45}{2}\log|1+3|)\right]$

$=5+\frac{5}{2}\log\frac{3}{2}-\frac{45}{2}\log\frac{5}{4}$

Question 17: $\int_0^{\frac{\pi}{4}}\left(2 \sec ^2 x+x^3+2\right) d x$

Solution : $I=\int_0^{\frac{\pi}{4}}\left(2 \sec ^2 x+x^3+2\right) d x$

$=\left[2 \tan x+\frac{x^4}{4}+2 x\right]_0^{\frac{\pi}{4}}$

$=\left[\left\{2 \tan \frac{\pi}{4}+\frac{1}{4}\left(\frac{\pi}{4}\right)^4+2\left(\frac{\pi}{4}\right)\right\}-\{2 \tan 0+0+0\}\right]$

$=2 \tan \frac{\pi}{4}+\frac{\pi^4}{4^5}+\frac{\pi}{2}$

$=2+\frac{\pi}{2}+\frac{\pi^4}{1024}$

Question 18: $\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{\pi}{2}\right) d x$

Solution: Let $I=\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{\pi}{2}\right) d x$

$=-\int_0^\pi\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{\pi}{2}\right) d x=-\int_0^\pi \cos x d x$

$=-[\sin x]=-[\sin \pi-\sin 0]=-[0-0]=0$

Question 19: $\int_0^2 \frac{6 x+3}{x^2+4} d x$

Solution: Let $I=\int_0^2 \frac{6 x+3}{x^2+4} d x$

$=3 \int_0^2 \frac{2 x+1}{x^2+4} d x$

$=3 \int_0^2 \frac{2 x}{x^2+4} d x+3 \int_0^2 \frac{1}{x^2+4}$

$=\left[3 \log \left(x^2+4\right)+\frac{3}{2} \tan ^{-1} \frac{x}{2}\right]_0^2$

$=\left\{3 \log \left(2^2+4\right)+\frac{3}{2} \tan ^{-1}\left(\frac{2}{2}\right)\right\}-\left\{3 \log (0+4)+\frac{3}{2} \tan ^{-1}\left(\frac{0}{2}\right)\right\}$

$=3 \log 8+\frac{3}{2} \tan ^{-1} 1-3 \log 4-\frac{3}{2} \tan ^{-1} 0$

$=3 \log 8+\frac{3}{2} \tan ^{-1}\left(\frac{\pi}{4}\right)-3 \log 4-0$

$=3 \log \left(\frac{8}{4}\right)+\frac{3 \pi}{4}$

$=3 \log 2+\frac{3 \pi}{4}$

Question 20: $\int_0^1\left(x e^x+\sin \frac{\pi x}{4}\right) d x$

Solution: Let $I=\int_0^1\left(x e^x+\sin \frac{\pi x}{4}\right) d x$

$\int\left(x e^x+\sin \frac{\pi x}{4}\right) d x=x \int e^x d x-\int\left\{\left(\frac{d}{d x} x\right) \int e^x d x\right\} d x+\left\{\frac{-\cos \frac{\pi x}{4}}{\frac{\pi}{4}}\right\}$