Question 1515 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\text{a}}\sqrt{\text{a}^2-\text{x}^2}\text{ dx}$
AnswerLet $\text{x}=\text{a}\sin\theta$
Differentiating w.r.t. x, we get
$\text{dx}=\text{a}\cos\theta\text{ d}\theta$
Now, $\text{x}=0\Rightarrow\theta=0$
$\text{x}=\text{a}\Rightarrow\theta=\frac{\pi}{2}$
$\therefore\ \int_{0}^\limits{\text{a}}\sqrt{\text{a}^2-\text{x}^2}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\text{a}^2(1-\sin^2\theta)}\text{a}\cos\theta\text{ d}\theta$
$=\text{a}^2\int_{0}^\limits{\frac{\pi}{2}}\cos^2\theta\text{ d}\theta$ $\Big[\because(1-\sin^2\theta)=\cos^2\theta\text{ and }\frac{1+\cos2\theta}{2}=\cos2\theta\Big]$
$=\frac{\text{a}^2}{2}\int_{0}^\limits{\frac{\pi}{2}}\big(1+\cos2\theta\big)\text{d}\theta$
$=\frac{\text{a}^2}{2}\Big[\frac{\pi}{2}+0-0-0\big]$
$=\frac{\pi\text{a}^2}{4}$
View full question & answer→Question 1525 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}\ ....(\text{i})$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\tan\text{x}}{\sec(\pi-\text{x})\text{ cosec}(\pi-\text{x})}\text{ dx}$ $\Bigg[\text{Using}\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\pi}_0\frac{-(\pi-\text{x})\tan\text{x}}{-\sec\text{x}\text{ cosec}\text{x}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}+\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\frac{\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0\pi\sin^2\text{x dx}$
$=\pi\int\limits^{\pi}_0\big(1-\cos^2\text{x}\big)\text{dx}$
$=\pi\big[\text{x}\big]^{\pi}_0-\frac{\pi}{2}\int\limits^{\pi}_0\big(1-\cos^2\text{x}\big)\text{dx}$
$=\frac{\pi}{2}\big[\text{x}\big]^{\pi}_0-\frac{\pi}{2}\Big[\frac{\sin2\text{x}}{2}\Big]^{\pi}_0$
$=\frac{\pi^2}{4}$
Hence, $\text{I}=\frac{\pi^2}{4}$
View full question & answer→Question 1535 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\cos^4\text{x}\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{\frac{\pi}{2}}\cos^4\text{x}\text{ dx}$
$=\frac{1}{4}\int_{0}^\limits{\frac{\pi}{2}}(1+\cos2\text{x})^2\text{dx}$ $\big[\because2\cos^2\text{x}=1+\cos2\text{x}\big]$
$=\frac{1}{4}\int_{0}^\limits{\frac{\pi}{2}}\big(1+\cos^22\text{x}+2\cos2\text{x}\big)\text{dx}$
$=\frac{1}{4}\int_{0}^\limits{\frac{\pi}{2}}\Big(1+\frac{1+\cos4\text{x}}{2}+2\cos2\text{x}\Big)\text{dx}$
$=\frac{1}{4}\Big[\text{x}+\frac{1}{2}\text{x}+\frac{\sin4\text{x}}{8}+\sin2\text{x}\Big]^{\frac{\pi}{2}}_0$ $\Big[\because\int\cos4\text{x dx}=\frac{\sin4\text{x}}{4}\Big]$
$=\frac{1}{4}\Big[\frac{\pi}{2}+\frac{\pi}{4}+0+0-0-0-0-0\Big]$
$=\frac{1}{4}\times\frac{3\pi}{4}$
$=\frac{3\pi}{16}$
View full question & answer→Question 1545 Marks
Evaluate the following integrals:
$\int\limits_{0}^{1}\text{x}\tan^{-1}\text{x}\text{ dx} $
AnswerWe have,
$\int_{0}^\limits{1}\text{x}\tan^{-1}\text{x}\text{ dx}=\tan^{-1}\text{x}\int_{0}^\limits{1}\text{x dx}-\int_{0}^\limits{1}\big(\int\text{dx}\big)\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)\text{dx}$
$=\Big[\frac{\text{x}^2}{2}\tan^{-1}\text{x}\Big]^1_0-\frac{1}{2}\int_{0}^\limits{1}\frac{\text{x}^2}{1+\text{x}^2}\text{ dx}$
$=\Big[\frac{\text{x}^2}{2}\tan^{-1}\text{x}\Big]^1_0-\frac{1}{2}\int_{0}^\limits{1}\frac{1+\text{x}^2-1}{1+\text{x}^2}\text{ dx}$
$=\frac{1}{2}\Big(\frac{\pi}{4}\Big)-\frac{1}{2}\Bigg[\int_{0}^\limits{1}\text{dx}-\int_{0}^\limits{1}\frac{\text{dx}}{1+\text{x}^2}\Bigg]$
$=\frac{\pi}{8}-\frac{1}{2}\big[\text{x}-\tan^{-1}\text{x}\big]^{1}_0$
$=\frac{\pi}{8}-\frac{1}{2}\Big[1-\frac{\pi}{4}\Big]$
$=\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}$
$=\frac{\pi}{4}-\frac{1}{2}$
View full question & answer→Question 1555 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin\text{x}\cos\text{x}}{1+\sin^4\text{x}}\text{ dx}$
AnswerLet $\sin^2\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$2\sin\text{x}\cos\text{x dx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=1$
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\text{x}\cos\text{x}}{1+\sin^4\text{x}}\text{ dx}$
$=\frac{1}{2}\int\limits^1_0\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{1}{2}\big[\tan^{-1}\text{t}\big]^1_0$
$=\frac{1}{2}\Big[\tan^{-1}(1)-\tan^{-1}(0)\Big]$
$=\frac{1}{2}\Big[\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)-\tan^{-1}(\tan0)\Big]$
$=\frac{1}{2}\times\frac{\pi}{4}$
$=\frac{\pi}{8}$
View full question & answer→Question 1565 Marks
Evaluate the following integrals:
$\int\limits^1_{-1}|2\text{x}+1|\text{dx}$
AnswerWe know that,
$\int^\limits1_{-1}|2\text{x}+1|\text{dx}$
$=\int^\limits\frac{1}{2}_{-1}-(2\text{x}+1)\text{dx}+\int\limits_{-\frac{1}{2}}^{1}(2\text{x}+1)\text{dx}$
$=-\Big[\frac{2\text{x}^2}{2}+\text{x}\Big]^{-\frac{1}{2}}_{-1}+\Big[\frac{2\text{x}^2}{2}+\text{x}\Big]^{1}_{-\frac{1}{2}}$
$=-\bigg[\Big(\frac{2}{8}-\frac{1}{2}\Big)-\Big(\frac{2}{2}-1\Big)\bigg]+\bigg[\Big(\frac{2}{2}+1\Big)-\Big(\frac{2}{8}-\frac{1}{2}\Big)\bigg]$
$=-\bigg[\Big(\frac{1}{4}-\frac{1}{2}\Big)-(1-1)\bigg]+\bigg[(1+1)+\Big(\frac{1}{4}-\frac{1}{2}\Big)\bigg]$
$=-\Big[-\frac{1}{4}\Big]++\Big[2+\frac{1}{4}\Big]$
$=\frac{1}{4}+2+\frac{1}{4}$
$=2\frac{1}{2}$
View full question & answer→Question 1575 Marks
Evaluate the following integrals:
$\int\limits^{\text{a}}_0\frac{1}{\text{x}+\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^{\text{a}}_0\frac{1}{\text{x}+\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}$
Putting $\text{x}=\text{a}\sin\theta$
$\text{dx}=\text{a}\cos\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$
And $\text{x}\rightarrow\text{a};\theta\rightarrow\frac{\pi}{2}$
$\therefore\ \text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\cos\theta}{\text{a}\sin\theta+\sqrt{\text{a}^2-(\text{a}\sin\theta)^2}}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\cos\theta}{\text{a}\sin\theta+\text{a}\cos\theta}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\cos\theta}{\sin\theta+\cos\theta}\text{ dx}\ ....(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\cos\big(\frac{\pi}{2}-\theta\big)}{\sin\big(\frac{\pi}{2}-\theta\big)+\cos\big(\frac{\pi}{2}-\theta\big)}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\theta}{\cos\theta+\sin\theta}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\theta}{\sin\theta+\cos\theta}\text{ d}\theta\ ....(\text{ii})$
By adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\cos\theta+\sin\theta}{\sin\theta+\cos\theta}\text{ d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{d}\theta$
$\Rightarrow2\text{I}=\Big[\theta\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{2}$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1585 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{2}\frac{1}{4+\text{x}-\text{x}^2}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{2}\frac{1}{4+\text{x}-\text{x}^2}\text{ dx}$ Then
$\text{I}=\int_{0}^\limits{2}\frac{1}{\text{x}^2-\text{x}-4}\text{ dx}$
$\Rightarrow\text{I}=-\int_{0}^\limits{2}\frac{1}{\Big(\text{x}^2-\text{x}+\frac{1}{4}\Big)-\frac{1}{4}-4}\text{ dx}$
$\Rightarrow\text{I}=-\int_{0}^\limits{2}\frac{1}{\Big(\text{x}-\frac{1}{2}\Big)^2-\frac{17}{4}}\text{ dx}$
$\Rightarrow\text{I}=-\int_{0}^\limits{2}\frac{1}{\Big(\text{x}-\frac{1}{2}\Big)^2-\Big(\frac{\sqrt{17}}{2}\Big)^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{2}\frac{1}{-\big(\frac{2\text{x}-1}{2}\big)^2+\big(\frac{\sqrt{17}}{2}\big)^2}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big[\log\Big(\frac{\sqrt{17}+2\text{x}-1}{\sqrt{17}-2\text{x}+1}\Big)\Big]^2_0$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{\sqrt{17}+3}{\sqrt{17}-3}-\log\frac{\sqrt{17}-1}{\sqrt{17}+1}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{26+6\sqrt{17}}{8}-\log\frac{18-2\sqrt{17}}{16}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{52+12\sqrt{17}}{18-2\sqrt{17}}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{52+12\sqrt{17}}{18-2\sqrt{17}}\times\frac{18+2\sqrt{17}}{18+2\sqrt{17}}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\log\frac{1344+320\sqrt{17}}{256}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\log\frac{21+5\sqrt{17}}{4}$
View full question & answer→Question 1595 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{0}\frac{\tan\text{x}}{1+\text{m}^2\tan^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\tan\text{x}}{1+\text{m}^2\tan^2\text{x}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\frac{\sin\text{x}}{\cos\text{x}}}{1+\text{m}^2\frac{\sin^2\text{x}}{\cos^2\text{x}}}\text{ dx}=\int^\limits{\frac{\pi}{2}}_{0}\frac{{\sin\text{x}}{\cos\text{x}}}{\cos^2\text{x}+\text{m}^2{\sin^2\text{x}}}\text{ dx}$
Put $\cos^2\text{x}+\text{m}^2\sin^2\text{x}=\text{z}$
$\therefore\ 2\cos\text{x}(-\sin\text{x})\text{dx}+\text{m}^2\times2\sin\text{x}\cos\text{x dx}=\text{ dz}$
$\Rightarrow2(\text{m}^2-1)\sin\text{x}\cos\text{x dx}=\text{dz}$
$\Rightarrow\sin\text{x}\cos\text{x dx}=\frac{\text{dz}}{2(\text{m}^2-1)}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow1$ $\big(\text{z}=\cos^2\text{x}+\text{m}^2\sin^2\text{x}=1+\text{m}^2\times0=1\big)$
When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow\text{m} ^2$ $\big(\text{z}=\cos^2\text{x}+\text{m}^2\sin^2\text{x}=0+\text{m}^2\times0=\text{m}^2\big)$
$\therefore\ \text{I}=-\frac{1}{2(\text{m}^2-1)}\int\limits^{\text{m}^2}_1\frac{\text{dz}}{\text{z}}$
$=\frac{1}{2(\text{m}^2-1)}\big[\log\text{z}\big]^{\text{m}^2}_1$
$=\frac{1}{2(\text{m}^2-1)}\big(\log\text{m}^2-\log1\big)$
$=\frac{1}{2(\text{m}^2-1)}\big(2\log|\text{m}|-0\big)$
$=\frac{\log|\text{m}|}{\text{m}^2-1}$
View full question & answer→Question 1605 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^{3}_{2}\text{x}^2\text{ dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=2,\text{ b}=3,\text{ f(x)}=\text{x}^2,\text{ h}=\frac{3-2}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^{3}_{2}\text{x}^2\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(2)+\text{f}(2+\text{h})\ ....\ +\text{f}\big\{2+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2^2+(2+\text{h})^2+\ ....+\ \big\{2(\text{n}-1)\text{h}\big\}^2\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2+4\text{h}\big\}\\+4\text{h}\big\{1+2+\ ....+\ (\text{n}-1)\text{h}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(\text{n}-2)}{6}+4\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{1}{\text{n}}\Big[4\text{n}+\frac{(\text{n}-1)(2\text{n}-1)}{6\text{n}}+2\text{n}-2\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}5\Big[6+\frac{1}{6}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{2}{\text{n}}\Big]$
$=6+\frac{1}{3}$
$=\frac{19}{3}$
View full question & answer→Question 1615 Marks
Evaluate the following integrals:
$\int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}$
Putting $\text{x}=\tan\theta$
$\text{dx}=\sec^2\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$
And $\text{x}\rightarrow1;\theta\rightarrow\frac{\pi}{4}$
Now, integral becomes,
$\text{I}=\int\limits^{\frac{\pi}{4}}_0\frac{\log(1+\tan\theta)}{\sec^2\theta}\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\big[\log(\tan\theta)\big]\text{d}\theta\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{1+\tan\Big(\frac{\pi}{4}-\theta\Big)\Big\}\bigg]\text{d}\theta$ $\Bigg[\because\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\Bigg[\log\bigg\{1+\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\tan\theta}\bigg\}\Bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{1+\frac{1-\tan\theta}{1+\tan\theta}\Big\}\bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{\frac{2}{1+\tan\theta}\Big\}\bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\big[\log2-\log(1+\tan\theta)\big]\text{d}\theta\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{4}}_0\big(\log2\big)\text{d}\theta$
$\Rightarrow2\text{I}=\big(\log2\big)\Big[\theta\Big]^{\frac{\pi}{4}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{4}\log2$
$\Rightarrow\text{I}=\frac{\pi}{8}\log2$
$\therefore\ \int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}=\frac{\pi}{8}\log2$
View full question & answer→Question 1625 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan\big[\frac{\pi}{3}+\big(-\frac{\pi}{3}\big)-\text{x}\big]}}\text{ dx}$ $\Bigg[\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan(-\text{x})}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{-\tan\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{\text{e}^{\tan\text{x}}}{\text{e}^{\tan\text{x}}+1}\text{ dx}\ ...{\text{(ii)}}$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{\text{e}^{1+\tan\text{x}}}{\text{e}^{1+\tan\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{\frac{\pi}{3}}_{-\frac{\pi}{3}}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\Big(-\frac{\pi}{3}\Big)$
$\Rightarrow2\text{I}=\frac{2\pi}{3}$
$\Rightarrow\text{I}=\frac{\pi}{3}$
View full question & answer→Question 1635 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}+3\cos\text{x}+2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}+3\cos\text{x}+2}\text{ dx}$ Then,
Let $\cos\text{x}=\text{t},$ Then, $-\sin\text{x dx}=\text{dt}$
When, $\text{x}=0,\text{ t}=1$ and $\text{x}=\frac{\pi}{2},\text{ t}=0$
$\therefore\ \text{I}=-\int^\limits0_1\frac{\text{t dt}}{\text{t}^2+3\text{t}+2}$
$\Rightarrow\text{I}=\int\limits^0_1\frac{-\text{t dt}}{\text{t}^3+3\text{t}+2}$
$\Rightarrow\text{I}=\int\limits^0_1\Big(\frac{1}{(\text{t}+1)}-\frac{2}{(\text{t}+2)}\Big)\text{dt}$
$\Rightarrow\text{I}=\Big[\log(\text{t}+1)-2\log(\text{t}+2)\Big]^0_1$
$\Rightarrow\text{I}=\bigg[\log\frac{(\text{t}+1)}{(\text{t}+2)^2}\bigg]^1_0$
$\Rightarrow\text{I}=\bigg[\log\Big(\frac{1}{4}\Big)-\log\Big(\frac{2}{9}\Big)\bigg]^1_0$
$\Rightarrow\text{I}=\log\frac{9}{8}$
View full question & answer→Question 1645 Marks
Evaluate the following integrals:
$\int\limits_{0}^{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}$
AnswerLet $1+\text{x}^2=\text{t}$
Differentiating w.r.t. x, we get
$2\text{xdx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=1$
$\text{x}=1\Rightarrow\text{t}=2$
$\int_{0}^\limits{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}=\int_{1}^\limits{2}\frac{12(\text{t}-1)}{\text{t}^4}\text{ dt}$
$=12\int_{1}^\limits{2}\Big(\frac{1}{\text{t}^3}-\frac{1}{\text{t}^4}\Big)\text{dt}$
$=12\Big[-\frac{1}{2\text{t}^2}-\frac{1}{3\text{t}^3}\Big]^2_1$
$=12\Big[-\frac{1}{8}+\frac{1}{24}+\frac{1}{2}-\frac{1}{3}\Big]$
$=12\Big[\frac{-3+1+12-8}{24}\Big]$
$=\frac{12\times2}{24}=1$
View full question & answer→Question 1655 Marks
If f(2a - x) = -f(x), prove that $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=0$
AnswerLet $\text{I}=\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=0$
Using additive property
$\text{I}=\int\limits^{\text{a}}_0\text{f(x)}\text{dx}+\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
Consider the integral $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
Let $\text{x}=2\text{a}-\text{t},$ Then $\text{dx}=-\text{dt}$
When $\text{x}=\text{a},\text{t}=\text{a}$ and $\text{x}=2\text{a},\text{t}=0$
Threrfore,
$=\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=-\int\limits^0_\text{a}\text{f}(2\text{a}-\text{t})\text{dt}$
$=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{t})\text{dt}$
$=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$ (Chang ing the variable)
We have
$\text{f}(2\text{a}-\text{x})=-\text{f(x)}$
Therefore,
$\text{I}=\int\limits^\text{a}_0\text{f(x)}\text{dx}-\int\limits^\text{a}_0\text{f(x)}\text{dx}=0$
View full question & answer→Question 1665 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}$
AnswerWe have,
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}=\frac{1}{\text{b}^2}\int_{0}^\limits{\infty}\frac{1}{\big(\frac{\text{a}}{\text{b}}\big)^2+\text{x}^2}\text{ dx}$
We have that,
$\int\frac{1}{\text{a}^2+\text{x}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}$
$\therefore\ \frac{1}{\text{b}^2}\int_{0}^\limits{\infty}\frac{1}{\big(\frac{\text{a}}{\text{b}}\big)^2+\text{x}^2}\text{ dx}=\frac{1}{\text{b}^2}\Big[\frac{\text{b}}{\text{a}}\tan^{-1}\Big(\frac{\text{bx}}{\text{a}}\Big)\Big]^{\infty}_0$
$=\frac{1}{\text{ab}}\Big[\tan^{-1}\Big(\frac{\text{bx}}{\text{a}}\Big)\Big]^{\infty}_0$
$=\frac{1}{\text{ab}}\big[\tan^{-1}\infty-\tan^{-1}0\big]$
$=\frac{1}{\text{ab}}\Big[\frac{\pi}{2}-0\Big]$
$=\frac{\pi}{2\text{ab}}$
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}=\frac{\pi}{2\text{ab}}$
View full question & answer→Question 1675 Marks
Evaluate the following integrals:
$\int\limits^{1}_0\big|\text{x}\sin\pi\text{x}\big|\text{dx}$
AnswerFor $0<\text{x}<1,\text{ x}>0$ and $\sin\pi\text{x}>0\Rightarrow\text{x}\sin\pi\text{x}>0$
$\therefore\ \int\limits^{1}_0|\text{x}\sin\pi\text{x}|\text{ dx}=\int\limits^{1}_0\text{x}\sin\pi\text{x}\text{ dx}$
Let $\text{I}=\int\text{x}\sin\pi\text{x}\text{ dx}$
$=\text{x}\int\sin\pi\text{x}\text{ dx}-\int\Big(\frac{\text{d}}{\text{dx}}\text{x}\int\sin\pi\text{x dx}\Big)\text{dx}$
$=\text{x}\Big(\frac{-\cos\pi\text{x}}{\pi}\Big)-\int\Big(\frac{-\text{cos}\pi\text{x}}{\pi}\Big)\text{dx}$
$=\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}$
Applying the limits, we get
$\int\limits^{1}_0|\text{x}\sin\pi\text{x}|\text{ dx}=\Big[\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}\Big]^1_0$
$=\Big(\frac{-\cos\pi}{\pi}+\frac{\sin\pi}{\pi^2}\big)-(0+0)$
$=\frac{1}{\pi}+0-0$
$=\frac{1}{\pi}$
View full question & answer→Question 1685 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^2_{0}\big(\text{x}^2+1\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}^2+1,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{0}\big(\text{x}^2+1\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big(0+(\text{n}-1)\text{h}\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+1)+(\text{h}^2+1)+\ ....\ +\big\{(\text{n}-1)^2\text{h}^2+1\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[\text{n}+\frac{2(\text{n}-1)(2\text{n}-1)}{3\text{n}}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{1+\frac{2}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=2+\frac{8}{3}$
$=\frac{14}{3}$
View full question & answer→Question 1695 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\frac{\text{x}\sin\text{x}}{1+\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\frac{\text{x}\sin\text{x}}{1+\sin\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin(\pi-\text{x})}{1+\sin(\pi-\text{x})}\text{ dx}$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin\text{x}}{1+\sin\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\frac{\sin\text{x}}{1+\sin\text{x}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi\sin\text{x}}{1+\sin\text{x}}\text{ dx}$
$=\pi\int\limits^{\pi}_0\frac{1+\sin\text{x}-1}{1+\sin\text{x}}\text{ dx}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\frac{(1-\sin\text{x})}{(1+\sin\text{x})(1-\sin\text{x})}\text{ dx}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\frac{(1-\sin\text{x})}{1-\sin}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\big(\sec^2\text{x}-\sec\text{x}\tan\text{x}\big)\text{dx}$
$=\pi\big[\text{x}\big]^{\pi}_0-\pi\big[\tan\text{x}-\sec\text{x}\big]^{\pi}_0$
$=\pi^2-\pi(0+1-0+1)$
$=\pi^2-2\pi$
Hence, $\text{I}=\pi\Big(\frac{\pi}{2}-1\Big)$
View full question & answer→Question 1705 Marks
Evaluate the following integrals:
$\int\limits_{0}^{1}\tan^{-1}\text{x dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\tan^{-1}\text{x dx}$ Then,
$\text{I}=\int_{0}^\limits{1}\tan^{-1}\text{x dx}$
Integrating by parts,
$\text{I}=\big[\text{x}\tan^{-1}\text{x}\big]^1_0-\int_{0}^\limits{1}\frac{\text{x}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\big[\text{x}\tan^{-1}\text{x}\big]^1_0-\frac{1}{2}\big[\log\big(\text{x}^2+1\big)\big]^1_0$
$\Rightarrow\text{I}=\frac{\pi}{4}-0-\frac{1}{2}\log2+0$
$\Rightarrow\text{I}=\frac{\pi}{4}-\frac{1}{2}\log2$
View full question & answer→Question 1715 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{4}}\Big(\sqrt{\tan\text{x}}+\sqrt{\cot}\text{x}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\Big(\sqrt{\tan\text{x}}+\sqrt{\cot}\text{x}\Big)\text{dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\bigg(\sqrt{\frac{\sin\text{x}}{\cos\text{x}}}+\sqrt{\frac{\cos\text{x}}{\sin\text{x}}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{\sin\text{x}\cos\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\sqrt{2}\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2\sin\text{x}\cos\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\sqrt{2}\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1-(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
Let $\sin\text{x}-\cos\text{x}=\text{t}$ Then, $\cos\text{x}+\sin\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=1$ and $\text{x}=\frac{\pi}{4},\text{t}=0$
$\therefore\ \text{I}=\sqrt{2}\int^\limits0_{-1}\frac{\text{dt}}{\sqrt{1-\text{t}^2}}$
$\Rightarrow\text{I}=\sqrt{2}\Big[\sin^{-1}\text{t}\Big]^0_{-1}$
$\Rightarrow\text{I}=\frac{\pi}{\sqrt{2}}$
View full question & answer→Question 1725 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{3}{2}}}\text{ dx}$
Answer$\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{3}{2}}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{3}{2}}}\text{ dx}\times\frac{\sqrt{1-\cos\text{x}}}{\sqrt{1-\cos\text{x}}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1-\cos^2\text{x}}}{(1-\cos\text{x})^2}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sin\text{x}}{(1-\cos\text{x})^2}\text{ dx}$
Let $1-\cos\text{x}=\text{t},$ Then $\sin\text{x dx}=\text{dt}$
When $\text{x}=\frac{\pi}{3},\text{ t}=\frac{1}{2}$ and $\text{x}=\frac{\pi}{2},\text{ t}=1$
Therefore the integral becomes
$=\int_{1}^{\frac{1}{2}}\frac{\text{dt}}{\text{t}^2}$
$=\Big[-\frac{1}{\text{t}}\Big]^1_\frac{1}{2}$
$=-1+2$
$=1$
View full question & answer→Question 1735 Marks
Evaluate the following integrals:
$\int\limits^1_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}$
AnswerLet $\text{x}=\cos2\theta$
Differentiating w.r.t. x, we get
$\text{dx}=-2\sin2\theta\text{ d}\theta$
Now, $\text{x}=0\Rightarrow\theta=\frac{\pi}{4}$
$\text{x}=1\Rightarrow\theta=0$
$\therefore\ \int^\limits1_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}=\int^0\limits_\frac{\pi}{4}\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\big(-2\sin2\theta\big)\text{d}\theta$
$=\int_0\limits^\frac{\pi}{4}\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\big(2\sin2\theta\big)\text{d}\theta$ $\Big[\because\sin2\theta=2\sin\theta\cos\theta;\text{ and }\sin^2\theta=\frac{1-\cos2\theta}{2}\Big]$
$=2=\int_0\limits^\frac{\pi}{4}\frac{\sin\theta}{\cos\theta}\cdot\sin2\theta\text{ d}\theta$
$=4\int_0\limits^\frac{\pi}{4}\sin^2\theta\text{ d}\theta$
$=2\int_0\limits^\frac{\pi}{4}\big(1-\cos2\theta\big)\text{d}\theta$
$=2\Big[\theta-\frac{\sin^2\theta}{2}\Big]^{\frac{\pi}{4}}_0$
$=2\Big[\frac{\pi}{4}-\frac{1}{2}\Big]$
$=\frac{\pi}{2}-1$
View full question & answer→Question 1745 Marks
Evaluate the following integrals:
$\int\limits^1_{-1}\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\limits^1_{-1}\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)\text{dx}$
Here $\text{f(x)}=\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)$
$\text{f}(-\text{x})=\log\Big(\frac{2+\text{x}}{2-\text{x}}\Big)$
$=-\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)$
$=-\text{f(x)}$
Hence f(x) is an odd function,
Therefore,
$\text{I}=0$
View full question & answer→Question 1755 Marks
Evaluate the following integrals:
$\int\limits^\pi_0\frac{\text{x}}{1+\sin\alpha\sin\text{x}}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^\pi_0\frac{\text{x}}{1+\sin\alpha\sin\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^\pi_0\frac{\pi-\text{x}}{1+\cos\alpha\sin(\pi-\text{x})}\text{ dx}$
$=\int\limits^\pi_0\frac{\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^\pi_0\frac{\text{x}+\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\frac{\pi}{2}\int\limits^\pi_0\frac{1}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{1}{1+\cos\alpha\sin\text{x}}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{1}{1+\cos\alpha\frac{2\tan\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{1+\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}+2\cos\alpha\tan^{2}\frac{\text{x}}{2}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{\sec^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}+2\cos\alpha\tan^{2}\frac{\text{x}}{2}}\text{ dx}$
Putting $\tan\frac{\text{x}}{2}=\text{t}$
$\Rightarrow\frac{1}{2}\sec^2\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0;\text{ t}\rightarrow0$
and $\text{x}\rightarrow\pi;\text{ t}\rightarrow\infty$
$\therefore\ \text{I}=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{1+\text{t}^2+2\cos\alpha+1}\text{ dt}$
$=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{(\text{t}+\cos\alpha)-\cos^2\alpha+1}\text{ dt}$
$={\pi}\int\limits^{\infty}_0\frac{1}{(\text{t}+\cos\alpha)+\sin^2\alpha}\text{ dt}$
$=\pi\Big[\frac{1}{\sin\alpha}\tan^{-1}\Big(\frac{1+\cos\alpha}{\sin\alpha}\Big)\Big]^1_0$
$=\frac{\pi}{\sin\alpha}\Big[\tan^{-1}(\infty)-\tan^{-1}(\cot\alpha)\Big]$
$=\frac{\pi}{\sin\alpha}\Big[\frac{\pi}{2}-\tan^{-1}\Big(\tan\Big(\frac{\pi}{2}-\alpha\Big)\Big)\Big]$
$=\frac{\pi\alpha}{\sin\alpha}$
View full question & answer→Question 1765 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^2_0(\text{x}+3)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}+3,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_0(\text{x}+3)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}(0+(\text{n}-1)\text{h})\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+3)+(0+\text{h}+3)+\ ....\ +(0+(\text{n}-1)\text{h}+3)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[3\text{n}+\text{h}\{1+2+3+\ ....\ +(\text{n}-1)\}\Big]$
$=\lim\limits_{\text{n}\rightarrow0}\text{h}\Big[3\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[3\text{n}+\text{n}-1\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big(4-\frac{1}{\text{n}}\Big)$
$=8$
View full question & answer→Question 1775 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^4\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^4\text{x}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin(\pi-\text{x})\cos^4(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin\text{x}\cos^4\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\sin\text{x}\cos^4\text{x dx}$
$=\pi\int\limits^{\pi}_0\sin\text{x}\cos^4\text{x dx}$
Let $\cos\text{x}=\text{t},$ Then $-\sin\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=1,\text{x}=\pi,\text{t}=-1$
Therefore, $2\text{I}=-\pi\int\limits^{-1}_1\text{t}^4\text{ dt}$
$=\pi\int\limits^{1}_{-1}\text{t}^4\text{ dt}$
$=\pi\Big[\frac{\text{t}^5}{5}\Big]^{1}_{-1}$
$=\frac{\pi}{5}+\frac{\pi}{5}$
$=\frac{2\pi}{5}$
Hence, $\text{I}=\frac{\pi}{5}$
View full question & answer→Question 1785 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{\text{dx}}{\text{a}\cos\text{x}+\text{b}\sin\text{x}}\text{ a},\text{b}>0$
Answer$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}\cos\text{x}+\text{b}\sin\text{x}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}\Bigg(\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)+\text{b}\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\big(1+\tan^{2}\frac{\text{x}}{2}\big)}{\text{a}-\text{a}\tan^{2}\frac{\text{x}}{2}+2\text{b}\tan\frac{\text{x}}{2}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sec^2\frac{\text{x}}{2}}{\text{a}-\text{a}\tan^{2}\frac{\text{x}}{2}+2\text{b}\tan\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Then, $\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi}{2},\text{t}=1$
Therefore the integral becomes
$\text{I}=\int^\limits1_0\frac{2\text{dt}}{\text{a}-\text{a}\text{t}^2+2\text{bt}}$
$=\int^\limits1_0\frac{2\text{dt}}{-\text{a}\big[\text{t}^2-\frac{2\text{bt}}{\text{a}-1}\big]}$
$=\frac{2}{\text{a}}\int^\limits1_0\frac{\text{dt}}{-\Big[\big(\text{t}-\frac{\text{b}}{\text{a}}\big)^2-1-\frac{\text{b}^2}{\text{a}^2}\Big]}$
$=\frac{2}{\text{a}}\int^\limits1_0\frac{\text{dt}}{\Big(\frac{\text{b}^2}{\text{a}^2}+1\Big)-\big(\text{t}-\frac{\text{b}}{\text{a}}\big)^2}$
$=\frac{2}{\text{a}}\begin{bmatrix}\frac{1}{2\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}}}\begin{pmatrix}\log\begin{vmatrix}\frac{2\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}}+\big(\text{t}-\frac{\text{b}}{\text{a}}\big)}{\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}-\big(\text{t}-\frac{\text{b}}{\text{a}}\big)}}\end{vmatrix} \end{pmatrix}^1_0\end{bmatrix}$
$=\frac{1}{\sqrt{\text{a}^2+\text{b}^2}}\log\bigg(\frac{\text{a}+\text{b}+\sqrt{\text{a}^2+\text{b}^2}}{\text{a}+\text{b}-\sqrt{\text{a}^2+\text{b}^2}}\bigg)$
View full question & answer→