Question 515 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_{\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\text{dx}$
$\Rightarrow2\text{I}=\Big[\text{x}\Big]^{\frac{\pi}{3}}_{\frac{\pi}{6}}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{12}$
View full question & answer→Question 525 Marks
Evaluate the following definite integrals:
$\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}$
AnswerWe have,
$\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}$
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{\sin^2\text{x}+\cot^2\text{x}}{\sin\text{x}\cos\text{x}}\Big)^2\text{dx}$
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{1}{\sin\text{x}\cos\text{x}}\Big)^2\text{dx}$
Multiplying numberator and denominator by 2
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{2}{2\sin\text{x}\cos\text{x}}\Big)^2\text{dx}$
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{2}{\sin2\text{x}}\Big)^2\text{dx}$ $\big[\because2\sin\text{x}\cos\text{x}=\sin2\text{x}\big]$
$=4\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\text{cosec}^2\text{x dx}$
$=4\Big[-\frac{\cot2\text{x}}{2}\Big]^{\frac{\pi}{4}}_\frac{\pi}{3}$
$=2\Big[-\cot\frac{\pi}{2}+\cot2\frac{\pi}{3}\Big]$
$=2\Big[\frac{-1}{\sqrt{3}}-0\Big]$
$=\frac{-2}{\sqrt{3}}$
$\therefore\ \int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}=\frac{-2}{\sqrt{3}}$
View full question & answer→Question 535 Marks
Evaluate the following integrals:
$\int\limits^2_{-2}\text{xe}^{|\text{x}|}\text{ dx}$
Answer$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}+\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}$
For
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}$
Using integration by parts
$\int\text{f}^\text{t}\text{g}=\text{fg}-\int\text{fg}^{\text{t}}$
$\text{f}^\text{t}=\text{e}^{-\text{x}},\text{g}=\text{x}$
$\text{f}=-\text{e}^{-\text{x}},\text{g}^\text{t}=1$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{-\text{xe}^{-\text{x}}\big\}^0_{-2}+\int\limits^0_{-2}\text{e}^{-\text{x}}\text{ dx}$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{-\text{xe}^{-\text{x}}-\text{e}^{-\text{x}}\big\}^0_{-2}$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{(-1)-\big(2\text{e}^2-\text{e}^2\big)\big\}$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{-1-\text{e}^2\big\}$
For
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}$
Using integration by parts
$\int\text{f}^\text{t}\text{g}=\text{fg}-\int\text{fg}^{\text{t}}$
$\text{f}^\text{t}=\text{e}^{\text{x}},\text{g}=\text{x}$
$\text{f}=-\text{e}^{-\text{x}},\text{g}^\text{t}=1$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=\big\{\text{xe}^{\text{x}}\big\}^2_{0}-\int\limits^2_{0}\text{e}^{\text{x}}\text{ dx}$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=\big\{\text{xe}^{\text{x}}-\text{e}^{\text{x}}\big\}^2_{0}$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=2\text{e}^2-\text{e}^2+1$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=\text{e}^2+1$
Hence answer is,
$\int\limits^2_{-2}\text{xe}^{|\text{x}|}\text{ dx}=-1-\text{e}^2+\text{e}^2+1=0$
View full question & answer→Question 545 Marks
Evaluate the following integrals:
$\int\limits_{\frac{1}{3}}^{1}\frac{\big(\text{x}-\text{x}^3\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int_\limits{\frac{1}{3}}^{1}\frac{\big(\text{x}-\text{x}^3\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\Bigg[\text{x}^3\Big(\frac{\text{x}}{\text{x}^3}-1\Big)\Bigg]^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\text{x}\big(\frac{1}{\text{x}^2}-1\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\text{x}\big(\frac{1}{\text{x}^2}-1\big)^{\frac{1}{3}}}{\text{x}^3}\text{ dx}$
Put $\Big(\frac{1}{\text{x}^2}-1\Big)=\text{Z}$
$\therefore\ -\frac{2}{\text{x}^3}\text{ dx}=\text{dz}$
$\Rightarrow\frac{\text{dx}}{\text{x}^3}=-\frac{\text{dz}}{2}$
When $\text{x}\rightarrow\frac{1}{3},\text{z}\rightarrow8$
When $\text{x}\rightarrow1,\text{z}\rightarrow0$
$\therefore\ \text{I}=-\frac{1}{2}\int^\limits0_8\text{z}^{\frac{1}{3}}\text{ dz}$
$=-\frac{1}{2}\times\Bigg[\frac{\text{z}^{\frac{4}{3}}}{\frac{4}{3}}\Bigg]^0_8$
$=-\frac{3}{8}\Big[0-(8)^{\frac{4}{3}}\Big]$
$=-\frac{3}{8}\times(-16)$
$=6$
View full question & answer→Question 555 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{3}}\frac{\cos\text{x}}{3+4\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{3}}\frac{\cos\text{x}}{3+4\sin\text{x}}\text{ dx}$
Let $\sin\text{x}=\text{t}$ Then, $\cos\text{x}\text{dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{3},\text{t}=\frac{\sqrt{3}}{2}$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{3}}\frac{\cos\text{x}}{3+4\sin\text{x}}\text{ dx}$
$=\int_{0}^\limits{\frac{\sqrt{3}}{2}}\frac{1}{3+4\text{t}}\text{ dt}$
$=\frac{1}{4}\big[\log\big(3+-4\text{t}\big)\big]^{\frac{\sqrt{3}}{2}}_0$
$=\frac{1}{4}\big(\log\big(3+2\sqrt{3}\big)-\log3\big)$
$=\frac{1}{4}\log\Big(\frac{3+2\sqrt{3}}{3}\Big)$
View full question & answer→Question 565 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\cot\big(\frac{\pi}{2}-\text{x}\big)}\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\tan\text{x}}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot\text{x}}+\frac{1}{1+\tan\text{x}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1+\tan\text{x}+1+\cot\text{x}}{(1+\cot\text{x})(1+\tan\text{x})}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\tan\text{x}+\cot\text{x}}{1+\tan\text{x}+\cot\text{x}+\tan\text{x}\cot\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\tan\text{x}+\cot\text{x}}{2+\tan\text{x}+\cot\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→Question 575 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^5_{3}(2-\text{x})\text{dx}$
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, $\text{a}=3,\text{ b}=5,\text{ f(x)}=2-\text{x},\text{ h}=\frac{5-3}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^5_{3}(2-\text{x})\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(2)+\text{f}(2+\text{h})+\ ....\ +\text{f}\big\{2+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(2-2)+(2-\text{h}-2)+\\\ ....\ +\big(2-(\text{n}-1)\text{h}-2\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[-\text{h}\big(1+2+3+\ ....\ +(\text{n}-1)\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[-2\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\big[-2\text{n}+2\big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big({-2}+\frac{2}{\text{n}}\Big)$
$=-4$
View full question & answer→Question 585 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}\ ....(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\big(\frac{\pi}{2}-\text{x}}\big)}\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\cot\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\text{x}}}+\frac{1}{1+\sqrt{\cot\text{x}}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}}{\big(1+\sqrt{\tan\text{x}}\big)\big(1+\sqrt{\cot\text{x}}\big)}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}}{1+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}+\sqrt{\tan\text{x}\cot\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0$
$=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→Question 595 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^3_{2}\big(2\text{x}^2+1\big)\text{ dx}$
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, $\text{a}=2,\text{ b}=3,\text{ f(x)}=2\text{x}^2+1,\text{ h}=\frac{3-2}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_{2}\big(2\text{x}^2+1\big)\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(2)+\text{f}(2+\text{h})+\ ....\ +\text{f}\big\{2+(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2(2.2)^2+1+\big\{2(2+\text{h})^2+1\big\}+\\\ ....\ +\big\{2((2+\text{n}-1)\text{h})^2+1\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+2\Big\{2^2+(2+\text{h})^2+\ .....\big((2+\text{n}-1\big)\text{h}\big)^2\Big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+8\text{n}+2\text{h}^2\Big\{1^2+2^2+3^3+\ .....\ +(\text{n}-1)^2\Big\}\\+8\text{h}\big\{1+2+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\text{h}\bigg[9\text{n}+\text{h}^2\frac{2\text{n}(\text{n}-1)(2\text{n}-1)}{6}+8\text{h}\frac{\text{n}(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[9\text{n}+\frac{(\text{n}-1)(2\text{n}-1)}{3\text{n}}+4\text{n}-4\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg\{13+\frac{1}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{4}{\text{n}}\bigg\}$
$=13+\frac{2}{3}$
$=\frac{41}{3}$
View full question & answer→Question 605 Marks
Evaluate the following integrals:
$\int\limits_{0}^{{\frac{\pi}{4}}}(\tan\text{x}+\cot\text{x})^{-2}\text{ dx}$
Answer$\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}(\tan\text{x}+\cot\text{x})^{-2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\frac{1}{(\tan\text{x}+\cot\text{x})^{2}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\frac{1}{\Big(\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin\text{x}\cos\text{x}}\Big)^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}(\sin\text{x}\cos\text{x})^2\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\sin^2\text{x}(1-\sin^2\text{x})\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\sin^2\text{x dx}-\int_{0}^\limits{{\frac{\pi}{4}}}\sin^4\text{x dx}$
We know that by reduction formula,
$\int\sin^{\text{n}}\text{x dx}=\frac{\text{n}-1}{\text{n}}\int\sin^{\text{n}-2}\text{x dx}-\frac{\cos\text{x}\sin^{\text{n}-1}}{\text{n}}$
For n = 2
$\Rightarrow\int\sin^2\text{x dx}=\frac{2-1}{2}\int1\text{ dx}-\frac{\cos\text{x}\sin\text{x}}{2}$
$\Rightarrow\int\sin^2\text{x dx}=\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}$
For n = 4
$\Rightarrow\int\sin^4\text{x dx}=\frac{4-1}{4}\int\sin^2\text{x dx}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
$\Rightarrow\int\sin^4\text{x dx}=\frac{3}{4}\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
Hence,
$\text{I}=\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}^{\frac{\pi}{4}}_0-\Big\{\frac{3}{4}\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}\Big\}^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\Big\{\frac{\pi}{8}-\frac{1}{4}\Big\}-\Big\{\frac{3}{4}\Big(\frac{\pi}{8}-\frac{1}{4}\Big)-\frac{1}{16}\Big\}$
$\Rightarrow\text{I}=\frac{\pi}{32}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}(\sin^{\text{x}}\cos\text{x})^2\text{dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x}(1-\sin^2\text{x})\text{dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x}-\sin^4\text{x dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x dx}-\int_{0}^\limits{\frac{\pi}{2}}\sin^4\text{x dx}$
We know, by reduction formula,
$\int\sin^{\text{n}}\text{x dx}=\frac{\text{n}-1}{\text{n}}\int\sin^{\text{n}-2}\text{x dx}-\frac{\cos\text{x }\sin^{\text{n}-1}\text{x}}{\text{n}}$
For n = 2
$\Rightarrow\int\sin^2\text{x dx}=\frac{2-1}{2}\int1\text{ dx}-\frac{\cos\text{x}\sin\text{x}}{2}$
$\Rightarrow\int\sin^2\text{x dx}=\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}$
For n = 4
$\Rightarrow\int\sin^4\text{x dx}=\frac{4-1}{4}\int\sin^2\text{x dx}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
$\Rightarrow\int\sin^4\text{x dx}=\frac{3}{4}\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
Hence,
$\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}^{\frac{\pi}{2}}_0-\Big\{\frac{3}{4}\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}\Big\}^{\frac{\pi}{2}}_0$
$\Rightarrow\frac{\pi}{4}-\frac{3}{4}\Big\{\frac{\pi}{4}\Big\}$
$\Rightarrow\frac{\pi}{16}$
View full question & answer→Question 615 Marks
Evaluate the following definite integrals:
$\int\limits_{1}^{2}\frac{\text{x}}{(\text{x}+1)(\text{x}+2)}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{2}\frac{\text{x}}{(\text{x}+1)(\text{x}+2)}\text{ dx}$ Then,
$\text{I}=\int_{1}^\limits{2}\Big(\frac{-1}{(\text{x}+1)}+\frac{2}{(\text{x}+2)}\Big)\text{dx}$
$\Rightarrow\text{I}=-\int_{1}^\limits{2}\frac{1}{(\text{x}+1)}\text{ dx}+2\int_{1}^\limits{2}\frac{1}{(\text{x}+2)}\text{ dx}$
$\Rightarrow\text{I}=\big[-\log(\text{x}+1)+2\log(\text{x}+2)\big]^2_1$
$\Rightarrow\text{I}=-\log3+2\log4+\log2-2\log3$
$\Rightarrow\text{I}=5\log2-3\log3$
$\Rightarrow\text{I}=\log2^5-\log3^3$
$\Rightarrow\text{I}=\log\frac{32}{27}$
View full question & answer→Question 625 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^2\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^2\text{x dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\pi}_0(\pi-\text{x})\sin(\pi-\text{x})\cos^2(\pi-\text{x})\text{dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin\text{x}\cos^2\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\pi-\text{x}+\text{x})\sin\text{x}\cos^2\text{x dx}$
$\Rightarrow2\text{I}=\pi\int\limits^{\pi}_0\sin\text{x}\cos^2\text{x dx}$
$\Rightarrow2\text{I}=-\pi\int\limits^{\pi}_0\cos^2\text{x}(-\sin\text{x})\text{dx}$
$\Rightarrow2\text{I}=-\pi\Big[\frac{\cos^3\text{x}}{3}\Big]^{\pi}_0$ $\Bigg[\int\big[\text{f(x)}\big]^{\text{n}}\text{f}'(\text{x})\text{dx}=\frac{\big[\text{f(x)}\big]^{\text{n}+1}}{\text{n}+1}+\text{C}\Bigg]$
$\Rightarrow2\text{I}=-\frac{\pi}{3}\big(\cos^3\text{x}-\cos^20\big)$
$\Rightarrow2\text{I}=-\frac{\pi}{3}(-1-1)=\frac{2\pi}{3}$
$\Rightarrow\text{I}=\frac{\pi}{3}$
View full question & answer→Question 635 Marks
Evaluate the following integrals:
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}$
AnswerWe know,
$\int\limits^{2\pi}_0\text{f(x)}\text{dx}=\int\limits^{2\pi}_0\text{f}(2\pi-\text{x})\text{dx}$
Hence,
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}=\int\limits^{2\pi}_0\log\big(\sec(2\pi-\text{x})+\tan(2\pi-\text{x})\big)\text{dx}$
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}=\int\limits^{2\pi}_0\log(\sec\text{x}-\tan\text{x})\text{dx}$
If
$\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}+\int\limits^{2\pi}_0\log(\sec\text{x}-\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}+\log(\sec\text{x}-\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec^2\text{x}-\tan^2\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(1)\text{dx}$
$2\text{I}=0$
$\text{I}=0$
View full question & answer→Question 645 Marks
Evaluate the following integrals:
$\int\limits^\pi_0\sin^{100}\text{x}\cos^{101}\text{x dx}$
AnswerLet $\text{I}=\int\limits^\pi_0\sin^{100}\text{x}\cos^{101}\text{x dx}$
Consider $\text{f(x)}=\sin^{100}\text{x}\cos^{101}\text{x}$
Now,
$\text{f}(2\pi-\text{x})=\sin^{100}(2\pi-\text{x})\cos^{101}(2\pi-\text{x})$
$=(-\sin\text{x})^{100}(\cos\text{x})^{101}=\sin^{100}\text{x}\cos^{101}\text{x}=\text{f}(\text{x)}$
$\therefore\ \text{I}=\int\limits^{2\pi}_0\sin^{100}\text{x}\cos^{101}\text{x dx}=2\int\limits^{\pi}_0\sin^{100}\text{ x}\cos^{101}\text{x dx}$$\begin{bmatrix}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx},&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\\0,&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\end{cases}\end{bmatrix}$
Again,
$\text{f}(\pi-\text{x})=\sin^{100}(\pi-\text{x})\cos^{101}(\pi-\text{x})$
$=(\sin\text{x})^{100}(-\cos\text{x})^{101}=-\sin^{100}\text{x}\cos^{101}\text{x}=-\text{f(x)}$
$\therefore\ \text{I}=2\times0=0$ $\begin{bmatrix}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx},&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\\0,&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\end{cases}\end{bmatrix}$
View full question & answer→Question 655 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\text{ dx}$
AnswerLet $\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\times\frac{\sqrt{1-\sin\text{x}}}{\sqrt{1-\sin\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sqrt{1-\sin^2\text{x}}}{\sqrt{1-\sin^2\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\cos\text{x}}{\sqrt{1-\sin\text{x}}}\text{ dx}$
Let $1-\sin\text{x}=\text{u}$
$\Rightarrow-\cos\text{x dx}=\text{du}$
$\therefore\ \text{I}=\int\frac{-\text{du}}{\sqrt{\text{u}}}$
$\Rightarrow\text{I}=\big[-2\sqrt{\text{u}}\big]$
$\Rightarrow\text{I}=\big[-2\sqrt{1-\sin\text{x}}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0+2$
$\Rightarrow\text{I}=2$
View full question & answer→Question 665 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^5_{0}(\text{x}+1)\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}=5,\text{ f(x)}=\text{x}+1,\text{ h}=\frac{5-1}{\text{n}}=\frac{5}{\text{n}}$
Therefore, $\text{I}=\int\limits^5_{0}(\text{x}+1)\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]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+1)+(\text{h}+1)+\ ....\ +\{(\text{n}-1)\text{h}+1\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{5}{\text{n}}\Big[\text{n}+\frac{5\text{n}-5}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}5\Big(\frac{7}{2}-\frac{5}{\text{n}}\Big)$
$=\frac{35}{2}$
View full question & answer→Question 675 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^{5}_{0}(\text{x}+1)\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}=5,\text{ f(x)}=\text{x}+1,\text{ h}=\frac{5-0}{\text{n}}=\frac{5}{\text{n}}$
Therefore, $\text{I}=\int\limits^{5}_{0}(\text{x}+1)\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}+1)+\ ....+\ \big\{(\text{n}-1)\text{h}+1\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{n}+\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{5}{\text{n}}\Big[\text{n}+\frac{5(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}5\Big[1+\frac{5}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big]$
$=5+\frac{25}{2}$
$=\frac{35}{2}$
View full question & answer→Question 685 Marks
Evaluate the following integrals:
$\int\limits_{0}^{1}\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{ dx}$ Then,
Let $\tan^{-1}\text{x}=\text{t}$ Then, $\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=1,\text{t}=\frac{\pi}{4}$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{4}}\text{t}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{\text{t}^2}{2}\Big]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\frac{\pi^2}{32}$
View full question & answer→Question 695 Marks
Evaluate the following integrals:
$\int\limits^{7}_0\frac{\sqrt[3]{\text{x}}}{\sqrt[3]{\text{x}}+\sqrt[3]{7}-\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{7}_0\frac{\sqrt[3]{\text{x}}}{\sqrt[3]{\text{x}}+\sqrt[3]{7-\text{x}}}\text{ dx}\ ...(\text{i})$
We know that $\int\limits^{\text{a}}_0\text{f(x)}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})$
Hence,
$\text{I}=\int\limits^{7}_0\frac{\sqrt[3]{7-\text{x}}}{\sqrt[3]{7-\text{x}}+\sqrt[3]{\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) & (ii)
$2\text{I}=\int\limits^{7}_0\frac{\sqrt[3]{\text{x}}}{\sqrt[3]{\text{x}}+\sqrt[3]{7-\text{x}}}\text{ dx}+\frac{\sqrt[3]{7-\text{x}}}{\sqrt[3]{7-\text{x}}+\sqrt[3]{\text{x}}}\text{ dx}$
$2\text{I}=\int\limits^{7}_0\frac{\sqrt[3]{\text{x}}+\sqrt[3]{7-\text{x}}}{\sqrt[3]{\text{x}}+\sqrt[3]{7-\text{x}}}\text{ dx}$
$2\text{I}=\int\limits^{7}_0\text{dx}$
$2\text{I}=\big[\text{x}\big]^7_0$
$\text{I}=\frac{7}{2}$
View full question & answer→Question 705 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2(1-\cos^2\text{x})\big)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(\text{b}^2+(\text{a}^2-\text{b}^2)\cos^2\text{x}\big)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\Bigg(\text{b}^2+\frac{\big(\text{a}^2-\text{b}^2\big)\big(1+\cos2\text{x}\big)}{2}\Bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\bigg[\text{b}^2\text{x}+\frac{\text{a}^2-\text{b}^2}{2}\Big(\text{x}+\frac{\sin2\text{x}}{2}\Big)\bigg]_0^{\frac{\pi}{2}}$
$\Rightarrow\text{I}=\frac{\text{b}^2\pi}{2}+\frac{\text{a}^2-\text{b}^2}{2}\frac{\pi}{2}+0$
$\Rightarrow\text{I}=\frac{\pi}{4}\big(\text{a}^2+\text{b}^2\big)$
View full question & answer→Question 715 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^{3}_{1}\big(3\text{x}^2+1\text{x}\big)\text{dx}$
AnswerWe have,
$\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx} = \lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}+\text{h})+(\text{a}+2\text{h})+\\...+\text{f}[(\text{a}+(\text{n}-1)\text{h})]\}$
Here, $\text{a}=1, \text{b}=3\text{ f}(\text{x})=3\text{x}^2+1$ and $\text{h}=\frac{3-1}{\text{n}}=\frac{2}{\text{n}}\Rightarrow \text{nh}=2$
$\therefore\int\limits^3_1(3\text{x}^2+1)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\{\text{f}(1)+\text{f}(1+\text{h})+\text{f}(1+2\text{h})...+\text{f}[1+(\text{n}-1)\text{h}]\}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\{[3\times1^2+1]+[3\times(1+\text{h})^2+1]\\+[3\times(1+2\text{h})^2+1]+.....+[3\times(1+(\text{n}-1)\text{h}^2+1]$
$=\lim\limits\text{h}\{3[1+(1+2\text{h}+\text{h}^2)+(1+4\text{h}+2^2\text{h}^2)+\\...+(1+2(\text{n}-1)\text{h}+(\text{n}-1)^2\text{h}^2)]+\text{n}\}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\{3[\text{n}+2(1+2+...+(\text{n}-1))\text{h}+(1^2+2^2+\\.....+(\text{n}-1)^2)\text{h}^2]+\text{n}\}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[4\text{n}+6\times\frac{\text{n}(\text{n}-1)}{2}\text{h}+3\times\frac{(\text{n}-1)\text{n}(2\text{n}-1)}{6}\text{h}^2\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[4\text{nh}+6\times\frac{\text{nh}(\text{nh}-\text{h)}}{2}+3\times\frac{(\text{nh}-\text{h})\text{nh}(2\text{nh}-\text{h}}{6}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[4\text{nh}+3\times\text{nh}(\text{nh}-\text{h}+3\times\frac{(\text{nh}-\text{h})\text{nh}(2\text{nh}-\text{h})}{6}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[4\times2+3\times2\times(2-\text{h})+3\times\frac{(2-\text{h)}\times2\times(2\times2-\text{h})}{6}\bigg]$
$= 8 + 6\times(2-0)+\frac{(2-0)\times2\times(4-0)}{2}$
$= 8+12+8$
$= 28$
View full question & answer→Question 725 Marks
Evaluate the following integrals:$\int\limits^3_{0}\big|3\text{x}-1\big|\text{dx}$
Answer$\int^\limits3_{0}\big|3\text{x}-1\big|\text{dx}=\int^\limits{\frac{1}{3}}_0-(3\text{x}-1)\text{dx}+\int^\limits{3}_\frac{1}{3}(3\text{x}-1)\text{dx}$
$=-\Big[\frac{3\text{x}^2}{2}-\text{x}\Big]^{\frac{1}{3}}_0+\Big[\frac{3\text{x}^2}{2}-\text{x}\Big]^3_{\frac{1}{3}}$
$=-\bigg[\Big(\frac{3}{9\times2}-\frac{1}{3}\Big)-(0)\Big]+\bigg[\Big(\frac{3\times9}{2}-3\Big)-\Big(\frac{3}{9\times2}-\frac{1}{3}\Big)\bigg]$
$=-\bigg[\Big(\frac{1}{6}-\frac{1}{3}\Big)\bigg]+\bigg[\Big(\frac{27}{2}-3\Big)-\Big(\frac{1}{6}-\frac{1}{3}\Big)\bigg]$
$=-\bigg[\Big(-\frac{1}{6}\Big)\bigg]+\bigg[10\frac{1}{2}+\frac{1}{6}\bigg]$
$=\frac{1}{6}+10\frac{1}{2}+\frac{1}{6}$
$=\frac{1}{3}+\frac{21}{2}$
$=\frac{2+63}{6}$
$=\frac{65}{6}$
View full question & answer→Question 735 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^3_1(3\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}=1,\text{ b}=3,\text{ f(x)}=3\text{x}-2,\text{ h}=\frac{3-1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_1(3\text{x}-2)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big(1+(\text{n}-1)\text{h}\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(3-2)+(3+3\text{h}-2)+(3+6\text{h}-2)\ +\\ ....\ +\big(3+(\text{n}-1)\text{h}+3-2\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+3\text{h}\big(1+2+3+\ ....\ + (\text{n}-1)\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+3\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[\text{n}+3\text{n}-3\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big(4-\frac{3}{\text{n}}\Big)$
$=8$
View full question & answer→Question 745 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_0\text{x}^2\sin\text{x dx}$
AnswerWe have, Using by parts, we get $\text{x}^2\int\sin\text{x dx}-\int\big(\sin\text{x dx}\big)\frac{\text{dx}^2}{\text{dx}}\text{ dx}$ $=\text{x}^2\cos\text{x}+\int\cos\text{x }2\text{x dx}$ Again applying by parts $=\text{x}^2\cos\text{x}+2\Big[\text{x}\int\cos\text{x dx}-\int\big(\int\cos\text{x dx}\big)\frac{\text{dx}}{\text{dx}}\text{ dx}\Big]$ $=\text{x}^2\cos\text{x}+2\big[\text{x}\sin\text{x}-\int\sin\text{x dx}\big]$ $=\Big[\text{x}^2\cos\text{x}+2\text{x }\sin\text{x}+2\cos\text{x}\Big]^{\frac{\pi}{2}}_0$ $=\pi+0-0-0-2$ $=\pi-2$
View full question & answer→Question 755 Marks
If $\int\limits_{0}^{\text{k}}\frac{1}{2+8\text{x}^2}\text{ dx}=\frac{\pi}{16},$ find the value of k.
AnswerWe have,
$\int_{0}^\limits{\text{k}}\frac{1}{2+8\text{x}^2}\text{ dx}=\frac{\pi}{16}$
$\Rightarrow\frac{1}{8}\int_{0}^\limits{\text{k}}\frac{\text{dx}}{\big(\frac{1}{2}\big)^2+\text{x}^2}=\frac{\pi}{16}$
$\Rightarrow\frac{1}{8}\big[2\tan^{-1}2\text{x}\big]^{\text{k}}_0=\frac{\pi}{16}$ $\Big[\because\int\frac{\text{dx}}{\text{a}^2-\text{x}^2}=2\tan^{-1}\frac{\text{x}}{\text{a}}\Big]$
$\Rightarrow\frac{1}{4}\big[\tan^{-1}2\text{k}-\tan^{-1}0\big]=\frac{\pi}{16}$
$\Rightarrow\tan^{-1}2\text{k}-0=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}2\text{k}=\frac{\pi}{4}$
$\Rightarrow\text{k}=\frac{1}{2}$
View full question & answer→Question 765 Marks
Evaluate the following integrals:
$\int\limits^{\text{a}}_0\text{x}\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\text{a}}_0\text{x}\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\text{ dx}$ Consider, $\text{x}^2=\text{a}^2\cos2\theta$ $\Rightarrow2\text{xdx}=-2\text{a}^2\sin2\theta\text{ d}\theta$ $\Rightarrow\text{xdx}=-\text{a}^2\sin2\theta\text{ d}\theta$ When, $\text{x}\rightarrow0;\ \theta\rightarrow\frac{\pi}{4}$ and $\text{x}\rightarrow\text{a};\ \theta\rightarrow0$ Now, integral becomes, $\text{I}=\int^\limits0_\frac{\pi}{4}-\text{a}^2\sin2\theta\sqrt{\frac{\text{a}^2-\text{a}^2\cos2\theta}{\text{a}^2+\text{a}^2\cos2\theta}}\text{ d}\theta$ $=\int^\limits0_\frac{\pi}{4}-\text{a}^2\sin2\theta\tan\theta\text{ d}\theta$ $=\text{a}^2\int^\limits{\frac{\pi}{4}}_02\sin\theta\cos\theta\frac{\sin\theta}{\cos\theta}\text{ d}\theta$ $=\text{a}^2\int^\limits{\frac{\pi}{4}}_02\sin\theta\text{ d}\theta$ $=\text{a}^2\int^\limits{\frac{\pi}{4}}_0\big[1-\cos\theta\big]\text{d}\theta$ $=\text{a}^2\Big[\theta-\frac{\sin2\theta}{2}\Big]^{\frac{\pi}{4}}_0$ $=\text{a}^2\Big[\frac{\pi}{4}-\frac{1}{2}\Big]$
View full question & answer→Question 775 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^2_{0}\big(\text{x}^2+4\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+4,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{0}\big(\text{x}^2+4\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+4)+(\text{h}^2+4)+\ ....\ +\big\{(\text{n}-1)^2\text{h}^2+4\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[4\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}\bigg[4\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\bigg[4\text{n}+\frac{2(\text{n}-1)(2\text{n}-1)}{3\text{n}}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{4+\frac{2}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=8+\frac{8}{3}$
$=\frac{32}{3}$
View full question & answer→Question 785 Marks
Evaluate the following integrals:
$\int\limits_{0}^{{2\pi}}\sqrt{1-\sin\frac{\text{x}}{2}}\text{ dx}$
Answer$\text{I}=\int_{0}^\limits{{2\pi}}\sqrt{1-\sin\frac{\text{x}}{2}}\text{ dx}$
$=\int_{0}^\limits{{2\pi}}\sqrt{\cos^2\frac{\text{x}}{4}+\sin^2\frac{\text{x}}{4}+2\sin\frac{\text{x}}{4}\cos\frac{\text{x}}{4}}\text{ dx}$
$=\int_{0}^\limits{{2\pi}}\sqrt{\Big(\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}\Big)^2}\text{ dx}$
$=\int_{0}^\limits{{2\pi}}\Big[\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}\Big]\text{dx}$
When $0\leq\text{x}\leq2\pi,0\leq\frac{\text{x}}{4}\leq\frac{\pi}{2}$
$\therefore\ \sin\frac{\text{x}}{4}\geq0,\cos\frac{\text{x}}{4}\geq0$
$\Rightarrow\Big[\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}\Big]=\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}$
$\therefore\ \text{I}=\int_{0}^\limits{{2\pi}}\Big(\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}\Big)\text{dx}$
$=\Bigg[\frac{\sin\frac{\text{x}}{4}}{\frac{1}{4}}\Bigg]^{2\pi}_0+\Bigg[\frac{\big(-\cos\frac{\text{x}}{4}\big)}{\frac{1}{4}}\Bigg]^{2\pi}_0$
$=4\Big(\sin\frac{\pi}{2}-\sin0\Big)-4\Big(\cos\frac{\pi}{2}-\cos0\Big)$
$=4(1-0)-4(0-1)$
$=4+4$
$=8$
View full question & answer→Question 795 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\pi}\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
Answer$\int_{0}^\limits{\pi}\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$=\frac{1}{2}\int_{0}^\limits{\pi}\frac{2\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$=\frac{1}{2}\int_{0}^\limits{\pi}\frac{(\sin\text{x}+\cos\text{x})-(\cos\text{x}-\sin\text{x})}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$=\frac{1}{2}\int_{0}^\limits{\pi}\text{dx}-\frac{1}{2}\int_{0}^\limits{\pi}\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$=\frac{1}{2}\big[\text{x}\big]^{\pi}_0-\frac{1}{2}\big[\log|\sin\text{x}+\cos\text{x}|\big]^{\pi}_0$
$=\frac{1}{2}\big[\pi-0\big]-\frac{1}{2}\big[\log1-\log1\big]$
$=\frac{\pi}{2}$
View full question & answer→Question 805 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\tan\text{x}}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\tan\text{x}}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\tan\big(\frac{\pi}{2}-\text{x}\big)}\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\cot\text{x}}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big(\frac{1}{1+\tan\text{x}}+\frac{1}{1+\cot\text{x}}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\Big[\frac{1+\cot\text{x}+1+\tan\text{x}}{(1+\tan\text{x})(1+\cot\text{x})}\Big]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\tan\text{x}+\cot\text{x}}{1+\tan\text{x}+\cot\text{x}+\tan\text{x}\cot\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\tan\text{x}+\cot\text{x}}{2+\tan\text{x}+\cot\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0$
$2\text{I}=\frac{\pi}{2}$
$\therefore\ \text{I}=\frac{\pi}{4}$
View full question & answer→Question 815 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\big(\frac{\pi}{2}-\text{x}\big)}{\sin^\text{n}\big(\frac{\pi}{2}-\text{x}\big)+\cos^\text{n}\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\cos^\text{n}\text{x}+\sin^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}+\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
i.e., $\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}=\frac{\pi}{4}$
View full question & answer→Question 825 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^5\phi\text{ d}\phi$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^5\phi\text{ d}\phi$
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^4\phi\cos\phi\text{ d}\phi$
Also, let $\sin\phi=\text{t}\Rightarrow\cos\phi\text{ d}\phi=\text{dt}$
When, $\phi=0,\text{t}=0$ and when $\phi=\frac{\pi}{2},\text{t}=1$
$\therefore\ \text{I}=\int_{0}^\limits{1}\sqrt{\text{t}}\big(1-\text{t}^2\big)^2\text{dt}$
$=\int_{0}^\limits{1}\text{t}^{\frac{1}{2}}\big(1+\text{t}^4-2\text{t}^2\big)\text{dt}$
$=\int_{0}^\limits{1}\Big[\text{t}^{\frac{1}{2}}+\text{t}^{\frac{9}{2}}-2\text{t}^{\frac{5}{2}}\Big]\text{dt}$
$=\Bigg[\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{\text{t}^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2\text{t}^{\frac{7}{2}}}{\frac{7}{2}}\Bigg]^1_0$
$=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$
$=\frac{64}{231}$
View full question & answer→Question 835 Marks
Evaluate the following integrals:
$\int\limits^{\text{a}}_{-\text{a}}\frac{1}{1+\text{a}^{\text{x}}}\text{ dx},\text{ a}>0$
AnswerLet $\text{I}=\int\limits^{\text{a}}_{-\text{a}}\frac{1}{1+\text{a}^{\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\text{a}}_{-\text{a}}\frac{1}{1+\text{a}^{[\text{a}+(-\text{a})-\text{x}]}}\text{ dx}$
$=\int\limits^{\text{a}}_{-\text{a}}\frac{1}{1+\text{a}^{-\text{x}}}\text{ dx}$
$\text{I}=\int\limits^{\text{a}}_{-\text{a}}\frac{1}{\text{a}^{\text{x}}+1}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\text{a}}_{-\text{a}}\frac{1+\text{a}^{\text{x}}}{1+\text{a}^{\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{\text{a}}_{-\text{a}}$
$\Rightarrow2\text{I}=\text{a}-(-\text{a})$
$\Rightarrow2\text{I}=2\text{a}$
$\Rightarrow\text{I}=\text{a}$
View full question & answer→Question 845 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^{\text{b}}_{\text{a}}\text{x}\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}=\text{a},\text{ b}=\text{b},\text{ f(x)}=\text{x},\text{ h}=\frac{\text{b}-\text{a}}{\text{n}}$
Therefore, $\text{I}=\int\limits^{\text{b}}_{\text{a}}\text{x}\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\ ....\ +\text{f}\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{a}+(\text{a}+\text{h})+(\text{a}+2\text{h})+\ ....+\ \big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{na}+\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{na}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{b}-\text{a}}{\text{n}}\Big[\text{na}\frac{[\text{b}-\text{a}](\text{n}-1)}{2}\Big]$
$=(\text{b}-\text{a})\text{a}+\frac{(\text{b}-\text{a})^2}{2}$
$=\frac{2\text{ab}-2\text{a}^2+\text{b}^2+\text{a}^2-2\text{ab}}{2}$
$=\frac{\text{b}^2-\text{a}^2}{2}$
View full question & answer→Question 855 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^4_{1}\big(\text{x}^2-\text{x}\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}=1,\text{ b}=4,\text{ f(x)}=\text{x}^2-\text{x},\text{ h}=\frac{4-1}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^4_{1}\big(\text{x}^2-\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(1-1)+\big\{(1+\text{h}^2)-(1+\text{h})\big\}+\\\ ....\ +\big\{(1+(\text{n}-1)\text{h})^2-(1+(\text{n}-1)\text{h})\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{h}^2\big\{1^2+2^2+3^2\ ....\ +\$\text{n}-1)^2\big\}+1+2\text{h}\big\{1+2+\ ....+\$\text{n}-1)\big\}-\text{n}-\text{h}\big\{1+2+\ .....\ +(\text{n}-1)\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+\text{h}\frac{(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{3}{\text{n}}\bigg[\frac{9(\text{n}-1)(2\text{n}-1)}{6\text{n}}+\frac{3(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}3\bigg[\frac{3}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)+\frac{3}{2}\Big(1-\frac{1}{\text{n}}\Big)\bigg]$
$=9+\frac{9}{2}$
$=\frac{27}{3}$
View full question & answer→Question 865 Marks
Evaluate the following integrals:
$\int\limits_{0}^{2}\text{x}\sqrt{\text{x}+2}\text{ dx}$
Answer$\int_{0}^\limits{2}\text{x}\sqrt{\text{x}+2}\text{ dx}$
Let $\text{x}+2=\text{t}^2\Rightarrow\text{dx}=2\text{tdt}$
When $\text{x}=0,\text{t}=\sqrt{2}$ and when $\text{x}=2,\text{t}=2$
$\therefore\ \int_{0}^\limits{2}\text{x}\sqrt{\text{x}+2}\text{ dx}=\int_{\sqrt{2}}^\limits{2}\big(\text{t}^2-2\big)\sqrt{\text{t}^2}2\text{t dt}$
$=2\int_{\sqrt{2}}^\limits{2}\big(\text{t}^2-2\big)\text{t}^2\text{ dt}$
$=2\int_{\sqrt{2}}^\limits{2}\big(\text{t}^4-2\text{t}^2\big)\text{dt}$
$=2\Big[\frac{\text{t}^5}{5}-\frac{2\text{t}^3}{3}\Big]^2_\sqrt{2}$
$=2\Big[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}\Big]$
$=2\Big[\frac{96-80-12\sqrt{2}+20\sqrt{2}}{15}\Big]$
$=2\Big[\frac{16+8\sqrt{2}}{15}\Big]$
$=\frac{16\big(2+\sqrt{2}\big)}{15}$
$=\frac{16\sqrt{2}\big(\sqrt{2}+1\big)}{15}$
View full question & answer→Question 875 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_02\sin\text{x }\cos\text{x}\tan^{-1}(\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_02\sin\text{x }\cos\text{x}\tan^{-1}(\sin\text{x})\text{dx}$ Then,
Let $\sin\text{x}=\text{t}$ Then, $\cos\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2},\text{t}=1$
$\Rightarrow\text{I}=2\Big[\frac{\text{t}^2\tan^{-1}\text{t}}{2}\Big]^1_0-\int^\limits1_0\frac{\text{t}^2}{1+\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=2\Big[\frac{\text{t}^2\tan^{-1}\text{t}}{2}\Big]^1_0-\int^\limits1_0\Big(\frac{1+\text{t}^3}{1+\text{t}^2}-\frac{1}{1+\text{t}^2}\Big)\text{dt}$
$\Rightarrow\text{I}=2\Big[\frac{\text{t}^2\tan^{-1}\text{t}}{2}\Big]^1_0-\big[\text{t}-\tan^{-1}\text{t}\big]^1_0$
$\Rightarrow\text{I}=1\tan^{-1}1-0-1+\tan^{-1}1+0$
$\Rightarrow\text{I}=\frac{\pi}{4}-1+\frac{\pi}{4}$
$\Rightarrow\text{I}=\frac{\pi}{2}-1$
View full question & answer→Question 885 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{4}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$
Answer$\int_{0}^\limits{\frac{\pi}{4}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\Big[\text{a}^2\Big(\frac{1+\cos2\text{x}}{2}\Big)+\text{b}^2\Big(\frac{1-\cos2\text{x}}{2}\Big)\Big]\text{dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\Big[\Big(\frac{\text{a}^2+\text{b}^2}{2}\Big)+\Big(\frac{\text{a}^2-\text{b}^2}{2}\Big)\cos2\text{x}\Big]\text{dx}$
$=\Big(\frac{\text{a}^2+\text{b}^2}{2}\Big)\int_{0}^\limits{\frac{\pi}{4}}\text{dx}+\Big(\frac{\text{a}^2-\text{b}^2}{2}\Big)\int_{0}^\limits{\frac{\pi}{4}}\cos2\text{x }\text{dx}$
$=\Big(\frac{\text{a}^2+\text{b}^2}{2}\Big)\Big(\frac{\pi}{4}-0\Big)+\Big(\frac{\text{a}^2-\text{b}^2}{2}\Big)\Big(\sin\frac{\pi}{2}-\sin0\Big)$
$=\Big(\frac{\text{a}^2+\text{b}^2}{2}\Big)\frac{\pi}{4}+\Big(\frac{\text{a}^2-\text{b}^2}{2}\Big)(1-0)$
$=\big(\text{a}^2+\text{b}^2\big)\frac{\pi}{8}+\frac{1}{4}\big(\text{a}^2-\text{b}^2\big)$
View full question & answer→Question 895 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\pi}\frac{1}{3+2\sin\text{x}+\cos\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\pi}\frac{1}{3+2\sin\text{x}+\cos\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\pi}\frac{1}{3+2\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\frac{1+\tan^{2}\frac{\text{x}}{2}}{2\tan^{2}\frac{\text{x}}{2}+4\tan\frac{\text{x}}{2}+4}\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$ and $\text{x}=\pi,\text{t}=\infty$
$\therefore\ \text{I}=\int\limits^{\infty}_0\frac{2\text{dt}}{2\text{t}^2+4\text{t}+4}$
$\Rightarrow\text{I}=\int\limits^{\infty}_0\frac{\text{dt}}{(\text{t}+1)^2+1}$
$\Rightarrow\text{I}=\Big[\tan^{-1}\big(\text{t}+1\big)\Big]^{\infty}_0$
$\Rightarrow\text{I}=\frac{\pi}{2}-\frac{\pi}{4}$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 905 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}\text{x}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\big(\frac{\pi}{2}-\text{x}\big)}{\sin^\text{n}\big(\frac{\pi}{2}-\text{x}\big)+\cos^\text{n}\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\cos^\text{n}\text{x}+\sin^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}+\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
i.e., $\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}=\frac{\pi}{4}$
$\therefore\ \int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}\text{x}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}\text{ dx}=\frac{\pi}{4}$
View full question & answer→Question 915 Marks
If $\text{f}(\text{a}+\text{b}-\text{x})=\text{f(x)},$ then prove that $\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=\frac{\text{a}+\text{b}}{2}\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}$
$=\int\limits^{\text{b}}_\text{a}(\text{a}+\text{b}-\text{x})\text{f}(\text{a}+\text{b}-\text{x})\text{dx}$$\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\text{b}}_\text{a}(\text{a}+\text{b}-\text{x})\text{f(x)}\text{dx}$$\Big[\text{f}(\text{a}+\text{b}-\text{x})=\text{f(x)}\Big]$
$\therefore\ \int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}(\text{a}+\text{b})\text{f(x)}\text{dx}-\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}$
$\Rightarrow\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}+\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=(\text{a}+\text{b})\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}$
$\Rightarrow2\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=(\text{a}+\text{b})\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}$
$\Rightarrow\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=\frac{\text{a}+\text{b}}{2}\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}$
View full question & answer→Question 925 Marks
If f(x) is a continuous function defind on [-a, a], then prove that:
$\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\big\{\text{f(x)}+\text{f}(-\text{x})\big\}\text{dx}$
AnswerLet $\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}$
By Additive property
$\text{I}=\int\limits^0_{-\text{a}}\text{f(x)}\text{dx}+\int\limits^{\text{a}}_0\text{f(x)}\text{dx}$
Let $\text{x}=-\text{t},$ then $\text{dx}=-\text{dt}$
When $\text{x}=-\text{a},\text{ t}=\text{a},\text{ x}=0,\text{ t}=0$
Hence, $\int\limits^0_{-\text{a}}\text{f(x)}\text{dx}=-\int\limits^0_{\text{a}}\text{f}(-\text{t})\text{dt}$
$=\int\limits_0^{\text{a}}\text{f}(-\text{t})\text{dt}=\int\limits_0^{\text{a}}\text{f}(-\text{x})\text{dx}$ (Changing the varible)
Therefore,
$\text{I}=\int\limits_0^{\text{a}}\text{f}(-\text{x})\text{dx}+\int\limits_0^{\text{a}}\text{f}(\text{x})\text{dx}$
$=\int\limits^{\text{a}}_0\big\{\text{f(x)}+\text{f}(-\text{x})\big\}\text{dx}$
Hence, proved.
View full question & answer→Question 935 Marks
Evaluate the following integrals:
$\int\limits^{2\pi}_0\frac{\text{e}^{\sin\text{x}}}{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{\sin\text{x}}}{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{\sin(2\pi-\text{x})}}{\text{e}^{\sin(2\pi-\text{x})}+\text{e}^{-\sin(2\pi-\text{x})}}\text{ dx}$ $\Bigg(\int\limits^\text{a}_0\text{f(x)}\text{dx}=\int\limits^\text{a}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg)$
$\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{-\sin\text{x}}}{\text{e}^{-\sin\text{x}}+\text{e}^{\sin\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{2\pi}_0\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{2\pi}_0$
$\Rightarrow2\text{I}=2\pi-0$
$\Rightarrow\text{I}=\pi$
View full question & answer→Question 945 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^\text{b}_{\text{a}}\text{e}^{\text{x}}\text{ dx}$
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, $\text{a}=\text{a},\text{ b}=\text{b},\text{ f(x)}=\text{e}^{\text{x}},\text{ h}=\frac{\text{b}-\text{a}}{\text{n}}$
Therefore, $\text{I}=\int\limits^\text{b}_{\text{a}}\text{e}^{\text{x}}\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\ ....\ +\text{f}\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{e}^\text{a}+\text{e}^{\text{a}+\text{h}}+\ .....\ +\text{e}^{\{\text{a}+(\text{n}-1)\text{h}\}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{e}^{\text{a}}\bigg\{\frac{(\text{e}^{\text{h}})^{\text{n}}-1}{\text{e}^{\text{h}}-1}\bigg\}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Big[\text{e}^\text{a}\frac{\text{e}^{\text{b}-\text{a}}-1}{\text{e}^{\text{h}}-1}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}=\Bigg[\frac{\text{e}^\text{b}-\text{e}^{\text{a}}}{\frac{\text{e}^\text{h}-1}{\text{h}}}\Bigg]$
$=\frac{\text{e}^\text{b}-\text{e}^\text{a}}{1}$
$=\text{e}^{\text{b}}-\text{e}^\text{a}$
View full question & answer→Question 955 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\cos^3\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\cos^3\text{x}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\cos^3\text{x}\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(1-\sin^2\text{x}\big)\cos\text{x}\text{ dx}$
Let $\text{u}=\sin\text{x},\text{ du}=\cos\text{x dx}$
$\Rightarrow\text{I}=\int(1-\text{u}^2)\text{du}$
$\Rightarrow\text{I}=\Big[\text{u}-\frac{\text{u}^3}{3}\Big]$
$\Rightarrow\text{I}=\Big[\sin\text{x}-\frac{\sin^3\text{x}}{3}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=1-\frac{1}{3}-0$
$\Rightarrow\text{I}=\frac{2}{3}$
View full question & answer→Question 965 Marks
Evaluate the following integrals:
$\int\limits_{0}^{1}\text{x}\log(1+2\text{x})\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\text{x}\log(1+2\text{x})\text{dx}$Apply integral by part
$\text{I}=\Big[\log(1+2\text{x})\frac{\text{x}^2}{2}\Big]^{1}_0-\int_{0}^\limits{1}\Big(\frac{2}{1+2\text{x}}\Big)\times\frac{\text{x}^2}{2}\text{ dx}$
$=\frac{1}{2}\big(\log3-0\big)-\int_{0}^\limits{1}\frac{\text{x}^2}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}\frac{4\text{x}^2-1+1}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}\frac{(2\text{x}+1)(2\text{x}-1)}{1+2\text{x}}\text{ dx}-\frac{1}{4}\int_{0}^\limits{1}\frac{2}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}(2\text{x}-1)\text{dx}-\frac{1}{4}\int_{0}^\limits{1}\frac{1}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\bigg[\frac{1}{4}\times\frac{(2\text{x}-1)^2}{2\times2}\bigg]^1_0-\bigg[\frac{1}{4}\times\frac{\log(1+2\text{x})}{2}\bigg]^1_0$
$=\frac{1}{2}\log3-\frac{1}{16}(1-1)-\frac{1}{8}\big(\log3-\log1\big)$
$=\frac{1}{2}\log3-0-\frac{1}{8}\log3$ $(\log1=0)$
$=\frac{3}{8}\log3$
View full question & answer→Question 975 Marks
Prove that:
$\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\frac{\pi}{2}\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
Answer$\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\int\limits^\pi_0(\pi-\text{x})\text{f}\big[\sin(\pi-\text{x})\big]\text{dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\int\limits^\pi_0(\pi-\text{x})\text{f}(\sin\text{x})\text{dx}$
$\Rightarrow\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\pi\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}-\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}$
$\Rightarrow2\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\pi\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
$\Rightarrow\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\frac{\pi}{2}\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
View full question & answer→Question 985 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^\limits{\frac{\pi}{2}}\text{x}^2\sin\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\sin\text{x}\text{ dx}$
Apply integration by part.
$\text{I}=\big[\text{x}^2(-\cos\text{x})\big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}2\text{x}(-\cos\text{x})\text{ dx}$
$\Rightarrow\text{I}=(0-0)+2\int_{0}^\limits{\frac{\pi}{2}}\text{x}\cos\text{x}\text{ dx}$ $\Big(\cos\frac{\pi}{2}=0\Big)$
Apply integration by part again,
$\text{I}=0+2\Bigg[\big[\text{x}\sin\text{x}\big]_0^{\frac{\pi}{2}}-\int_{0}^\limits{\frac{\pi}{2}}1\times\sin\text{x}\text{ dx}\Bigg]$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}\sin\frac{\pi}{2}-0\Big)-2\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x dx}$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}-0\Big)-\big[2(-\cos\text{x})\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=\pi+2\Big(\cos\frac{\pi}{2}-\cos0\Big)$
$\Rightarrow\text{I}=\pi+2(0-1)$
$\Rightarrow\text{I}=\pi-2$
View full question & answer→Question 995 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^1_{0}\big(3\text{x}^2+5\text{x}\big)\text{dx}$
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, $\text{a}=0,\text{ b}=1,\text{ f(x)}=3\text{x}^2+5\text{x},\text{ h}=\frac{1-0}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^1_{0}\big(3\text{x}^2+5\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(0+0)+(3\text{h}^2+5\text{h})+\ \\ .....\ +\big\{3(\text{n}-1)^2\text{h}^2+5(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[5\text{h}(1+2+\ ....\ +\text{n})\\+3\text{h}^2\big\{1^2+2^2+3^2\ ....\ +(\text{n}-1)^2\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[5\text{h}\frac{\text{n}(\text{n}-1)}{2}+\text{h}^2\frac{3\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[\frac{5(\text{n}-1)}{2}+\frac{(\text{n}-1)(2\text{n}-1)}{2\text{n}}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg[\frac{5}{2}\Big(1-\frac{1}{\text{n}}\Big)+\frac{1}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg]$
$=\frac{5}{2}+1$
$=\frac{7}{2}$
View full question & answer→Question 1005 Marks
Evaluate the following integrals:
$\int\limits^{\infty}_0\frac{\log\text{x}}{1+\text{x}^2}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^{\infty}_0\frac{\log\text{x}}{1+\text{x}^2}\text{ dx}$
Putting $\text{x}=\tan\theta$
$\text{dx}=\sec^2\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$
And $\text{x}\rightarrow\infty;\theta\rightarrow\frac{\pi}{2}$
Now, integral becomes,
$\therefore\ \text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\log(\tan\theta)}{1+\tan^2\theta}\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log(\tan\theta)\text{d}\theta\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\Big[\tan\Big(\frac{\pi}{2}-\theta\Big)\Big]\text{d}\theta$ $\Bigg[\because\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log(\cot\theta)\text{d}\theta\ ....(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\log(\tan\theta)\text{d}\theta+\int\limits^{\frac{\pi}{2}}_0\log(\cot\theta)\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\big[\log(\tan\theta)+\log(\cot\theta)\big]\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\big[\log(\tan\theta\times\cot\theta)\big]\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\big(\log1\big)\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0(0)\text{d}\theta$
$\Rightarrow2\text{I}=0$
$\Rightarrow\text{I}=0$
$\therefore\ \int\limits^{\infty}_0\frac{\log\text{x}}{1+\text{x}^2}\text{ dx}=0$
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