Question 1515 Marks
Integrate the function in Exercise: $\frac{\sqrt{\text{x}^{2}+1}\big[\log\text{(x}^{2}+1)-2\log\text{x}\big]}{\text{x}^{4}}$
Answer$\frac{\sqrt{\text{x}^{2}+1}\big[\log\text{(x}^{2}+1)-2\log\text{x}\big]}{\text{x}^{4}}=\frac{\sqrt{\text{x}^{2}+1}}{\text{x}^{4}}\big[\log\text{(x}^{2}+1)-\log\text{x}^{2}\big]$
$=\frac{\sqrt{\text{x}^{2}+1}}{\text{x}^{4}}\bigg[\log\bigg(\frac{\text{x}^{2}+1}{\text{x}^{2}}\bigg)\bigg]$
$=\frac{\sqrt{\text{x}^{2}+1}}{\text{x}^{4}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)$
$=\frac{1}{\text{x}^{3}}\sqrt{1+\frac{\text{x}^{2}+1}{\text{x}^{2}}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)$
$=\frac{1}{\text{x}^{3}}\sqrt{1+\frac{1}{\text{x}^{2}}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)$
$\text{Let}\ 1+\frac{1}{\text{x}^{2}}=\text{t}\Rightarrow\frac{-2}{\text{x}^{3}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{1}{\text{x}^{3}}\sqrt{1+\frac{1}{\text{x}^{2}}}\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)\text{dx}$
$=-\frac{1}{2}\int\sqrt{\text{t}}\log\text{t dt}$
$=-\frac{1}{2}\int\text{t}^{\frac{1}{2}}.\log\text{t dt}$
Integrating by parts, we obtain
$\text{I}=-\frac{1}{2}\Bigg[\log\text{t}.\int\text{t}^{\frac{1}{2}}\text{dt}-\left\{\bigg(\frac{\text{d}}{\text{dt}}\log\text{t}\bigg)\int\text{t}^{\frac{1}{2}}\text{dt}\right\}\text{dt}\Bigg]$
$=-\frac{1}{2}\begin{bmatrix}\log\text{t}.\frac{\text{t}\frac{3}{2}}{\frac{3}{2}}-\int\frac{1}{\text{t}}.\frac{\text{t}\frac{3}{2}}{\frac{3}{2}}\text{dt} \\ \end{bmatrix}$
$=-\frac{1}{2}\Bigg[\frac{2}{3}\text{t}^{\frac{3}{2}}\log\text{t}-\frac{2}{3}\int\text{t}^{\frac{1}{2}}\text{dt}\Bigg]$
$=-\frac{1}{2}\Bigg[\frac{2}{3}\text{t}^{\frac{3}{2}}\log\text{t}-\frac{4}{9}\int\text{t}^{\frac{3}{2}}\text{dt}\Bigg]$
$=-\frac{1}{3}\text{t}^{\frac{3}{2}}\log\text{t}+\frac{2}{9}\text{t}^{\frac{3}{2}}$
$=-\frac{1}{3}\text{t}^{\frac{3}{2}}\bigg[\log\text{t}-\frac{2}{3}\bigg]$
$=-\frac{1}{3}\bigg(1+\frac{1}{\text{x}^{2}}\bigg)^{\frac{3}{2}}\Bigg[\log\bigg(1+\frac{1}{\text{x}^{2}}\bigg)-\frac{2}{3}\Bigg]+\text{C}$
View full question & answer→Question 1525 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{3+\sin2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{3+\sin2\text{x}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{4-(1-\sin2\text{x})}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{4-(\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x})}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{4-(\sin\text{x}-\cos\text{x})^2}\text{ dx}$
Put $\sin\text{x}-\cos\text{x}=\text{z}$
$\therefore\ (\cos\text{x}+\sin\text{x})\text{dx}=\text{dz}$
When $\text{x}\rightarrow0,\text{z}\rightarrow-1$ $(\text{z}=\sin0-\cos0=0-1=-1)$
When $\text{x}\rightarrow\frac{\pi}{4},\text{z}\rightarrow0$ $\Big(\text{z}=\sin\frac{\pi}{4}-\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\Big)$
$\therefore\ \text{I}=\int_{-1}^\limits{0}\frac{\text{dz}}{2^2-\text{z}^2}$
$=\frac{1}{2\times2}\log\Big[\log\Big(\frac{2+\text{z}}{2-\text{z}}\Big)\Big]^0_{-1}$
$=\frac{1}{4}\Big(\log1-\log\frac{1}{3}\Big)$
$=\frac{1}{4}\big[0-\big(\log1-\log3\big)\big]$
$=-\frac{1}{4}\big(0-\log3\big)$
$=\frac{1}{4}\log3$
View full question & answer→Question 1535 Marks
Evaluate the following integrals:
$\int\sin^7\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\sin^7\text{x}\text{ dx}$ Then
$\text{I}=\int\sin^6\text{x }\sin\text{x}\text{ dx}$
$=\int\big(\sin^2\text{x}\big)^3\sin\text{x}\text{ dx}$
$=\int\big(1-\cos^2\text{x}\big)^3\sin\text{x}\text{ dx}$
$=\int\big(1-\cos^6\text{x}+3\cos^4\text{x}-3\cos^2\text{x}\big)\sin\text{x}\text{ dx}$
$\text{I}=\int\sin\text{x}\text{ dx}-\int\cos^6\text{x }\sin\text{x}\text{ dx}+3\int\cos^4\text{x }\sin\text{x}\text{ dx}-3\int\cos^2\text{x }\sin\text{x}\text{ dx}$
Putting $\cos\text{x}=\text{t}$ and $-\sin\text{x}\text{dx}=\text{dt}$ in $2^{nd}, 3^{rd} $ and $4^{th} $ integral we get
$\text{I}=\int\sin\text{x}\text{ dx}-\int\text{t}^6(-\text{dt})+3\int\text{t}^4(-\text{dt})-3\int\text{t}^2(-\text{dt})$
$=-\cos\text{x}+\frac{\text{t}^7}{7}-\frac{3}{5}\text{t}^5+\frac{3}{3}\text{t}^3+\text{C}$
$=-\cos\text{x}+\frac{\cos^7\text{x}}{7}-\frac{3}{5}\cos^5\text{x}+\cos^3\text{x}+\text{C}$
$\therefore\ \text{I}=-\cos\text{x}+\cos^3\text{x}-\frac{3}{5}\cos^5\text{x}+\frac{1}{7}\cos^7\text{x}+\text{C}$
View full question & answer→Question 1545 Marks
Evaluate the following integrals:
$\int4\text{x}^3\sqrt{5-\text{x}^2}\text{ dx}$
Answer$\int4\text{x}^3\sqrt{5-\text{x}^2}\text{ dx}$
$=4\int\text{x}^2\times\text{x}\sqrt{5-\text{x}^2}\text{ dx}$
Let $5-\text{x}^2=\text{t}$
$\Rightarrow\text{x}^2=5-\text{t}$
$\Rightarrow2\text{x}=-\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{x dx}=-\frac{\text{dt}}{2}$
Now, $4\int\text{x}^2\times\text{x}\sqrt{5-\text{x}^2}\text{ dx}$
$=\frac{4}{-2}\int(5-\text{x})\sqrt{\text{t}}\text{ dt}$
$=-2\int5\text{t}^{\frac{1}{2}}+2\int\text{t}^{\frac{3}{2}}\text{ dt}$
$=-10\Bigg[\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\Bigg]+2\Bigg[\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]+\text{C}$
$=-\frac{20}{3}\text{t}^{\frac{3}{2}}+\frac{4}{5}\text{t}^{\frac{5}{2}}+\text{C}$
$=-\frac{20}{3}\big(5-\text{x}^2\big)^{\frac{3}{2}}+\frac{4}{5}\big(5-\text{x}^2\big)^{\frac{5}{2}}+\text{C}$
View full question & answer→Question 1555 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}$
Answer$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}$ Let $\text{x}^2=\text{t}$ $\Rightarrow2\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ Now, $\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}$$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}+\text{a}^2}+\sqrt{\text{t}-\text{a}^2}}$
$=\frac{1}{2}\int\frac{\text{dt}}{\Big(\sqrt{\text{t}+\text{a}^2}+\sqrt{\text{t}-\text{a}^2}\Big)}\times\frac{\Big(\sqrt{\text{t}+\text{a}^2}-\sqrt{\text{t}-\text{a}^2}\Big)}{\Big(\sqrt{\text{t}+\text{a}^2}-\sqrt{\text{t}-\text{a}^2}\Big)}$
$=\frac{1}{2}\int\frac{\Big(\sqrt{\text{t}+\text{a}^2}-\sqrt{\text{t}-\text{a}^2}\Big)}{(\text{t}+\text{a}^2)-(\text{t}-\text{a}^2)}\text{ dt}$
$=\frac{1}{4\text{a}^2}\int\Big(\text{t}+\text{a}^2\Big)^{\frac12}\text{dt}-\frac{1}{4\text{a}^2}\big(\text{t}-\text{a}^2\big)^{\frac12}\text{dt}$
$=\frac{1}{4\text{a}^2}\begin{bmatrix}\frac{\big(\text{t}+\text{a}^2\big)^{\frac12+1}}{\frac{1}2+1}\end{bmatrix}-\frac{1}{4\text{a}^2}\begin{bmatrix}\frac{\big(\text{t}-\text{a}^2\big)^{\frac12+1}}{\frac12+1}\end{bmatrix}+\text{C}$
$=\frac{1}{6\text{a}^2}\begin{bmatrix}(\text{t}+\text{a}^2)^{\frac{3}{2}}-(\text{t}-\text{a}^2)^{\frac32}\end{bmatrix}+\text{C}$
$=\frac{1}{6\text{a}^2}\begin{bmatrix}(\text{x}^2+\text{a}^2)^{\frac32}-(\text{x}^2-\text{a}^2)^{\frac32}\end{bmatrix}+\text{C}$
View full question & answer→Question 1565 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 1575 Marks
Evaluate the following definite integrals:
$\int_{\frac{\pi}{3}}^\limits{\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 1585 Marks
Evalute the following integrals:
$\int\frac{-\sin\text{x}+2\cos\text{x}}{2\sin\text{x}+\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{-\sin\text{x}+2\cos\text{x}}{2\sin\text{x}+\cos\text{x}}\text{dx}\ .....\text{(i)}$
Let $2\sin\text{x}+\cos\text{x}=\text{t}$ then,
$\text{d}(2\sin\text{x}+\cos\text{x})=\text{dt}$
$\Rightarrow(2\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{-\sin\text{x}+2\cos\text{x}}$
Putting $2\sin\text{x}+\cos\text{x}=\text{t and dx}=\frac{\text{dt}}{-\sin\text{x}+2\cos\text{x}}$ in equation (i), we get,
$\text{I}=\int\frac{-\sin\text{x}+2\cos\text{x}}{\text{t}}\times\frac{\text{dt}}{-\sin\text{x}+2\cos\text{x}}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|2\sin\text{x}+\cos\text{x}|+\text{C}$
$\therefore\text{I}=\log|2\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 1595 Marks
Integrate the rational function in exercise: $\frac{1}{(\text{e}^\text{x}-1)} [$Hint: Put $e^x = t]$
Answer$\text{I}=\int\frac{1}{\text{e}^\text{x}-1}\text{dx}\dots(\text{i})$ Putting $e^x = t$
$\Rightarrow \ \text{e}^\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow \text{e}^{\text{x}} \text{dx = dt} \Rightarrow \text{x}$
$\Rightarrow \ \text{dx}=\frac{\text{dt}}{\text{e}^\text{x}}$
$\therefore$ From eq. $(i), \text{I}=\int\frac{1}{\text{t}-1}\frac{\text{dt}}{\text{e}^\text{x}}=\int\frac{1}{\text{t}-1}\frac{\text{dt}}{\text{t}}=\int\frac{1}{\text{t}(\text{t}-1)}\text{dt}=\int\frac{\text{t}-(\text{t}-1)}{\text{t}(\text{t}-1)}\text{dt}$
$=\int\Bigg(\frac{\text{t}}{\text{t}(\text{t}-1)}-\frac{(\text{t}-1)}{\text{t}(\text{t}-1)}\Bigg)\text{dt}$
$=\int\Bigg(\frac{1}{(\text{t}-1)}-\frac{1}{\text{t}}\Bigg)\text{dt}$
$=\int\frac{1}{\text{t}-1}\text{dt}-\int\frac{1}{\text{t}}\text{dt}$
$=\text{log}|\text{t}-1|-\text{log}|\text{t}|+\text{c}$
$=\text{log}\Bigg|\frac{\text{t}-1}{\text{t}}\Bigg|+\text{c}$
$=\text{log}\Bigg|\frac{\text{e}^\text{x}-1}{\text{e}^\text{x}}\Bigg|+\text{c}$
View full question & answer→Question 1605 Marks
Evaluate the following definite integral as limit of sum:
$\int_\limits{0}^{5} \ \text{(x}+1) \ \text{d}\text{x}$
Answer$\text{we}\ \ \text{know}\ \text{that}\int_\limits{\text{a}}^\text{b}\ \text{f} \ \text{(x)} \ \text{dx}=\lim_\limits{\text{h}\rightarrow0} \text{h}[\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})]$
$\text{where}\ \text{nh}=\text{b}-\text{a}$
$\text{a}=0,\text{b}=5,\text{n}\text{h}= 5\ \text{and}\ \text{f}\text{(x)}=\text{x}+1$
$ \therefore \ \ \int_\limits{0}^{5}\text{(x}+1) \ \text{dx}=\lim_\limits{\text{h}\rightarrow0} \ \text{h}[1+\text{(h}+1)+(2\text{h}+1)+.......+\text{((n}-1)\text{h}+1)] $
$\Rightarrow\int_\limits{0}^{5}\text{(x}+1) \ \text{dx}=\lim_\limits{\text{h}\rightarrow0}\text{h}[\text{n}+\text{h}(1+2+3+.......+\text{(n}-1)]$
$\Rightarrow\int_\limits{0}^{5}\text{(x}+1) \ \text{dx}=\lim_\limits{\text{h}\rightarrow0}\Bigg[\text{n}\text{h}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Bigg]=\lim_\limits{\text{h}\rightarrow0}\Bigg[\text{n}\text{h}+\frac{\text{(nh}-\text{h)}}{2}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[5+\frac{5(5-\text{h)}}{2}\bigg]=\bigg[5+\frac{5(5-0)}{2}\bigg]=5+\frac{25}{2}=\frac{35}{2}$
View full question & answer→Question 1615 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 1625 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 1635 Marks
Evalute the following integrals:
$\int\frac{1-\cot\text{x}}{1+\cot\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1-\cot\text{x}}{1+\cot\text{x}}\text{dx}$ then,
$\text{I}=\int\frac{1-\frac{\cos\text{x}}{\sin\text{x}}}{1+\frac{\cos\text{x}}{\sin\text{x}}}\text{dx}$
$=\int\frac{\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}}}{\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}}}\text{dx}$
$\Rightarrow\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{dx}\ .....(\text{i})$
Let $\sin\text{x}+\cos\text{x}=\text{t},$ then,
$\text{d}(\sin\text{x}+\cos\text{x})=\text{dt}$
$\Rightarrow(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\Rightarrow-(\sin\text{x}-\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=-\frac{\text{dt}}{\sin\text{x}-\cos\text{x}}$
Putting $\sin\text{x}+\cos\text{x}=\text{t and dx}=-\frac{\text{dt}}{\sin\text{x}-\cos\text{x}}$ in equation (i), we het
$\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\text{t}}\times\frac{-\text{dt}}{\sin\text{x}-\cos\text{x}}$
$=\int\frac{-\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$=-\log|\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 1645 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 1655 Marks
$\int\limits_{0}^{1}\text{x}\log(1+2\text{x})\text{dx}$
AnswerLet $\text{I}=\int\limits_{0}^{1}\text{x}\log(1+2\text{x})\text{dx}$
$=\Bigg[\log(1+2\text{x})\frac{\text{x}^2}{2}\Bigg]_{0}^{1}-\int\frac{1}{1+2\text{x}}\cdot2\cdot\frac{\text{x}^2}{2}\text{dx}$
$=\frac{1}{2}\big[\text{x}^2\log(1+2\text{x})\big]_{0}^{1}-\int\frac{\text{x}^2}{1+2\text{x}}\text{dx}$
$=\frac{1}{2}[\log3-0]-\Bigg[\int\limits_{0}^{1}\bigg(\frac{\text{x}}{2}-\frac{\frac{\text{x}}{2}}{1+2\text{x}}\bigg)\text{dx}\Bigg]$
$=\frac{1}{2}\log3-\frac{1}{2}\int\limits^{1}_{0}\text{x dx}+\frac{1}{2}\int\limits^{1}_{0}\frac{\text{x}}{1+2\text{x}}\text{dx}$
$=\frac{1}{2}\log3-\frac{1}{2}\bigg[\frac{\text{x}^2}{2}\bigg]^{1}_{0}+\frac{1}{2}\int\limits^{1}_{0}\frac{\frac{1}{2}(2\text{x}+1-1)}{(2\text{x}+1)}\text{dx}$
$=\frac{1}{2}\log3-\frac{1}{2}\Big[\frac{1}{2}-0\Big]+\frac{1}{4}\int^1_0\text{dx}-\frac{1}{4}\int^1_0\frac{1}{1+2\text{x}}\text{dx}$
$=\frac{1}{2}\log3-\frac{1}{4}+\frac{1}{4}\big[\text{x}\big]^{1}_{0}-\frac{1}{8}\big[\log|(1+2\text{x})|\big]^{1}_{0}$
$=\frac{1}{2}\log3-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}\big[\log3-\log1\big]$
$=\frac{1}{2}\log3-\frac{1}{8}\log3$
$=\frac{3}{8}\log3$
View full question & answer→Question 1665 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 1675 Marks
Evalute the following integrals:
$\int\tan2\text{x}\tan3\text{x}\tan5\text{x dx}$
AnswerLet $\text{I}=\int1+\tan\text{x}\tan(\text{x}+\theta)\text{dx}$
$=\int1+\tan\text{x}\Big(\frac{\tan\text{x}+\tan\theta}{1-\tan\text{x}\tan\theta}\Big)\text{dx}$
$=\int\frac{1+\tan^2\text{x}}{1-\tan\text{x}\tan\theta}\text{dx}$
$=\int\frac{\sec^2\text{x dx}}{1-\tan\text{x}\tan\theta}$
Putting $\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$
$\therefore\text{I}\approx\int\frac{1}{1-\text{t}\tan\theta}\text{dt}$
$=\frac{-1}{\tan\theta}\text{ ln}|1-\text{t}\tan\theta|+\text{C}$
$\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\text{ ln}|\text{ax}+\text{b}|+\text{C}\Big]$
$=-\cot\theta\text{ ln}|1-\tan\text{ x }\tan\theta|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{1}{1-\tan\text{ x }\tan\theta}\Big|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{\cos\text{ x }\cos\theta}{\cos\text{x}\cos\theta-\sin\text{x}\sin\theta}\Big|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{\cos\text{x}}{\cos(\text{x}+\theta)}\Big|+\text{C }\big[\text{Let C}'=\text{C}+\cot\theta\text{ ln}\cos\theta\big]$
View full question & answer→Question 1685 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 1695 Marks
Evaluate the following integrals:
$\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{dx}$
Answer$\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{dx}$
$=\int\frac{\sqrt{\tan\text{x}}}{\frac{\sin\text{x}}{\cos\text{x}}\times\cos^2\text{x}}\text{dx}$
$=\int\frac{\sqrt{\tan\text{x}}}{\tan\text{x}}\times\sec^2\text{x dx}$
$=\int\frac{1}{\sqrt{\tan\text{x}}}\times\sec^2\text{x dx}$
$=\int(\tan\text{x})^{-\frac{1}{2}}\sec^2\text{x dx}$
$\text{Let }\tan\text{x}=t$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\text{Now,}\int(\tan\text{x})^{-\frac{1}{2}}\sec^2\text{x dx}$
$=\int\text{t}^{{-\frac{1}{2}}}\text{dt}$
$=\Bigg[\frac{-{\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\tan\text{x}}+\text{C}$
View full question & answer→Question 1705 Marks
Evaluate the following integrals:
$\int_{0}^\limits{{\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 1715 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 1725 Marks
Evaluate the following integrals: $\int\frac{1}{\text{x}^2(\text{x}^4+1)^{\frac{3}{4}}}\text{ dx}$
Answer$\int\frac{1}{\text{x}^2(\text{x}^4+1)^{\frac{3}{4}}}\text{ dx}$
Multiplying and dividing by $x^{-3},$ we obtain
$\frac{\text{x}^{-3}}{\text{x}^2.\text{x}^{-3}(\text{x}^4+1)^{\frac{3}{4}}}=\frac{\text{x}^{-3}(\text{x}^4+1)^{\frac{-3}{4}}}{\text{x}^2.\text{x}^{-3}}$
$=\frac{(\text{x}^4+1)^{\frac{-3}{4}}}{\text{x}^5.(\text{x}^{4})^{-\frac{3}{4}}}$
$=\frac{1}{\text{x}^5}\Big(\frac{\text{x}^4+1}{\text{x}^4}\Big)^{-\frac{3}{4}}$
$=\frac{1}{\text{x}^5}\Big(1+\frac{1}{\text{x}^4}\Big)^{-\frac{3}{4}}$
Let, $\frac{1}{\text{x}^4}=\text{t}$
$\Rightarrow-\frac{4}{\text{x}^5}\text{ dx}=\text{dt}$
$\Rightarrow\frac{1}{\text{x}^5}\text{ dx}=-\frac{\text{dt}}{4}$
View full question & answer→Question 1735 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 1745 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\text{ dx}$
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 1755 Marks
Evaluate the definite integral in Exercise:
$\int^{\frac{\pi}{2}}_{0}\sin2\text{x}\tan^{-1}(\sin\text{x})\text{dx}$
Answer$\text{Let I}=\int^{\frac{\pi}{2}}_{0}\sin2\text{x}\tan^{-1}(\sin\text{x})\text{dx}=\int^{\frac{\pi}{2}}_{0}2\sin\text{x}\cos\text{x}\tan^{-1}(\sin\text{x})\text{dx} $
Also, let $\sin\text{x}=\text{t}=\ \Rightarrow\cos\text{x}\ \text{dx}=\text{dt}$
when $\text{x}=0,\text{t}=0$ and when $\text{x}=\frac{\pi}{2},=1$
$\Rightarrow \text{I}=2\int^{1}\limits_{0}\text{t}\cdot\tan^{-1}\text{t dt}$
Consider $\int\text{t}.\tan^{-1}\text{dt}=\tan^{-1}\text{t}.\int\text{t}\ \text{dt}-\int\left\{\frac{\text{d}}{\text{dt}}(\tan^{-1}\text{t)}\int\text{t}\ \text{dt}\right\}\text{dt}$
$=\tan^{-1}\text{t}.\frac{\text{t}^{2}}{2}-\int\frac{1}{1+\text{t}^{2}}.\frac{\text{t}^{2}}{2}\text{dt}$
$=\frac{\text{t}^{2}\tan^{-1}\text{t}}{2}-\frac{1}{2}\int\frac{\text{t}^{2}+1-1}{1+\text{t}^{2}}\text{dt}$
$=\frac{\text{t}^{2}\tan^{-1}\text{t}}{2}-\frac{1}{2}\int1\text{dt}+\frac{1}{2}\int\frac{1}{1+\text{t}^{2}}\text{dt}$
$=\frac{\text{t}^{2}\tan^{-1}\text{t}}{2}-\frac{1}{2}.\text{t}+\frac{1}{2}\tan^{-1}\text{t}$
$\Rightarrow\int^{1}\limits_{0}\text{t}.\tan^{-1}\text{t}\ \text{dt}=\bigg[\frac{\text{t}^{2}.\tan^{-1}\text{t}}{2}-\frac{\text{t}}{2}+\frac{1}{2}\tan^{-1}\text{t}\bigg]^{1}_{0}$
$=\frac{1}{2}\bigg[\frac{\pi}{4}-1+\frac{\pi}{4}\bigg]$
$=\frac{1}{2}\bigg[\frac{\pi}{4}-1\bigg]=\frac{\pi}{4}-\frac{1}{2}$
From equation (1), we obtain
$\text{I}=2\bigg[\frac{\pi}{4}-\frac{1}{2}\bigg]=\frac{\pi}{2}-1$
View full question & answer→Question 1765 Marks
Evaluate the following:
$\int\sqrt{\frac{\text{a}+\text{x}}{\text{a}-\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\frac{\text{a}+\text{x}}{\text{a}-\text{x}}}\text{dx}$
Put $\text{x}=\text{a}\cos2\theta$ $\Rightarrow\ =\text{a}\cdot\sin2\theta\cdot2\cdot\text{d}\theta$
$\therefore\ =-2\int\sqrt{\frac{\text{a}+\text{a}\cos2\theta}{\text{a}-\text{a}\cos2\theta}}\cdot\text{a}\sin2\theta\text{ d}\theta$
$=-2\text{a}\int\sqrt{\frac{1+\cos2\theta}{1-\cos2\theta}}\sin2\theta\text{ d}\theta$
$=-2\int\sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}}\sin2\theta\text{ d}\theta$
$=-2\text{a}\int\cot\theta\cdot\sin2\theta\text{ d}\theta$
$=-2\text{a}\int\frac{\cos\theta}{\sin\theta}\cdot2\sin\theta\cos\theta\text{ d}\theta$
$=-4\text{a}\int\cos^2\theta\text{ d}\theta$
$=-2\text{a}\int(1+\cos2\theta)\text{d}\theta$
$=-2\text{a}\Big[\theta+\frac{1}{2}\sin2\theta\Big]+\text{C}$
$=-2\text{a}\bigg[\frac{1}{2}\cos^{-1}\frac{\text{x}}{\text{a}}+\frac{1}{2}\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\bigg]+\text{C}$
$=-\text{a}\bigg[\cos^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\sqrt{1-\frac{\text{x}^2}{\text{a}^2}}\bigg]+\text{C}$
View full question & answer→Question 1775 Marks
Integrate the following integrals:
$\int\sin\text{x}\cos2\text{x}\sin3\text{x dx}$
Answer$\int\sin\text{x}\cos2\text{x}\sin3\text{x dx}$
$=\frac{1}{2}\int(2\sin\text{x}\cos2\text{x})\sin3\text{x dx}$
$=\frac{1}{2}\int\big[\sin(\text{x}+2\text{x})+\sin(\text{x}-2\text{x})\big]\sin(3\text{x) dx}$
$=\frac{1}{2}\int\big[\sin(3\text{x})-\sin(\text{x})\big]\sin(3\text{x) dx}$
$=\frac{1}{2}\big[\int\sin^2(3\text{x})\text{dx}-\int\sin(\text{x})\sin(3\text{x})\text{dx}\big]$
$=\frac{1}{4}\big[\int2\sin^2(3\text{x})\text{dx}-\int2\sin(\text{x})\sin(3\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big\{\int\big[1-\cos(6\text{x})\big]\text{dx}-\int\big[\cos(\text{x}-3\text{x})-\cos(\text{x}+3\text{x})\big]\text{dx}\Big\}$
$=\frac{1}{4}\big[\int1\text{dx}-\int\cos(6\text{x})\text{dx}-\int\cos(2\text{x})\text{dx}+\int\cos(4\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big[\text{x}-\frac{\sin(6\text{x})}{6}-\frac{\sin(2\text{x})}{2}+\frac{\sin(4\text{x})}{4}\Big]+\text{C}$
$=\frac{\text{x}}{4}-\frac{\sin(6\text{x})}{24}-\frac{\sin(2\text{x})}{8}+\frac{\sin(4\text{x})}{16}+\text{C}$
Hence, $\int\sin\text{x}\cos2\text{x}\sin3\text{x}\text{ dx}$ $=\frac{\text{x}}{4}-\frac{\sin(6\text{x})}{24}-\frac{\sin(2\text{x})}{8}+\frac{\sin(4\text{x})}{16}+\text{C}$
View full question & answer→Question 1785 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$
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 1795 Marks
Evaluate the following intregals:
$\int\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}\text{dx}$
Answer$\int\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}=\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}$
Let $\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}=\frac{\text{A}}{(\text{x}+1)}+\frac{\text{B}}{(\text{x}+2)}+\frac{\text{C}}{(\text{x}-2)}$
$5\text{x}=\text{A}(\text{x}+2)(\text{x}-2)+\text{B}(\text{x}+1)(\text{x}-2)\\+\text{C}(\text{x}+1)(\text{x}+2)\ \dots(1)$
Substituting x = -1, -2 and 2 respectively in equation (1), we obtain
$\text{A}=\frac{5}{3},\text{B}=\frac{5}{2},\text{and }\text{C}=\frac{5}{6}$
$\therefore\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}=\frac{5}{3(\text{x}+1)}-\frac{5}{2(\text{x}+2)}+\frac{5}{6(\text{x}-2)}$
$\Rightarrow\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}\ \text{dx}=\frac{5}{3}\int\frac{1}{(\text{x}+1)}\ \text{dx}-\frac{5}{2}\int\frac{1}{(\text{x}+2)}\\\ \text{dx}+\frac{5}{6}\int\frac{1}{(\text{x}-2)}\ \text{dx}$
$=\frac{5}{3}\log|\text{x}+1|-\frac{5}{2}\log|\text{x}+2|+\frac{5}{6}\log|\text{x}-2|+\text{C}$
View full question & answer→Question 1805 Marks
Evaluate the following integrals:
$\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}$
AnswerLet $\text{I}=\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}$
Let $\text{x}=\cos2\theta$
On differentiating both sides, we get
$\text{dx}=-2\sin2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\cos2\theta}{1-\cos2\theta}}\Big\}2\sin2\theta\text{ d}\theta$
$=-2\int\cos\Big\{2\cot^{-1}\sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}}\Big\}\sin2\theta\text{ d}\theta$
$=-2\int\cos\{2\cot^{-1}(\cot\theta)\}\sin2\theta\text{ d}\theta$
$=-2\int\cos2\theta\sin2\theta\text{ d}\theta$
$=\frac{\cos4\theta}{4}+\text{C}_1$
$=-\int\sin4\theta\text{ d}\theta$
$=\frac{1}{4}(2\cos^2\theta-1)+\text{C}_1$
$=\frac{1}{2}\text{x}^2-\frac{1}{4}+\text{C}_1$
$=\frac{1}{2}\text{x}^2+\text{C},$ where $\text{C}=-\frac{1}{4}+\text{C}_1$
Hence, $\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}=\frac{1}{2}\text{x}^2+\text{C}$
View full question & answer→Question 1815 Marks
Evaluate the definite integral in Exercise:
$\int^{\frac{\pi}{4}}_{0}\frac{\sin\text{x}+\cos\text{x}}{9+16\sin2\text{x}}\text{dx}$
Answer$\text{Let I}=\int^{\frac{\pi}{4}}_{0}\frac{\sin\text{x}+\cos\text{x}}{9+16\sin2\text{x}}\text{dx}$
Also, let $\sin\text{x}-\cos\text{x}=\text{t}\Rightarrow(\cos\text{x}+\sin\text{x})\text{dx}=\text{dt}$
when $\text{x}=0,\text{t}=-1$ and when $\text{x}=\frac{\pi}{4},\text{t}=0$
$\Rightarrow(\sin\text{x}-\cos\text{x})^{2}=\text{t}^{2}$
$\Rightarrow\sin^{2}\text{x}+\cos^{2}\text{x}-2\sin\text{x}\cos\text{x}=\text{t}^{2}$
$\Rightarrow1-\sin2\text{x}=\text{t}^{2}$
$\Rightarrow\sin2\text{x}=1-\text{t}^{2}$
$\therefore\text{I}=\int^{0}\limits_{-1}\frac{\text{dt}}{9+16(1-\text{t}^{2})}$
$=\int^{0}\limits_{-1}\frac{\text{dt}}{9+16-16\text{t}^{2}}$
$=\int^{0}\limits_{-1}\frac{\text{dt}}{25-16\text{t}^{2}}=\int^{0}\limits_{-1}\frac{\text{dt}}{(5)^{2}-(4\text{t})^{2}}$
$=\frac{1}{4}\Bigg[\frac{1}{2(5)}\log\Bigg|\frac{5+4\text{t}}{5-4\text{t}}\Bigg|\Bigg]^{0}_{-1}$
$=\frac{1}{40}\bigg[\log(1)-\log\bigg|\frac{1}{9}\bigg|\bigg]^{0}_{-1}$
$=\frac{1}{40}\log9$
View full question & answer→Question 1825 Marks
Integrate the function in Exercise:
$\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}$
Answer$\text{Let I }=\int\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx} \ \ \ \ \ ...\text{(i)}$
$\text{Let Linear}=\text{A}\frac{\text{d}}{\text{dx}}(\text{Quadratic})+\text{B}$
$\Rightarrow\ \ 5\text{x}+3=\text{A}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+4\text{x}+10\big)+\text{B}$
$\Rightarrow\ \ 5\text{x}+3=\text{A}(2\text{x}+4)+\text{B} \ \ \ \ ...\text{(ii)}$
$\Rightarrow\ \ 5\text{x}+3=2\text{A}\text{x}+4\text{A}+\text{B}$
Comparing coefficients of x,
$2\text{A}=5\ \ \Rightarrow\ \ \text{A}=\frac{5}{2}$
Comparing constants,
$4\text{A}+\text{B}=3$
On solving, we get
$\text{A}=\frac{5}{2}, \ \text{B}=-7$
Putting the values of $A$ and $B$ in eq. $(ii),$
$5\text{x}+3=\frac{5}{2}(2\text{x}+4)-7$
Putting this value of $5x + 3$ in eq. $(i),$
$\text{I}=\int\frac{\frac{5}{2}(2\text{x}+4)-7}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}$
$\text{I}=\frac{5}{2}\int\frac{2\text{x}+4}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}-7\int\frac{1}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}$
$\Rightarrow\ \ \text{I}=\frac{5}{2}\text{I}_1-7\ \text{I}_2\ \ \ \ ...\text{(iii)}$
$\text{Now I}_1=\int\frac{2\text{x}+4}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}$
$\text{Putting }\text{ x}^2+4\text{x}+10=\text{t}\ \ \Rightarrow\ \ \ 2\text{x}+4=\frac{\text{dt}}{\text{dx}}\ \ \Rightarrow\ \ \ (2\text{x}+4)\text{ dx}=\text{dt}$
$\therefore\ \ \ \text{I}_1=\int\frac{\text{dt}}{\sqrt{\text{t}}}=\int\text{t}^{\frac{-1}{2}}\text{ dt}=\frac{\text{t}^{\frac{1}{2}}}{\frac{1}{2}}$
$2\sqrt{\text{t}}=2\sqrt{\text{x}^2+4\text{x}+10} \ \ \ \ ...\text{(iv)}$
$\text{Again I}_2=\int\frac{1}{\sqrt{\text{x}^2+4\text{x}+10}}\text{ dx}$
$=\int\frac{1}{\sqrt{\text{x}^2+4\text{x}+4+6}}$
$=\int\frac{1}{\sqrt{(\text{x}+2)^2+\big(\sqrt{6}\big)^2}}\text{ dx}$
$=\log\begin{vmatrix}\text{x}+2+\sqrt{(\text{x}+2)^2+(6)^2}\end{vmatrix}$
$=\log\begin{vmatrix}\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\end{vmatrix} \ \ \ \ ...\text{(v)}$
Putting values of $I_1$ and $I_2$ in eq. $(iii),$
$\text{I}=5\sqrt{\text{x}^2+4\text{x}+10}-7\log\begin{vmatrix}\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\end{vmatrix}+\text{c}$
View full question & answer→Question 1835 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 1845 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 1855 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 1865 Marks
Evaluate the following integrals:
$\int_{0}^\limits{{2\pi}}\sqrt{1-\sin\frac{\text{x}}{2}}\text{ dx}$
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 1875 Marks
Evalute the following integrals:
$\int\frac{\cos4\text{x}-\cos2\text{x}}{\sin4\text{x}-\sin2\text{x}}\text{dx}$
Answer$\int\Big(\frac{\cos4\text{x}-\cos2\text{x}}{\sin4\text{x}-\sin2\text{x}}\Big)\text{dx}$
$=\int\frac{-2\sin\Big(\frac{4\text{x}+2\text{x}}{2}\Big)\sin\Big(\frac{4\text{x}-2\text{x}}{2}\Big)}{2\cos\Big(\frac{4\text{x}+2\text{x}}{2}\Big)\sin\Big(\frac{4\text{x}-2\text{x}}{2}\Big)}\text{dx}$
$\bigg[\because\cos\text{A}-\cos\text{B}=-2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\ \& \\ \sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\bigg]$
$=-\int\frac{\sin3\text{x}}{\cos3\text{x}}\text{dx}$
$=-\int\tan3\text{x dx}$
$=\frac{-\text{In}|\sec3\text{x}|}{3}+\text{C}$
$=\frac{1}{3}\text{ln}\big(|\sec3\text{x}|\big)^{-1}+\text{C}$
$=\frac{1}{3}\text{ln}|\cos3\text{x}|+\text{C}$
View full question & answer→Question 1885 Marks
Integrate the function in Exercise:
$\frac{1}{\text{x}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}}$
$\Bigg[\text{Hint:}\frac{1}{\text{x}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}}=\frac{1}{\text{x}^{\frac{1}{3}}\bigg(1+\text{x}^{\frac{1}{6}}\bigg)},\text{put}\ \text{x}=\text{t}^{6}\Bigg]$
Answer$\frac{1}{\text{x}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}}=\frac{1}{\text{x}^{\frac{1}{3}}\bigg(1+\text{x}^{\frac{1}{6}}\bigg)}$ $\text{Let}\ \text{x}=\text{t}^{6}\Rightarrow\text{dx}=6\text{t}^{5}\text{dt}$ $\therefore\int\frac{1}{\text{x}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}}\text{dx}=\int\frac{1}{\text{x}^{\frac{1}{3}}\bigg(1+\text{x}\frac{1}{6}\bigg)}\text{dx}$ $=\int\frac{6\text{t}^{5}}{\text{t}^{2}(1+\text{t)}}\text{dt}$ $=6\int\frac{\text{t}^{3}}{(1+\text{t)}}\text{dt}$ On dividing, we obtain$\int\frac{1}{\text{x}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}}\text{dx}=6\int\left\{(\text{t}^{2}-\text{t}+1)-\frac{1}{1+\text{t}}\right\}\text{dt}$
$=6\bigg[\bigg(\frac{\text{t}^{3}}{3}\bigg)-\bigg(\frac{\text{t}^{2}}{2}\bigg)+\text{t}-\log|1+\text{t}|\bigg]$
$=2\text{x}^{\frac{1}{2}}-3\text{x}^{\frac{1}{3}}+6\text{x}^{\frac{1}{6}}-6\log\bigg(1+\text{x}^{\frac{1}{6}}\bigg)+\text{C}$
$=2\sqrt{\text{x}}-3\text{x}^{\frac{1}{3}}+6\text{x}^{\frac{1}{6}}-6\log\bigg(1+\text{x}^{\frac{1}{6}}\bigg)+\text{C}$
View full question & answer→Question 1895 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{ dx}$
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 1905 Marks
Find the integrals of the functions in Exercises:
$\frac{1}{\cos(\text{x}-\text{a})\cos(\text{x}-\text{b})}$
Answer$\frac{1}{\cos(\text{x}-\text{a})\cos(\text{x}-\text{b})}$
$=\frac{1}{\sin(\text{a}-\text{b})}\bigg[\frac{\sin(\text{a}-\text{b})}{\cos(\text{x}-\text{a})\cos(\text{x}-\text{b})}\bigg]$
$=\frac{1}{\sin(\text{a}-\text{b})}\Bigg[\frac{\sin\big[(\text{x}-\text{b})-(\text{x}-\text{a})\big]}{\cos(\text{x}-\text{a})\cos(\text{x}-\text{b})}\Bigg]$
$=\frac{1}{\sin(\text{a}-\text{b})}\frac{\big[\sin(\text{x}-\text{b})\cos(\text{x}-\text{a})-\cos(\text{x}-\text{b})\sin(\text{x}-\text{a})\big]}{\cos(\text{x}-\text{a})\cos(\text{x}-\text{b})}$
$=\frac{1}{\sin(\text{a}-\text{b})}\big[{\tan(\text{x}-\text{b})-\tan(\text{x}-\text{a})}\big]$
$\Rightarrow\int\frac{1}{\cos(\text{x}-\text{a})\cos(\text{x}-\text{b})}\text{ dx}=\frac{1}{\sin(\text{a}-\text{b})}\int\big[{\tan(\text{x}-\text{b})-\tan(\text{x}-\text{a})}\big]\text{dx}$
$=\frac{1}{\sin(\text{a}-\text{b})}\Big[-\log\big|\cos(\text{x}-\text{b})\big|+\log\big|\cos(\text{x}-\text{a})\big|\Big]$
$=\frac{1}{\sin(\text{a}-\text{b})}\Bigg[\log\bigg|\frac{\cos(\text{x}-\text{a})}{\cos(\text{x}-\text{b})}\bigg|\Bigg]+\text{C}$
View full question & answer→Question 1915 Marks
Integrate the function in Exercise:
$\frac{1}{\sqrt{(\text{x}-\text{a})(\text{x}-\text{b})}}$a
Answer$\int\frac{1}{\sqrt{(\text{x}-\text{a})(\text{x}-\text{b})}}\text{ dx}$
$=\int\frac{1}{\sqrt{\text{x}^2-\text{bx}-\text{ax}+\text{ab}}}\text{ dx}$
$=\int\frac{1}{\sqrt{\text{x}^2-\text{x}(\text{a}+\text{b})+\text{ab}}}\text{ dx}$
$=\int\frac{1}{\sqrt{\text{x}^2-\text{x}(\text{a}+\text{b})+\bigg(\frac{\text{a}+\text{b}}{2}\bigg)^2-\bigg(\frac{\text{a}+\text{b}}{2}\bigg)^2+\text{ab}}}\text{ dx}$
$=\int\frac{1}{\sqrt{\Bigg[\bigg\{\text{x}-\Big(\frac{\text{a+b}}{2}\Big)\bigg\}^2-\bigg\{\frac{(\text{a+b})^2}{4}-\text{ab}\bigg\}\Bigg]}}\text{ dx}$
$=\int\frac{1}{\sqrt{\Bigg[\bigg\{\text{x}-\Big(\frac{\text{a+b}}{2}\Big)\bigg\}^2-\bigg\{\frac{(\text{a+b})^2-4\text{ab}}{4}\bigg\}\Bigg]}}\text{ dx}$
$=\int\frac{1}{\sqrt{\Bigg[\bigg\{\text{x}-\Big(\frac{\text{a+b}}{2}\Big)\bigg\}^2-\bigg\{\frac{(\text{a}-\text{b})^2}{4}\bigg\}\Bigg]}}\text{ dx}$
$=\int\frac{1}{\sqrt{\Bigg[\bigg\{\text{x}-\Big(\frac{\text{a+b}}{2}\Big)\bigg\}^2-\bigg\{\bigg(\frac{\text{a}-\text{b}}{2}\bigg)^2\bigg\}\Bigg]}}\text{ dx}$
$=\log\begin{vmatrix}\text{x}-\bigg(\frac{\text{a}+\text{b}}{2}\bigg)+\sqrt{\bigg\{\Big(\text{x}-\frac{\text{a+b}}{2}\Big)^2-\bigg(\frac{\text{a}-\text{b}}{2}\bigg)^2\bigg\}}\end{vmatrix}+\text{c}$
$=\log\begin{vmatrix}\text{x}-\bigg(\frac{\text{a}+\text{b}}{2}\bigg)+\sqrt{\text{x}^2-\text{x}(\text{a}+\text{b})+\text{ab}}\end{vmatrix}+\text{c}$
View full question & answer→Question 1925 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\pi}\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
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 1935 Marks
Evaluate the following intergrals:
$\int\text{e}^\text{ax}\cos\text{bx dx}$
AnswerLet $\text{I}=\int\text{e}^\text{ax}\cos\text{bx dx}$
Intergrating by parts,
$\text{I}=\text{e}^\text{ax}\frac{\sin\text{bx}}{\text{b}}-\text{a}\int\text{e}^\text{ax}\frac{\sin\text{bx}}{\text{x}}\text{dx}$
$=\frac{1}{\text{b}}\text{e}^\text{ax}\sin\text{bx}-\frac{\text{a}}{\text{b}}\int\text{e}^\text{ax}\sin\text{bx dx}$
$=\frac{1}{\text{a}}\text{e}^\text{ax}\sin\text{bx}-\frac{\text{a}}{\text{b}}\Big[-\text{e}^\text{ax}\frac{\cos\text{bx}}{\text{b}}+\int\text{ae}^\text{ax}\frac{\cos\text{bx}}{\text{b}}\text{dx}\Big]$
$=\frac{1}{\text{a}}\text{e}^\text{ax}\sin\text{bx}-\frac{\text{a}}{\text{b}^2}\text{e}^\text{ax}\cos\text{bx}-\frac{\text{a}^2}{\text{b}^2}\int\text{e}^\text{ax}\cos\text{bx dx}$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big[\text{b}\sin\text{bx}+\text{a}\cos\text{bx}\big]-\frac{\text{a}^2}{\text{b}^2}\text{I}+\text{C}$
$\Rightarrow\text{I}.\Big\{\frac{\text{a}^2+\text{b}^2}{\text{b}^2}\Big\}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big[\text{b}\cos\text{bx}+\text{a}\cos\text{bx}\big]+\text{C}$
Thus,
$\text{I}=\frac{\text{e}^\text{ax}}{\text{a}^2+\text{b}^2}\big[\text{a}\cos\text{bx}+\text{a}\cos\text{bx}\big]+\text{C}$
View full question & answer→Question 1945 Marks
Prove the following Exercise:
$\int^{3}\limits_{1}\frac{\text{dx}}{\text{x}^{2}(\text{x}+1)}=\frac{2}{3}+\log\frac{2}{3}$
Answer$\text{Let I}=\int^{3}\limits_{1}\frac{\text{dx}}{\text{x}^{2}(\text{x}+1)}$
Also, $\text{Let I}\int^{3}\limits_{1}\frac{1}{\text{x}^{2}(\text{x}+1)}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}^{2}}+\frac{\text{C}}{\text{x}+1}$
$\Rightarrow1=\text{Ax(x}+1)+\text{B}\text{(x}+1)+\text{C}(\text{x}^{2})$
$\Rightarrow1=\text{Ax}^{2}+\text{Ax}+\text{Bx}+\text{B}+\text{Cx}^{2}$
Equating the Coefficient of $\text{x}^{2},\text{x}$ and constant term, we obtain
$\text{A}+\text{C}=0$
$\text{A}+\text{B}=0$
$\text{B}=1$
On solving these equation, we obtain
$\text{A}=-1,\text{C}=1,\ \text{and B}=1$
$\therefore\frac{1}{\text{x}^{2}(\text{x}+1)}=-\frac{1}{\text{x}}+\frac{1}{\text{x}^{2}}+\frac{1}{(\text{x}+1)}$
$\Rightarrow\text{I}=\int^{3}\limits_{1}\left\{-\frac{1}{\text{x}}+\frac{1}{\text{x}^{2}}+\frac{1}{\text{(x}+1)}\right\}\text{dx}$
$=\bigg[-\log\text{x}-\frac{1}{\text{x}}+\log(\text{x}+1)\bigg]^{3}_{1}$
$=\bigg[-\log\bigg(\frac{\text{x}+1}{\text{x}}\bigg)-\frac{1}{\text{x}}\bigg]^{3}_{1}$
$=\log4-\log3-\log2+\frac{2}{3}$
$=-\log2-\log3+\frac{2}{3}$
$=\log\Big(\frac{2}{3}\Big)+\frac{2}{3}$
Hence, the given result is proved
View full question & answer→Question 1955 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 1965 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^5\phi\text{ d}\phi$
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 1975 Marks
Evaluate the following:
$\int\frac{(\cos5\text{x}+\cos4\text{x})}{1-2\cos3\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{(\cos5\text{x}+\cos4\text{x})}{1-2\cos3\text{x}}\text{dx}$
$=\int\frac{2\cos\frac{9\text{x}}{2}\cdot\cos\frac{\text{x}}{2}}{1-2\Big(2\cos^2\frac{3\text{x}}{2}-1\Big)}\text{dx}$
$=-\int\frac{2\cos\frac{9\text{x}}{2}\cdot\cos\frac{\text{x}}{2}}{4\cos^2\frac{3\text{x}}{2}-3}\text{dx}$
$=-\int\frac{2\cos\frac{9\text{x}}{2}\cdot\cos\frac{\text{x}}{2}\cdot\cos\frac{3\text{x}}{2}}{4\cos^3\frac{3\text{x}}{2}-3\cos\frac{3\text{x}}{2}}\text{dx}$ $\big[\because\cos3\text{x}=4\cos^3\text{x}-3\cos\text{x}\big]$
$=-\int\frac{2\cos\frac{9\text{x}}{2}\cdot\cos\frac{\text{x}}{2}\cdot\cos\frac{3\text{x}}{2}}{\cos3\cdot\frac{3\text{x}}{2}}\text{dx}$
$=-\int2\cos\frac{3\text{x}}{2}\cdot\cos\frac{\text{x}}{2}\text{dx}$
$=\int\bigg\{\cos\Big(\frac{3\text{x}}{2}+\frac{\text{x}}{2}\Big)+\cos\Big(\frac{3\text{x}}{2}-\frac{\text{x}}{2}\Big)\bigg\}\text{dx}$
$=-\int(\cos2\text{x}+\cos\text{x})\text{dx}=-\frac{1}{2}\sin2\text{x}-\sin\text{x}+\text{C}$
View full question & answer→Question 1985 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 1995 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 2005 Marks
Evaluate the following:
$\int\frac{\text{x}}{\sqrt{\text{x}}+1}\text{dx}$
Hint: Put $\sqrt{\text{x}}=\text{z}$
AnswerLet $\text{I}=\int\frac{\text{x}}{\sqrt{\text{x}}+1}\text{dx}$
Put $\sqrt{\text{x}}=\text{t}$
$\Rightarrow\ \frac{1}{2\sqrt{\text{x}}}\text{dx}=\text{dt}$
$\Rightarrow\ \text{dx}=2\sqrt{\text{x}}\text{dt}$
$\Rightarrow\ \text{dx}=2\text{t}\text{dt}$
Substituting $\sqrt{\text{x}}=\text{t}$ and dx = 2t in I, we get
$\text{I}=2\int\frac{\text{t}^2\cdot\text{t}}{\text{t}+1}\text{dt}$ $\Big[\because\sqrt{\text{x}}=\text{t}\Rightarrow\text{x}=\text{t}^2\Big]$
$=2\int\frac{\text{t}^3}{\text{t}+1}\text{dt}$
$=2\int\Big[(\text{t}^2-\text{t}+1)-\frac{1}{\text{t}+1}\Big]\text{dt}$
$=2\int(\text{t}^2-\text{t}+1)\text{dt}-2\int\frac{1}{\text{t}+1}\text{dt}$
$=2\Big[\frac{\text{t}^3}{3}-\frac{\text{t}^2}{3}+\text{t}-\log\big|(\text{t}-1)\big|\Big]+\text{C}$ $\bigg[\because\int\frac{1}{\text{x}}\text{dx}=\log|\text{x}|+\text{C and}\int\text{x}^\text{n}\text{dx}=\frac{\text{x}^{\text{n}+1}}{\text{n}}+\text{C}\bigg]$
$=2\bigg[\frac{\text{x}\sqrt{\text{x}}}{3}-\frac{\text{x}}{2}+\sqrt{\text{x}}-\log\big|(\sqrt{\text{x}}-1)\big|\bigg]+\text{C}$ $\Big[\because\ \text{t}=\sqrt{\text{x}}\Big]$
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