Question 1515 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 1525 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 1535 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 1545 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 1555 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 1565 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 1575 Marks
Evaluate the following integrals:
$\int\limits^{\text{a}}_{-\text{a}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$
AnswerLet $\text{I}=\int\limits^{\text{a}}_{-\text{a}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$
Consider $\text{f}(\theta)=\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)$
$\text{f}(-\theta)=\log\Big(\frac{\text{a}-\sin(-\theta)}{\text{a}+\sin(-\theta)}\Big)$
$=\log\Big(\frac{\text{a}+\sin\theta}{\text{a}-\sin\theta}\Big)$ $[\sin(-\text{x})=-\sin\text{x}\big]$
$=\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)^{-1}$
$=-\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)$ $\Big[\log\text{a}^{\text{b}}=\text{b}\log\text{a}\Big]$
$=-\text{f}(\theta)$
$\therefore\ \text{f}(-\theta)=-\text{f}(\theta)$
$\Rightarrow\text{I}=\int\limits^{\text{a}}_{-\text{a}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta=0$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
View full question & answer→Question 1585 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 1595 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 1605 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 1615 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 1625 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 1635 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 1645 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 1655 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 1665 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 1675 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 1685 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\sin^3\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin^3\text{x}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x }\sin^2\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x}(1-\cos^2\text{x})\text{dx}$
Let $\text{u}=\cos\text{x},\text{ du}=-\sin\text{x dx}$
$\therefore\ \text{I}=\int-(1-\text{u}^2)\text{du}$
$\Rightarrow\text{I}=\Big[\frac{\text{u}^3}{3}-\text{u}\Big]$
$\Rightarrow\text{I}=\Big[\frac{\cos^3\text{x}}{3}-\cos\text{x}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0-\frac{1}{3}+1$
$\Rightarrow\text{I}=\frac{2}{3}$
View full question & answer→Question 1695 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\text{a}}\frac{\text{x}}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}$
AnswerLet $\text{a}^2+\text{x}^2=\text{t}^2$
Differentiating w.r.t. x, we get
$2\text{xdx}=2\text{tdt}$
$\text{xdx}=\text{tdt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\text{a}\Rightarrow\text{t}=\sqrt{2}\text{a}$
$\therefore\ \int_{0}^\limits{\text{a}}\frac{\text{x}}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\int_{\text{a}}^\limits{\sqrt{2\text{a}}}\frac{\text{t dt}}{\text{t}}$
$=\int_{\text{a}}^\limits{\sqrt{2\text{a}}}\text{dt}$
$=\big[\text{t}\big]^{\sqrt{2}\text{a}}_\text{a}$
$=\big[\sqrt{2}\text{a}-\text{a}\big]$
$=\text{a}\big(\sqrt{2}-1\big)$
View full question & answer→Question 1705 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sin^2\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sin^2\text{x dx}$
Here $\text{f(x)}=\sin^2\text{x}$
$\text{f}(-\text{x})=\sin^2(-\text{x})=\sin^2\text{x}=\text{f(x)}$
Hence $\sin^2\text{x}$ is an even function
Therefore,
$\text{I}=2\int\limits^{\frac{\pi}{4}}_{0}\sin^2\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_{0}\Big(\frac{1-\cos2\text{x}}{2}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{4}}_{0}(1-\cos2\text{x})\text{dx}$
$=\Big[\text{x}-\frac{\sin^2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0$
$=\frac{\pi}{4}-\frac{1}{2}$
View full question & answer→Question 1715 Marks
Evaluate the following integrals:
$\int\limits^2_1\frac{1}{\text{x}(1+\log\text{x})^2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits2_1\frac{1}{\text{x}(1+\log\text{x})^2}\text{ dx}$ Then,
Let $(1+\log\text{x})=\text{t}$ Then, $\frac{1}{\text{x}}\text{ dx}=\text{dt}$
When, $\text{x}=1,\text{t}=1$ and $\text{x}=2,\text{t}=1+\log2$
$\therefore\ \text{I}=\int^\limits{(1+\log2)}_{1}\frac{1}{\text{t}}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{-1}{\text{t}}\Big]^{(1+\log2)}_1$
$\Rightarrow\text{I}=-\frac{1}{1+\log2}+1$
$\Rightarrow\text{I}=\frac{\log2}{1+\log2}$
View full question & answer→Question 1725 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\cos^{2}\text{x}}{1+\text{e}^{\text{x}}}\text{ dx}$
AnswerWe know,$\int\limits_{\text{a}}^{\text{b}}\text{f}\text{(x)}\text{dx}=\int\limits_{\text{a}}^{\text{b}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}$
Hence,
$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{(-x)}}{1+\text{e}^\text{-x}}\text{dx}$
$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$
If,
$\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
Then
$\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
So,
$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}+\frac{\cos^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$
$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}+\frac{\text{e}^\text{x}\cos^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$
$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{(1+\text{e}^\text{x})\cos^2\text{x}}{1+\text{e}^\text{x}}$
$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2\text{x}\text{dx}$
$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1+\cos2\text{x}}{2}\text{dx}$
$\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{4}\bigg\{\text{x}+\frac{\sin2\text{x}}{2}\bigg\}^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$\text{I}=\frac{1}{4}\bigg\{\bigg(\frac{\pi}{2}\bigg)-\bigg(-\frac{\pi}{2}\bigg)\bigg\} $
$\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1735 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\tan^{2}\text{x}}{1+\text{e}^{\text{x}}}\text{ dx}$
AnswerWe now $\int_\limits{a}^{b}\text{f}\text{(x)}\text{dx}=\int_\limits{a}^{b} \text{f}(\text{a}+\text{b}-\text{x}) \text{dx}$
Hence,
$\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}\text{dx}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{(-x)}}{1-\text{e}^\text{-x}}\text{dx}$
$\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}\text{dx}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{1-\text{e}^\text{-x}}\text{dx}$
If,
$\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}\text{dx}$
Then
$\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{-x}}\text{dx}$
So,
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}+\frac{\tan^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}+\frac{\tan^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}+\frac{\text{e}^x\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{1+\text{e}^\text{x}}+\frac{\text{e}^\text{x}\tan^2\text{x}}{1+\text{e}^2}\text{dx}$
$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}+\text{e}^\text{x}\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{(1+\text{e}^\text{x})\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}+\text{e}^\text{x}\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{(1+\text{e}^\text{x})\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\tan^2\text{x}\text{dx}$.
$\text{I}=\frac{1}{2}\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\tan^2\text{x}\text{dx}$
We know
If f(x)is even
$\int_\limits{-a}^{a} \text{f}\text{(x)}\text{dx}=2\int_\limits{0}^{a}\text{f}\text{(x)}\text{dx}$
If f(x)is odd
$\int_\limits{-a}^{a} \text{f}\text{(x)}\text{dx}=0$
Here
$\text{f}\text{(x)}=\tan^2\text{x}$
f(x)is even,hence
$\text{I}=\int\limits_{0}^{\frac{\pi}{4}}\tan^2\text{x}\text{dx}$
$\text{I}=\int\limits_{0}^{\frac{\pi}{4}}\sec^2\text{x}-1\text{dx}$.
$\text{I}=\big\{\tan\text{x}-\text{x}\big\}\frac{\frac{\pi}{4}}{0 }$
$\text{I}=1-\frac{\pi}{4}$
View full question & answer→Question 1745 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5+1}{\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5+1}{\cos^2\text{x}}\text{ dx}$$=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}+\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{1}{\cos^2\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}+\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sec^2\text{x}\text{ dx}$
$=\text{I}_1+\text{I}_2$
Now,
Consider, $\text{f(x)}=\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}$
$\therefore\ \text{f}(-\text{x})=\frac{(-\text{x})^{11}-3(-\text{x})^9+5(-\text{x})^7-(-\text{x})^5}{\cos^2(-\text{x})}$
$=\frac{-\text{x}^{11}+3\text{x}^9-5\text{x}^7+\text{x}^5}{\cos^2\text{x}}$
$=-\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}=\text{f(x)}$
$\Rightarrow\text{I}_1=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}=0$ $\begin{bmatrix}\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{\text{a}}_0\text{f(x)}\text{dx},&\text{ if }\text{ f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{ f}(-\text{x})=-\text{f(x)}\end{cases}\end{bmatrix}$
$\text{g(x)}=\sec^2\text{x}$
Let $\text{g}(-\text{x})=\sec^2(-\text{x})=\sec^2\text{x}=\text{g}(\text{x})$
$\Rightarrow\text{I}_1=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sec^2\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_0\sec^2\text{x dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{\text{a}}_0\text{f(x)}\text{dx},&\text{ if }\text{ f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{ f}(-\text{x})=-\text{f(x)}\end{cases}\end{bmatrix}$
$=2\times\Big[\tan\text{x}\Big]^{\frac{\pi}{4}}_0$
$=2\Big(\tan\frac{\pi}{4}-\tan0\Big)$
$=2\times(1-0)$
$=2$
$\therefore\ \text{I}=\text{I}_1+\text{I}_2$
$\text{I}=0+2$
$\text{I}=2$
View full question & answer→Question 1755 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 1765 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\sin^3\text{x}\text{ dx}$
Answer Let $\text{I}=\int\limits^{\pi}_0\text{x}\sin^3\text{x}\text{ dx}\ ...(\text{i})$$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3\text{x}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\sin^3\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\sin^3\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\frac{3\sin\text{x}-\sin3\text{x}}{4}\text{ dx}$
$=\frac{\pi}{4}\int\limits^{\pi}_0\big(3\sin\text{x}-\sin3\text{x}\big)\text{dx}$
$=\frac{\pi}{4}\Big[-3\cos\text{x}+\frac{\cos3\text{x}}{3}\Big]^{\pi}_0$
$=\frac{\pi}{4}\Big[-3\cos\pi+3\cos0+\frac{\cos3\pi}{3}-\frac{\cos0}{3}\Big]$
$=\frac{\pi}{4}\Big[3+3+\frac{-1}{3}-\frac{1}{3}\Big]$
$=\frac{\pi}{4}\Big[3-\frac{1}{3}\Big]$
$=\frac{\pi}{2}\times\frac{8}{3}$
$=\frac{4\pi}{3}$
$\therefore\ \text{I}=\frac{2\pi}{3}$
View full question & answer→Question 1775 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 1785 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→