Question 1515 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 1525 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 1535 Marks
If $\int_{0}^\limits{\text{k}}\frac{1}{2+8\text{x}^2}\text{ dx}=\frac{\pi}{16},$ 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 1545 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 1555 Marks
Evaluate the following integrals:$\int\frac{\sin2\text{x}}{\sqrt{\sin^4\text{x}+4\sin^2\text{x}-2}}\text{ dx}$
Answer$\int\frac{\sin(2\text{x})\text{dx}}{\sqrt{\sin^4\text{x}+4\sin^2\text{x}-2}}$
Let $\sin^2\text{x}=\text{t}$
$\Rightarrow2\sin\text{x}\cos\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\sin(2\text{x})\text{ dx}=\text{dt}$
Now, $\int\frac{\sin(2\text{x})\text{dx}}{\sqrt{\sin^4\text{x}+4\sin^2\text{x}-2}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+4\text{t}-2}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+4\text{t}+4-4-2}}$
$=\int\frac{\text{dt}}{\sqrt{(\text{t}+2)^2-\big(\sqrt6\big)^2}}$
$=\log\Big|\text{t}+2+\sqrt{(\text{t}+2)^2-6}\Big|+\text{C}$
$=\log\Big|\text{t}+2+\sqrt{\text{t}^2+4\text{t}-2}\Big|+\text{C}$
$=\log\Big|\sin^2\text{x}+2\sqrt{\sin^4\text{x}+4\sin^2\text{x}-2}\Big|+\text{C}$
View full question & answer→Question 1565 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 1575 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 1585 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 1595 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 1605 Marks
Evaluate the following intregals:
$\int\frac{5\text{x}^2+20\text{x}+6}{\text{x}^2+2\text{x}^2+\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{5\text{x}^2+20\text{x}+6}{\text{x}^2+2\text{x}^2+\text{x}}\ \text{dx}$$=\int\frac{5\text{x}^2+20\text{x}+6}{\text{x}(\text{x}+1)^2}\ \text{dx}$
Now,
Let $\frac{5\text{x}^2+20\text{x}+6}{\text{x}(\text{x}+1)^2}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}+1}+\frac{\text{C}}{(\text{x}+1)^2}$
$\Rightarrow5\text{x}^2+20\text{x}+6=\text{A}(\text{x}+1)^2+\text{Bx}(\text{x}+1)+\text{Cx}$
Equating similar terms, we get,
A + B = 5, 2A + B + C = 20, A = 6
Solving, we get, B = -1, C = 9
Thus,
$\text{I}=\int\frac{6\text{dx}}{\text{x}}-1\int\frac{\text{dx}}{\text{x}+1}+9\int\frac{\text{dx}}{(\text{x}+1)^2}$
$\therefore\text{I}=6\log|\text{x}|-\log|\text{x}+1|-\frac{9}{\text{x}+1}+\text{C}$
View full question & answer→Question 1615 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 1625 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 1635 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 1645 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x}}{\log(\sec\text{x}+\tan\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{\sec\text{x}}{\log(\sec\text{x}+\tan\text{x})}\text{dx}$
Putting $\log(\sec\text{x}+\tan\text{x})=\text{t}$
$\Rightarrow\frac{\sec\text{x}\tan\text{x}+\sec^2\text{x}}{\sec\text{x}+\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x}\frac{(\sec\text{x}+\tan\text{x})}{\sec\text{x}+\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x }\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\sec\text{x}+\tan\text{x})|+\text{C}$
View full question & answer→Question 1655 Marks
Evalute the following integrals:
$\int\frac{\cos\text{x}}{2+3\sin\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{2+3\sin\text{x}}\text{dx}\ .....\text{(i)}$
Let $2+3\sin\text{x}=\text{t}$ then,
$\text{d}(2+3\sin\text{x})=\text{dt}$
$\Rightarrow3\cos\text{x dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{3\cos\text{x}}$
Putting $2+3\sin\text{x}=\text{t and dx}=\frac{\text{dt}}{3\cos\text{x}}$ in equation (i), we get,
$\text{I}=\int\frac{\cos\text{x}}{\text{t}}\times\frac{\text{dt}}{3\cos\text{x}}$
$=\frac{1}{3}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{3}\log|\text{t}|+\text{C}$
$=\frac{1}{3}\log|2+3\sin\text{x}|+\text{C}$
View full question & answer→Question 1665 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 1675 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 1685 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 1695 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 1705 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 1715 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 1725 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]$
View full question & answer→Question 1735 Marks
Evaluate the following intregals:
$\int\frac{1}{\text{x}(\text{x}^4+1)}\ \text{dx}$
AnswerLet $\frac{1}{\text{x}(\text{x}^4+1)}+\frac{\text{A}}{\text{x}}+\frac{\text{Bx}^3+\text{Cx}^2+\text{Dx}+\text{E}}{\text{x}^4+1}$$\Rightarrow1=\text{A}(\text{x}^4+1)+(\text{Bx}^3+\text{Cx}^2+\text{Dx}+\text{E})\text{x}$
$=(\text{A}+\text{B})\text{x}^4+\text{Cx}^3+\text{Dx}^2+\text{Ex}+\text{A}$
Equating similar terms, we get,
$\text{A}+\text{B}+0,\text{C}=0,\text{E}=0,\text{A}=1$
$\therefore\text{B}=-1$
Thus,
$\text{I}=\int\frac{\text{dx}}{\text{x}}+\int-\frac{\text{x}^3\text{dx}}{\text{x}^4+1}$
$=\log|\text{x}|-\frac{1}{4}\log|\text{x}^4+1|+\text{C}$
$\text{I}=\frac{1}{4}\log\Big|\frac{\text{x}^4}{\text{x}^4+1}\Big|+\text{C}$
View full question & answer→Question 1745 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}.\frac{\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+1}{\sqrt{1-\text{x}^2}}\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\bigg[\frac{\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+1}{\sqrt{1-\text{x}^2}}\bigg]\text{dx}$
$=\int\text{e}^{\text{x}}\Big[\sin^{-1}\text{x}+\frac{1}{\sqrt{1-\text{x}^2}}\Big]\text{dx}$
Here, $\text{f(x)}=\sin^{-1}\text{x}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\sqrt{1-\text{x}^2}}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\sin^{-1}\text{x}=\text{t}$
Diff both sides w.r.t x
$\Big(\text{e}^{\text{x}}\sin^{-1}\text{x}+\text{e}^{\text{x}}\times\frac{1}{\sqrt{1-\text{x}^2}}\Big)\text{dx = dt}$
$\because\text{I}=\int\text{dt}$
$=\text{t + C}$
$=\text{e}^{\text{x}}\sin^{-1}\text{x + C}$
View full question & answer→Question 1755 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_02\sin\text{x }\cos\text{x}\tan^{-1}(\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_02\sin\text{x }\cos\text{x}\tan^{-1}(\sin\text{x})\text{dx}$ Then,
Let $\sin\text{x}=\text{t}$ Then, $\cos\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2},\text{t}=1$
$\Rightarrow\text{I}=2\Big[\frac{\text{t}^2\tan^{-1}\text{t}}{2}\Big]^1_0-\int^\limits1_0\frac{\text{t}^2}{1+\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=2\Big[\frac{\text{t}^2\tan^{-1}\text{t}}{2}\Big]^1_0-\int^\limits1_0\Big(\frac{1+\text{t}^3}{1+\text{t}^2}-\frac{1}{1+\text{t}^2}\Big)\text{dt}$
$\Rightarrow\text{I}=2\Big[\frac{\text{t}^2\tan^{-1}\text{t}}{2}\Big]^1_0-\big[\text{t}-\tan^{-1}\text{t}\big]^1_0$
$\Rightarrow\text{I}=1\tan^{-1}1-0-1+\tan^{-1}1+0$
$\Rightarrow\text{I}=\frac{\pi}{4}-1+\frac{\pi}{4}$
$\Rightarrow\text{I}=\frac{\pi}{2}-1$
View full question & answer→Question 1765 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{4}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$
Answer$\int_{0}^\limits{\frac{\pi}{4}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\Big[\text{a}^2\Big(\frac{1+\cos2\text{x}}{2}\Big)+\text{b}^2\Big(\frac{1-\cos2\text{x}}{2}\Big)\Big]\text{dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\Big[\Big(\frac{\text{a}^2+\text{b}^2}{2}\Big)+\Big(\frac{\text{a}^2-\text{b}^2}{2}\Big)\cos2\text{x}\Big]\text{dx}$
$=\Big(\frac{\text{a}^2+\text{b}^2}{2}\Big)\int_{0}^\limits{\frac{\pi}{4}}\text{dx}+\Big(\frac{\text{a}^2-\text{b}^2}{2}\Big)\int_{0}^\limits{\frac{\pi}{4}}\cos2\text{x }\text{dx}$
$=\Big(\frac{\text{a}^2+\text{b}^2}{2}\Big)\Big(\frac{\pi}{4}-0\Big)+\Big(\frac{\text{a}^2-\text{b}^2}{2}\Big)\Big(\sin\frac{\pi}{2}-\sin0\Big)$
$=\Big(\frac{\text{a}^2+\text{b}^2}{2}\Big)\frac{\pi}{4}+\Big(\frac{\text{a}^2-\text{b}^2}{2}\Big)(1-0)$
$=\big(\text{a}^2+\text{b}^2\big)\frac{\pi}{8}+\frac{1}{4}\big(\text{a}^2-\text{b}^2\big)$
View full question & answer→Question 1775 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\pi}\frac{1}{3+2\sin\text{x}+\cos\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\pi}\frac{1}{3+2\sin\text{x}+\cos\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\pi}\frac{1}{3+2\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\frac{1+\tan^{2}\frac{\text{x}}{2}}{2\tan^{2}\frac{\text{x}}{2}+4\tan\frac{\text{x}}{2}+4}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Then, $\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=\pi,\text{t}=\infty$
$\therefore\ \text{I}=\int\limits^{\infty}_0\frac{2\text{dt}}{2\text{t}^2+4\text{t}+4}$
$\Rightarrow\text{I}=\int\limits^{\infty}_0\frac{\text{dt}}{(\text{t}+1)^2+1}$
$\Rightarrow\text{I}=\Big[\tan^{-1}\big(\text{t}+1\big)\Big]^{\infty}_0$
$\Rightarrow\text{I}=\frac{\pi}{2}-\frac{\pi}{4}$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1785 Marks
Evaluate the following integrals:$\int\sin\text{x}\log(\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\sin\text{x}\cdot\log(\cos\text{x})\text{dx}$
Let $\cos\text{x = t}$
$\Rightarrow-\sin\text{x dx =}\text{ dt}$
$\Rightarrow\sin\text{x dx =}-\text{dt}$
$\therefore\text{I}=-\int\log\text{t dt}$
$=-\int1\cdot\log\text{t dt}$
Taking log t as the first function and 1 as the second function.
$=\log\text{t}\int1\text{dt}-\int\big\{\frac{\text{d}}{\text{dt}}(\log\text{t})\int1\text{dt}\big\}\text{dt}$
$=-[\log\text{t}\cdot\text{t}-\int\frac{1}{\text{t}}\times\text{t dt}]$
$=-[\log\text{t}\cdot\text{t}-\text{t}]+\text{C}$
$=-\text{t}(\log\text{t}-1)+\text{C} \dots(1)$
Substituting the value of t in eq (1)
$=-\cos\text{x}\{\log(\cos\text{x})-1\}+\text{C}$
$=\cos\text{x}\{1-\log(\cos\text{x})\}+\text{C}$
View full question & answer→Question 1795 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{(\text{x}-\alpha)(\beta-\text{x})}}\text{ dx},(\beta>\alpha)$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{(\text{x}-\alpha)(\beta-\text{x})}}\text{ dx},(\beta>\alpha)$
$=\int\frac{1}{-\text{x}^2-\text{x}(\alpha+\beta)-\alpha\beta}\text{ dx}$
$=\int\frac{1}{\sqrt{-\Big[\text{x}^2-2\text{x}\big(\frac{\alpha+\beta}{2}\big)+\big(\frac{\alpha+\beta}{2}\big)^2-\big(\frac{\alpha+\beta}{2}\big)^2+\alpha\beta\Big]}}\text{ dx}$
$=\int\frac{1}{\sqrt{-\Big[\big(\text{x}-\frac{\alpha+\beta}{2}\big)^2-\big(\frac{\alpha+\beta}{2}\big)^2\Big]}}\text{ dx}$
$=\int\frac{1}{\sqrt{\Big[\big(\frac{\beta-\alpha}{2}\big)^2-\big(\text{x}-\frac{\alpha+\beta}{2}\big)^2\Big]}}\text{ dx}$ $[\therefore\ \beta>\alpha]$
Let $\Big(\text{x}-\frac{\alpha+\beta}{2}\Big)=\text{t}$
$\Rightarrow\text{dx}=\text{dt}$
$\text{I}=\int\frac{1}{\sqrt{\big(\frac{\beta-\alpha}{2}\big)^2-\text{t}^2}}\text{ dt}$
$\text{I}=\sin^{-1}\bigg(\frac{\text{t}}{\frac{\beta-\alpha}{2}}\bigg)+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}=\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\sin^{-1}\Bigg(\frac{2\big(\text{x}-\frac{\alpha+\beta}{2}\big)}{\beta-\alpha}\Bigg)+\text{C}$
$\text{I}=\sin^{-1}\Big(\frac{2\text{x}-\alpha-\beta}{\beta-\alpha}\Big)+\text{C}$
View full question & answer→Question 1805 Marks
Evaluate the following integrals:
$\int\text{x}\cos^3\text{x}^2\sin\text{x}^2\text{ dx}$
Answer$\int\text{x}\cos^3\text{x}^2\sin\text{x}^2\text{ dx}$
Let $\text{x}^2=\text{t}$
$2\text{x}\text{dx}=\text{dt}$
$\text{x}\text{ dx}=\frac{\text{dt}}{2}$
Now, $\int\text{x}\cos^3\text{x}^2\sin\text{x}^2\text{ dx}$
$=\frac{1}{2}\int\cos^3\text{t}\cdot\sin\text{t}\text{ dt}$
Again let $\cos\text{t}=\text{p}$
$-\sin\text{t}\text{ dt}=\text{dp}$
$\sin\text{t}\text{ dt}=-\text{dp}$
So, $\frac{1}{2}\int\cos^3\text{t}\cdot\sin\text{t}\text{ dt}$
$=-\frac{1}{2}\text{p}^3\text{ dp}$
$=-\frac{1}{2}\Big(\frac{\text{p}^4}{4}\Big)+\text{C}$
$=-\frac{\text{p}^4}{8}+\text{C}$
$=-\frac{\cos^4\text{t}}{8}+\text{C}$
$=-\frac{\cos^4\text{x}^2}{8}+\text{C}$
View full question & answer→Question 1815 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}\text{x}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\big(\frac{\pi}{2}-\text{x}\big)}{\sin^\text{n}\big(\frac{\pi}{2}-\text{x}\big)+\cos^\text{n}\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\cos^\text{n}\text{x}+\sin^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}+\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
i.e., $\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}=\frac{\pi}{4}$
$\therefore\ \int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}\text{x}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}\text{ dx}=\frac{\pi}{4}$
View full question & answer→Question 1825 Marks
If $\text{f}(\text{a}+\text{b}-\text{x})=\text{f(x)},$ then prove that $\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=\frac{\text{a}+\text{b}}{2}\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}$
$=\int\limits^{\text{b}}_\text{a}(\text{a}+\text{b}-\text{x})\text{f}(\text{a}+\text{b}-\text{x})\text{dx}$$\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\text{b}}_\text{a}(\text{a}+\text{b}-\text{x})\text{f(x)}\text{dx}$$\Big[\text{f}(\text{a}+\text{b}-\text{x})=\text{f(x)}\Big]$
$\therefore\ \int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}(\text{a}+\text{b})\text{f(x)}\text{dx}-\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}$
$\Rightarrow\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}+\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=(\text{a}+\text{b})\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}$
$\Rightarrow2\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=(\text{a}+\text{b})\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}$
$\Rightarrow\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=\frac{\text{a}+\text{b}}{2}\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}$
View full question & answer→Question 1835 Marks
Evaluate the following definite integral as limit of sum:
$\int_\limits{2}^{5} \ \text{x}^{2} \ \text{dx}$
Answer$\text{we}\ \text{know}\ \text{that}\ \int_\limits{\text{a}}^{\text{b}}\ \text{f}\ \text{(x)}\ \text{dx}=\lim_\limits{\text{h}\rightarrow0}\ \text{h}[\text{f}\text{(a)}+\text{f}\text{(a}+\text{h)+}\text{f} \ \text{(a}+2\text{h)}+..........+\text{f}\text{(a}+\text{(n}-1)\text{h)}]$ $\text{Here},\ \ \text{a}=2,\text{b}=3,\text{nh}=1 \ \text{and} \ \text{f}\text{(x)}=\text{x}^{2}$ $\therefore\ \ \ \ \int\limits_{2}^{3}\text{x}^{2} \ \text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}[4+(4+4\text{h}+\text{h}^{2})+(4+8\text{h}+2^{2}\text{h}^{2})+.......+(4+4(\text{n}-1)\text{h}+(\text{n}-1)^{2}\text{h}^{2})]$ $=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[4\text{n}+4\text{h}(1+2+3+.....+\text{(n}-1)+\text{h}^{2}(1^{2}+2^{2}+......\text{(n}-1)^{2}\big)\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\bigg[4\text{nh}+4\text{hh}\frac{\text{n}(\text{n}-1)}{2}+\text{hhh}\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\bigg[4\text{n}\text{h}+4\text{n}\text{h}\frac{\text{(nh}-\text{h)}}{2}+\frac{\text{nh}(\text{nh}-\text{h)}(2\text{nh}-\text{h)}}{6}\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\bigg[4+2(1-\text{h)}+\frac{(1-\text{h)}(2-\text{h)}}{6}\bigg]=\bigg[4+2(1-0)+\frac{(1-0)(2-0)}{6}\bigg]$$=6+\frac{1}{3}=\frac{19}{3}$
View full question & answer→Question 1845 Marks
If f(x) is a continuous function defind on [-a, a], then prove that:
$\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\big\{\text{f(x)}+\text{f}(-\text{x})\big\}\text{dx}$
AnswerLet $\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}$
By Additive property
$\text{I}=\int\limits^0_{-\text{a}}\text{f(x)}\text{dx}+\int\limits^{\text{a}}_0\text{f(x)}\text{dx}$
Let $\text{x}=-\text{t},$ then $\text{dx}=-\text{dt}$
When $\text{x}=-\text{a},\text{ t}=\text{a},\text{ x}=0,\text{ t}=0$
Hence, $\int\limits^0_{-\text{a}}\text{f(x)}\text{dx}=-\int\limits^0_{\text{a}}\text{f}(-\text{t})\text{dt}$
$=\int\limits_0^{\text{a}}\text{f}(-\text{t})\text{dt}=\int\limits_0^{\text{a}}\text{f}(-\text{x})\text{dx}$ (Changing the varible)
Therefore,
$\text{I}=\int\limits_0^{\text{a}}\text{f}(-\text{x})\text{dx}+\int\limits_0^{\text{a}}\text{f}(\text{x})\text{dx}$
$=\int\limits^{\text{a}}_0\big\{\text{f(x)}+\text{f}(-\text{x})\big\}\text{dx}$
Hence, proved.
View full question & answer→Question 1855 Marks
Evaluate the following integrals:
$\int\limits^{2\pi}_0\frac{\text{e}^{\sin\text{x}}}{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{\sin\text{x}}}{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{\sin(2\pi-\text{x})}}{\text{e}^{\sin(2\pi-\text{x})}+\text{e}^{-\sin(2\pi-\text{x})}}\text{ dx}$ $\Bigg(\int\limits^\text{a}_0\text{f(x)}\text{dx}=\int\limits^\text{a}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg)$
$\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{-\sin\text{x}}}{\text{e}^{-\sin\text{x}}+\text{e}^{\sin\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{2\pi}_0\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{2\pi}_0$
$\Rightarrow2\text{I}=2\pi-0$
$\Rightarrow\text{I}=\pi$
View full question & answer→Question 1865 Marks
Evalute the following integrals:
$\int\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\text{dx}$
Putting x - b = t
⇒ x = b + t
& dx = dt
$\therefore\text{I}=\int\frac{\sin(\text{b}+\text{t}-\text{a})}{\sin\text{t}}\text{dt}$
$=\int\frac{\sin\big\{(\text{b}-\text{a})+\text{t}\big\}}{\sin\text{t}}\text{dt}$
$=\int\frac{\sin(\text{b}-\text{a})\cos\text{t}}{\sin\text{t}}+\int\frac{\cos(\text{b}-\text{a})\sin\text{t}}{\sin\text{t}}\text{dt}$
$=\int\sin(\text{b}-\text{a})\cot\text{t dt}+\int\cos(\text{b}-\text{a})\text{dt}$
$=\sin(\text{b}-\text{a})\text{in}|\sin\text{t}\big|+\text{t}\cos(\text{b}-\text{a})+\text{C}$
$=\sin(\text{b}-\text{a})\text{in}|\sin(\text{x}-\text{b})|+(\text{x}-\text{b})\cos(\text{b}-\text{a})+\text{C}$
$\big[\because\text{t}=\text{x}-\text{b}\big]$
View full question & answer→Question 1875 Marks
Integrate the function in Exercise:
$\frac{1}{\text{x}-\text{x}^{3}}$
Answer$\frac{1}{\text{x}-\text{x}^{3}}=\frac{1}{\text{x}(1-\text{x}^{2})}=\frac{1}{\text{x}(1-\text{x})(1+\text{x})}$ $\text{Let}\frac{1}{\text{x}(1-\text{x})(1+\text{x)}}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{(1-\text{x})}+\frac{\text{C}}{1+\text{x}}$ $\Rightarrow1=\text{A}(1-\text{x}^{2})+\text{B}\text{x}(1+\text{x})+\text{c}\text{x}(1-\text{x})$$\Rightarrow1=\text{A}-\text{A}\text{x}^{2}+\text{B}\text{x}+\text{B}\text{x}^{2}+\text{C}\text{x}-\text{C}\text{x}^{2}$
Equating the coefflclents of $\text{x}^{2},\text{x}$ and constant term, we obtain $-\text{A}+\text{B}-\text{C}=0$ $\text{B}+\text{C}=0$ $\text{A}=1$ on solving these equations, we obtain$\text{A}=1,\text{B}=\frac{1}{2},\text{and}\ \text{C}=\frac{1}{2}$
from eqution (1), we obtain $\frac{1}{\text{x}(1-\text{x})(1+\text{x})}=\frac{1}{\text{x}}+\frac{1}{2(1-\text{x})}-\frac{1}{2(1+\text{x})}$ $\Rightarrow\int\frac{1}{\text{x}(1-\text{x})(1+\text{x}}\text{dx}=\int\frac{1}{\text{x}}\text{dx}+\frac{1}{2}\int\frac{1}{1-\text{x}}\text{dx}-\frac{1}{2}\int\frac{1}{1+\text{x}}\text{dx}$ $=\log|\text{x}|-\frac{1}{2}\log|(1-\text{x})|-\frac{1}{2}\log|(1+\text{x)}|$ $=\log|\text{x}|-\log\bigg|(1-\text{x})^{\frac{1}{2}}\bigg|-\log|(1+\text{x)}^{\frac{1}{2}}\bigg|$ $=\log\left|\frac{\text{x}}{(1-\text{x)}^{\frac{1}{2}}(1+\text{x)}^{\frac{1}{2}}}\right|+\text{C}$ $=\log\left|\bigg(\frac{\text{x}^{2}}{1-\text{x}^{2}}\bigg)^{\frac{1}{2}}\right|+\text{C}$ $=\frac{1}{2}\log\left|\frac{\text{x}^{2}}{1-\text{x}}\right|+\text{C}$
View full question & answer→Question 1885 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^\text{b}_{\text{a}}\text{e}^{\text{x}}\text{ dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=\text{a},\text{ b}=\text{b},\text{ f(x)}=\text{e}^{\text{x}},\text{ h}=\frac{\text{b}-\text{a}}{\text{n}}$
Therefore, $\text{I}=\int\limits^\text{b}_{\text{a}}\text{e}^{\text{x}}\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\ ....\ +\text{f}\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{e}^\text{a}+\text{e}^{\text{a}+\text{h}}+\ .....\ +\text{e}^{\{\text{a}+(\text{n}-1)\text{h}\}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{e}^{\text{a}}\bigg\{\frac{(\text{e}^{\text{h}})^{\text{n}}-1}{\text{e}^{\text{h}}-1}\bigg\}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Big[\text{e}^\text{a}\frac{\text{e}^{\text{b}-\text{a}}-1}{\text{e}^{\text{h}}-1}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}=\Bigg[\frac{\text{e}^\text{b}-\text{e}^{\text{a}}}{\frac{\text{e}^\text{h}-1}{\text{h}}}\Bigg]$
$=\frac{\text{e}^\text{b}-\text{e}^\text{a}}{1}$
$=\text{e}^{\text{b}}-\text{e}^\text{a}$
View full question & answer→Question 1895 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\frac{(1-\text{x})^2}{(1+\text{x}^2)^2}\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\bigg[\frac{(1-\text{x})^2}{(1+\text{x}^2)^2}\bigg]\text{dx}$
$=\int\text{e}^\text{x}\bigg[\frac{1+\text{x}^2-2\text{x}}{(1+\text{x}^2)^2}\bigg]\text{dx}$
$=\int\text{e}^{\text{x}}\bigg[\frac{1}{1+\text{x}^2}-\frac{2\text{x}}{(1+\text{x}^2)^2}\bigg]\text{dx}$
Here, $\text{f(x)}=\frac{1}{1+\text{x}^2}$
$\Rightarrow\text{f}'(\text{x})=\frac{-2\text{x}}{(1+\text{x}^2)^2}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\frac{1}{1+\text{x}^2}=\text{t}$
Diff. both sides w.r.t w
$\text{e}^{\text{x}}\frac{1}{1+\text{x}^2}+\text{e}^{\text{x}}\frac{-1}{(1+\text{x}^2)^2}2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}\bigg[\frac{1}{1+\text{x}^2}-\frac{2\text{x}}{(1+\text{x}^2)^2}\bigg]\text{dx = dt}$
$\therefore\int\text{e}^{\text{x}}\bigg[\frac{1}{1+\text{x}^2}-\frac{2\text{x}}{(1+\text{x}^2)^2}\bigg]\text{dx}=\int\text{dt}$
$\Rightarrow\text{I}=\text{t}+\text{C}$
$=\frac{\text{e}^{\text{x}}}{1+\text{x}^2}+\text{C}$
View full question & answer→Question 1905 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2}{(\text{a}-\text{x}^2)^{\frac{3}{2}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{(\text{a}-\text{x}^2)^{\frac{3}{2}}}\text{ dx}$
Let $\text{x}=\text{a}\cos\theta$
On differentiating both sides, we get
$\text{dx}=-\text{a}\sin\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{\text{a}^2\cos^2\theta}{(\text{a}^2-\text{a}^2\cos^2\theta)^\frac{3}{2}}\times-\text{a}\sin\theta\text{ d}\theta$
$=-\int\frac{\text{a}^3\cos^2\theta\sin\theta}{\text{a}^3(1-\cos^2\theta)^{\frac{3}{2}}}\text{ d}\theta$
$=-\int\frac{\cos^2\theta\sin\theta}{\sin^3\theta}\text{ d}\theta$
$=-\int\cot^2\theta\text{ d}\theta$
$=-\int(\text{cosec}^2\theta-1)\text{d}\theta$
$=-(-\cot\theta-\theta)+\text{C}$
$=\cot\theta+\theta+\text{C}$
$=\cot\Big(\cos^{-1}\frac{\text{x}}{\text{a}}\Big)+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
$=\cot\Big(\cos^{-1}\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big)+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
$=\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
Hence, $\int\frac{\text{x}^2}{(\text{a}^2-\text{x}^2)^{\frac{3}{2}}}\text{ dx}=\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
View full question & answer→Question 1915 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\text{x}\log(1+2\text{x})\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\text{x}\log(1+2\text{x})\text{dx}$Apply integral by part
$\text{I}=\Big[\log(1+2\text{x})\frac{\text{x}^2}{2}\Big]^{1}_0-\int_{0}^\limits{1}\Big(\frac{2}{1+2\text{x}}\Big)\times\frac{\text{x}^2}{2}\text{ dx}$
$=\frac{1}{2}\big(\log3-0\big)-\int_{0}^\limits{1}\frac{\text{x}^2}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}\frac{4\text{x}^2-1+1}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}\frac{(2\text{x}+1)(2\text{x}-1)}{1+2\text{x}}\text{ dx}-\frac{1}{4}\int_{0}^\limits{1}\frac{2}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}(2\text{x}-1)\text{dx}-\frac{1}{4}\int_{0}^\limits{1}\frac{1}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\bigg[\frac{1}{4}\times\frac{(2\text{x}-1)^2}{2\times2}\bigg]^1_0-\bigg[\frac{1}{4}\times\frac{\log(1+2\text{x})}{2}\bigg]^1_0$
$=\frac{1}{2}\log3-\frac{1}{16}(1-1)-\frac{1}{8}\big(\log3-\log1\big)$
$=\frac{1}{2}\log3-0-\frac{1}{8}\log3$ $(\log1=0)$
$=\frac{3}{8}\log3$
View full question & answer→Question 1925 Marks
Evaluate the following:
$\int\frac{\sin^6\text{x}+\cos^6\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin^6\text{x}+\cos^6\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}$ $=\int\frac{(\sin^2\text{x})^3+(\cos^2\text{x})^3}{\sin^2\text{x}\cdot\cos^2\text{x}}\text{dx}$
$=\int\frac{(\sin^2\text{x}+\cos^2\text{x})(\sin^4\text{x}-\sin^2\text{x}\cos^2\text{x}+\cos^4\text{x})}{\sin^2\text{x}\cdot\cos^2\text{x}}\text{dx}$
$=\int\frac{\sin^4\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{dx}+\int\frac{\cos^4\text{x}}{\sin^2\text{x}\cdot\cos^2\text{x}}\text{dx}-\int\frac{\sin^2\text{x}\cos^2\text{x}}{\sin^2\text{x}\cdot\cos^2\text{x}}\text{dx}$
$=\int\tan^2\text{xdx}+\int\cos^2\text{xdx}-\int1\text{dx}$
$=\int(\sec^2\text{x}-1)\text{dx}+\int(\cos\text{ec}^2\text{x}-1)\text{dx}-\int1\text{dx}$
$=\int\sec^2\text{xdx}+\int\cos\text{ec}^2\text{xdx}-3\int\text{dx}$
$\text{I}=\tan\text{x}-\cot\text{x}-3\text{x}+\text{C}$
View full question & answer→Question 1935 Marks
$\int\text{x}(1-\text{x})^{23}\text{dx}$
AnswerLet $\text{I}=\int\text{x}(1-\text{x})^{23}\text{dx}$
Substituting 1 - x = t and dx = -dt, we get
$\text{I}=\int(1-\text{t})^{23}\text{dt}$
$=-\int(\text{t}^{23}-\text{t}^{24})\text{dt}$
$=-\int\Big(\frac{\text{t}^{24}}{24}-\frac{\text{t}^{25}}{25}\Big)+\text{C}$
$=\frac{\text{t}^{25}}{24}-\frac{\text{t}^{24}}{25}+\text{C}$
$=\frac{(1-\text{x})^{25}}{25}-\frac{(1-\text{x})^{24}}{24}+\text{C}$
$\therefore\ \text{I}=\frac{(1-\text{x})^{25}}{25}-\frac{(1-\text{x})^{24}}{24}+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\big[24(1-\text{x})-25\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\big[24-24\text{x}-25\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\big[-1-24\text{x}\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\times-\big[1+24\text{x}\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}(1+24\text{x})+\text{C}$
View full question & answer→Question 1945 Marks
Evaluate the following integrals:
$\int\cot^5\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\cot^5\text{x}\text{ dx}$ Then
$\text{I}=\int\cot^3\text{x}\times\big(\text{cosec}^2-1\big)\text{dx}$
$=\int\cot^3\text{x}\times\big(\text{cosec}^2\text{x}-1\big)\text{dx}$
$=\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}-\int\cot^3\text{x}\text{dx}$
$=\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}-\int\big(\text{cosec}^2\text{x}-1\big)\cot\text{dx}$
$=\int\cot^3\text{x}\text{cosec}^2\text{x}\text{ dx}-\int\text{cosec}^2\text{x}\cot\text{x}\text{ dx}+\int\cot\text{x}\text{ dx}$
$\text{I}=\int\cot^3\text{x}\text{cosec}^2\text{x}\text{ dx}-\int\text{cosec}^2\text{x}\cot\text{x}\text{ dx}+\int\cot\text{x}\text{ dx}$
Substituting $\cot\text{x}=\text{t}$ and $-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$ in first two integral, we get
$\text{I}=\int\text{t}^3(-\text{dt})-\int\text{t}\times(-\text{dt})+\int\cot\text{x}\text{ dx}$
$=-\frac{\text{t}^4}{4}+\frac{\text{t}^2}{2}+\log|\sin\text{x}|+\text{C}$
$=-\frac{1}{4}\cot^4\text{x}+\frac{1}{2}\cot^2\text{x}+\log|\sin\text{x}|+\text{C}$
$\therefore\ \text{I}=-\frac{1}{4}\cot^4\text{x}+\frac{1}{2}\cot^2\text{x}+\log|\sin\text{x}|+\text{C}$
View full question & answer→Question 1955 Marks
Evaluate the following definite integral as limit of sum:
$\int\limits^{\text{b}}_{\text{a}}\text{x dx}$
AnswerWe know that,$\int\limits^{b}_\text{a}\text{f}\big(\text{x}\big) \text{dx}=\lim\limits_{\text{h} \rightarrow 0}\text{h}\big[\text{f}(\text{a)}+\text{a}(\text{a+h})+\text{f}(\text{a+2h})+.....+\text{f}(\text{a}+\text+{(\text{n}-1)\text{h})}\big]\text{where nh = b - a}$Here, $\text{a = a, b = b}\text{ and } \text{f}(\text{x})=\text{x}$
$\therefore \int_\limits\text{a}^{b} \text{x} \text { dx}=\lim_\limits{\text{h}\rightarrow0} \text{h} \big[\text{a+(a+h)+(a+2h)+......+(a+(n-1)h}\big]$
$\Rightarrow \int_\limits{a}^{b} \text{x} \text{ dx}=\lim_\limits{\text{h}\rightarrow0} \text{h} \big[\text{na}+\text{h}(1+2+3+..........+(\text{n}-1))\big]$
$\Rightarrow \int_\limits{a}^{b}\text{x}\text{ dx}\lim\limits_{\text{h}\rightarrow0}\bigg[\text{anh}+\text{h}\frac{\text{n}\text{(n}-1)}{2}\bigg]=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{anh}+\frac{\text{nh}{\text{(nh-h)}}}{2}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{a}(\text{b}-\text{a})+\frac{(\text{b}-\text{a})(\text{b}-\text{a}-\text{h})}{2}\bigg][\because\text{nh}=\text{b}-\text{a}]$ $=\bigg[\text{a}\text{(b}-\text{a)}+\frac{\text{(b}-\text{a)}\text{(b}-\text{a)}}{2}\bigg]=\text{(b}-\text{a)}\bigg[\text{a}+\frac{\text{(b}-\text{a)}}{2}\bigg]=\text{(b}-\text{a)}\bigg[\frac{2\text{a}+\text{b}-\text{a}}{2}\bigg]$$=\frac{\text{(b}-\text{a)}\text{(b}+\text{a)}}{2}=\frac{\text{b}^{2}-\text{a}^{2}}{2} $
View full question & answer→Question 1965 Marks
Prove that:
$\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\frac{\pi}{2}\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
Answer$\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\int\limits^\pi_0(\pi-\text{x})\text{f}\big[\sin(\pi-\text{x})\big]\text{dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\int\limits^\pi_0(\pi-\text{x})\text{f}(\sin\text{x})\text{dx}$
$\Rightarrow\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\pi\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}-\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}$
$\Rightarrow2\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\pi\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
$\Rightarrow\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\frac{\pi}{2}\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
View full question & answer→Question 1975 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\sin\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\sin\text{x}\text{ dx}$
Apply integration by part.
$\text{I}=\big[\text{x}^2(-\cos\text{x})\big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}2\text{x}(-\cos\text{x})\text{ dx}$
$\Rightarrow\text{I}=(0-0)+2\int_{0}^\limits{\frac{\pi}{2}}\text{x}\cos\text{x}\text{ dx}$ $\Big(\cos\frac{\pi}{2}=0\Big)$
Apply integration by part again,
$\text{I}=0+2\Bigg[\big[\text{x}\sin\text{x}\big]_0^{\frac{\pi}{2}}-\int_{0}^\limits{\frac{\pi}{2}}1\times\sin\text{x}\text{ dx}\Bigg]$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}\sin\frac{\pi}{2}-0\Big)-2\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x dx}$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}-0\Big)-\big[2(-\cos\text{x})\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=\pi+2\Big(\cos\frac{\pi}{2}-\cos0\Big)$
$\Rightarrow\text{I}=\pi+2(0-1)$
$\Rightarrow\text{I}=\pi-2$
View full question & answer→Question 1985 Marks
Integrate the function in exercise.
$\text{x}\ \sin^{-1}\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\sin^{-1}\text{x dx}$
Taking $\sin^{-1}\text{x}$ as first function and x as second function and integrating by parts, we obtain.
$\text{I}=\sin^{-1}\text{x}\int\text{x} \ \text{dx}-\int\Bigg\{\Big(\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}\Big)\int\text{x} \ \text{dx}\Bigg\}\text{dx}$
$=\sin^{-1}\text{x}\Big(\frac{\text{x}^2}{2}\Big)-\int\frac{1}{\sqrt{1-\text{x}^2}}.\frac{\text{x}^2}{2}\text{dx}$
$=\sin^{-1}\text{x}\Big(\frac{\text{x}^2}{2}\Big)-\int\frac{1}{\sqrt{1-\text{x}^2}}.\frac{\text{x}^2}{2}+\text{dx}$
$=\frac{\text{x}^2\sin^{-1}\text{x}}{2}+\frac{1}{2}\int\frac{-\text{x}^2}{\sqrt{1-\text{x}^2}}\text{dx}$
$=\frac{\text{x}^2\sin^{-1}\text{x}}{2}+\frac{1}{2}\int\Bigg\{\frac{1-\text{x}^2}{\sqrt{1-\text{x}^2}}-\frac{1}{\sqrt{1-\text{x}^2}}\Bigg\}\text{dx}$
$=\frac{\text{x}^2\sin^{-1}\text{x}}{2}+\frac{1}{2}\int\Bigg\{\sqrt{1-\text{x}^2}-\frac{1}{\sqrt{1-\text{x}^2}}\Bigg\}\text{dx}$
$=\frac{\text{x}^2\sin^{-1}\text{x}}{2}+\frac{1}{2}\Bigg\{\int\sqrt{1-\text{x}^2}\text{dx}-\int\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}\Bigg\}$
$=\frac{\text{x}^2\sin^{-1}\text{x}}{2}+\frac{1}{2}\Bigg\{\frac{\text{x}}{2}\sqrt{1-\text{x}^2}+\frac{1}{2}\sin^{-1}\text{x}-\sin^{-1}\text{x}\Bigg\}+\text{C}$
$=\frac{1}{4}(2\text{x}^2-1)\sin^{-1}\text{x}+\frac{\text{x}}{4}\sqrt{1-\text{x}^2}+\text{C}$
View full question & answer→Question 1995 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^1_{0}\big(3\text{x}^2+5\text{x}\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=1,\text{ f(x)}=3\text{x}^2+5\text{x},\text{ h}=\frac{1-0}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^1_{0}\big(3\text{x}^2+5\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(0+0)+(3\text{h}^2+5\text{h})+\ \\ .....\ +\big\{3(\text{n}-1)^2\text{h}^2+5(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[5\text{h}(1+2+\ ....\ +\text{n})\\+3\text{h}^2\big\{1^2+2^2+3^2\ ....\ +(\text{n}-1)^2\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[5\text{h}\frac{\text{n}(\text{n}-1)}{2}+\text{h}^2\frac{3\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[\frac{5(\text{n}-1)}{2}+\frac{(\text{n}-1)(2\text{n}-1)}{2\text{n}}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg[\frac{5}{2}\Big(1-\frac{1}{\text{n}}\Big)+\frac{1}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg]$
$=\frac{5}{2}+1$
$=\frac{7}{2}$
View full question & answer→Question 2005 Marks
Evaluate the following integrals:
$\int\limits^{\infty}_0\frac{\log\text{x}}{1+\text{x}^2}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^{\infty}_0\frac{\log\text{x}}{1+\text{x}^2}\text{ dx}$
Putting $\text{x}=\tan\theta$
$\text{dx}=\sec^2\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$
And $\text{x}\rightarrow\infty;\theta\rightarrow\frac{\pi}{2}$
Now, integral becomes,
$\therefore\ \text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\log(\tan\theta)}{1+\tan^2\theta}\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log(\tan\theta)\text{d}\theta\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\Big[\tan\Big(\frac{\pi}{2}-\theta\Big)\Big]\text{d}\theta$ $\Bigg[\because\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log(\cot\theta)\text{d}\theta\ ....(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\log(\tan\theta)\text{d}\theta+\int\limits^{\frac{\pi}{2}}_0\log(\cot\theta)\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\big[\log(\tan\theta)+\log(\cot\theta)\big]\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\big[\log(\tan\theta\times\cot\theta)\big]\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\big(\log1\big)\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0(0)\text{d}\theta$
$\Rightarrow2\text{I}=0$
$\Rightarrow\text{I}=0$
$\therefore\ \int\limits^{\infty}_0\frac{\log\text{x}}{1+\text{x}^2}\text{ dx}=0$
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