Question 1513 Marks
Integrate the following integrals:
$\int\cos3\text{x}\cos4\text{x dx}$
AnswerLet I $=\int\cos3\text{x}\cos4\text{x dx}.$ Then,
$\text{I}=\frac{1}{2}\int(2\cos3\text{x}\cos4\text{x})\text{dx}$
$=\frac{1}{2}\int(\cos7\text{x}+\cos(-\text{x}))\text{dx}$
$=\frac{1}{2}\int\cos7\text{x}+\frac{1}{2}\int\cos\text{dx}$ $[\because\cos(-0)=\cos0]$
$=\frac{\sin7\text{x}}{2\times7}+\frac{\sin\text{x}}{2}+\text{C}$
$=\frac{1}{14}\times\sin7\text{x}+\frac{1}{2}\sin\text{x}+\text{C}$
$\therefore\text{I}=\frac{1}{14}\times\sin7\text{x}+\frac{1}{2}\times\sin\text{x}+\text{C}$
View full question & answer→Question 1523 Marks
Evaluate the following integrals:
$\int\sqrt{16\text{x}^2+25}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{16\text{x}^2+25}\text{dx}$
$=\int\sqrt{(4\text{x})^2+5^2}\text{dx}$
$=4\int\sqrt{\text{x}^2+\Big(\frac{5}{4}\Big)^2}\text{dx}$
$=4\begin{Bmatrix}\frac{\text{x}}{2}\sqrt{\text{x}^2+\Big(\frac{5}{4}\Big)^2}+\frac{\big(\frac{5}{4}\big)^2}{2}\log\Bigg|\text{x}+\sqrt{\text{x}^2+\Big(\frac{5}{4}\Big)^2}\Bigg|+\text{C}\end{Bmatrix}$
$\therefore\ \text{I}=2\text{x}\sqrt{\text{x}^2+\frac{25}{16}}+\frac{25}{8}\log\bigg|\text{x}+\sqrt{\text{x}^2+\frac{25}{16}}\bigg|+\text{C}$
View full question & answer→Question 1533 Marks
Evaluate the following integrals:
$\int_{1}^\limits{3}\frac{\cos(\log\text{x})}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{3}\frac{\cos(\log\text{x})}{\text{x}}\text{ dx}$
Let $\log\text{x}=\text{t}$ Then, $\frac{1}{\text{x}}\text{ dx}=\text{dt}$
When $\text{x}=1,\text{t}=0$ and $\text{x}=3,\text{t}=\log3$
$\therefore\ \text{I}=\int_{0}^\limits{\log3}\cos\text{t dt}$
$=\big[\sin\text{t}\big]^{\log3}_0$
$=\sin(\log3)$
View full question & answer→Question 1543 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{1+\cos\text{x}}}\text{dx}$
AnswerWe have,
$\int\frac{1}{\sqrt{1+\cos\text{x}}}\text{dx}$
$\int\frac{1}{\sqrt{2\cos^2\frac{\text{x}}{2}}}\text{dx}$
$=\int\frac{1}{\sqrt{2}\cos\frac{\text{x}}{2}}\text{dx}$
$=\frac{1}{\sqrt{2}}\int\sec\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{\sqrt{2}}\int\text{cosec}\Big(\frac{\pi}{2}+\frac{\text{x}}{2}\Big)\text{dx}$
$=\frac{2}{\sqrt{2}}\log\Big|\tan\Big(\frac{\pi}{4}+\frac{\text{x}}{4}\Big)\Big|+\text{C}$
$\because\int\frac{1}{\sqrt{1+\cos\text{x}}}\text{dx}=\sqrt{2}\log\Big|\tan\Big(\frac{\pi}{4}+\frac{\text{x}}{4}\Big)\Big|+\text{C}$
View full question & answer→Question 1553 Marks
Evaluate the following definite integrals:
$\int_{1}^\limits{3}\frac{\log\text{x}}{(\text{x}+1)^2}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{3}\frac{\log\text{x}}{(\text{x}+1)^2}\text{ dx}$ Then,
$\text{I}=\Big[\frac{-1}{1+\text{x}}\log\text{x}\Big]^3_1-\int_{1}^\limits{3}\Big(\frac{1}{\text{x}+1}\Big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{-1}{1+\text{x}}\log\text{x}\Big]^3_1+\int_{1}^\limits{3}\frac{1}{\text{x}(\text{x}+1)}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{-1}{1+\text{x}}\log\text{x}\Big]^3_1+\int_{1}^\limits{3}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}+1}\Big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{-1}{1+\text{x}}\log\text{x}\Big]^3_1+\big[\log\text{x}-\log(\text{x}+1)\big]^3_1$
$\Rightarrow\text{I}=\frac{-1}{4}\log3+\log3-\log4+\log2$
$\Rightarrow\text{I}=\frac{3}{4}\log3-\log2$
View full question & answer→Question 1563 Marks
Evaluate the following:
$\int\limits^1_0\frac{\text{x}}{\sqrt{1+\text{x}^2}}\text{dx}$
AnswerLet $\text{I}=\int\limits^1_0\frac{\text{x}}{\sqrt{1+\text{x}^2}}\text{dx}$
Put $1+\text{x}^2=\text{t}^2$
$\Rightarrow\ 2\text{xdx}=2\text{tdt}$
$\Rightarrow\ \text{xdx}=\text{tdt}$
As $\text{x}\rightarrow0,$ then $\text{t}\rightarrow1$
and $\text{x}\rightarrow\pi,$ then $\text{t}\rightarrow\sqrt{2}$
Substituting $1+\text{x}^2=\text{t}^2$ and $\text{xdx}=\text{tdt}$ in I, we get
$\therefore\ \text{I}=\int\limits^{\sqrt{2}}_0\frac{\text{tdt}}{\text{t}}$
$=[\text{t}]^{\sqrt{2}}_1=\sqrt{2}-1$
View full question & answer→Question 1573 Marks
Evaluate the following integrals:
$\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Answer$\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$ Let $\sqrt{\text{x}}=\text{t}$ $\Rightarrow\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\frac{\text{dx}}{\sqrt{\text{x}}}=2\text{dt}$ Now, $\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$$=2\int\sec^2\text{t dt}$
$=2\tan(\text{t})+\text{C}$
$=2\tan\big(\sqrt{\text{x}}\big)+\text{C}$
View full question & answer→Question 1583 Marks
Write a value of $\int\frac{\text{a}^{\text{x}}}{3+\text{a}^{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{a}^{\text{x}}}{3+\text{a}^{\text{x}}}\text{ dx}$
Let $3+\text{a}^{\text{x}}=\text{t}$
$\text{a}^{\text{x}}\log\text{a dx}=\text{dt}$
$\text{a}^{\text{x}}\text{dx}=\frac{\text{dt}}{\log\text{a}}$
$\text{I}=\int\frac{\text{dt}}{\log\text{a}\cdot\text{t}}=\frac{1}{\log\text{a}}\log\text{t}+\text{C}$
$\text{I}=\frac{1}{\log\text{a}}\log(3+\text{a}^{\text{x}})+\text{C}$
View full question & answer→Question 1593 Marks
Evalute the following integrals:
$\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$$=\int\frac{\cos^2\text{x}-\sin^2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}+\sin\text{x}}\text{dx}$
Putting $\cos\text{x}+\sin\text{x}=\text{t}$
$\Rightarrow-\sin\text{x}+\cos\text{x}=\frac{\text{dt}}{\text{dt}}$
$\Rightarrow(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}|\cos\text{x}+\sin\text{x}|+\text{C}\ \big[\because\text{t}=\cos\text{x}+\sin\text{x}\big]$
View full question & answer→Question 1603 Marks
Evaluate the following integrals:
$\int\limits^\pi_0\cos\text{x}|\cos\text{x}|\text{dx}$
AnswerConsider $\text{f(x)}=\cos\text{x}|\cos\text{x}|$
Now,
$\text{f}(\pi-\text{x})=\cos(\pi-\text{x})|\cos(\pi-\text{x})|$
$=-\cos\text{x}|-\cos\text{x}|=-\cos\text{x}|\cos\text{x}|$
$=-\text{f(x)}$
$\therefore\ \int\limits^\pi_0\cos\text{x}|\cos\text{x}|\text{dx}=0$ $\begin{bmatrix}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx},&\text{ if }\text{ f}(2\text{a}-\text{x})=\text{f(x)}\\0,&\text{ if }\text{ f}(2\text{a}-\text{x})=\text{f(x)}\end{cases}\end{bmatrix}$
View full question & answer→Question 1613 Marks
Evaluate the following integrals:
$\int\big\{\sqrt{\text{x}}\big(\text{ax}^2+\text{bx}+\text{c}\big)\big\}\text{dx}$
Answer$\int\sqrt{\text{x}}\Big(\text{ax}^2+\text{bx}+\text{c}\Big)\text{dx}$
$=\int\text{x}^{\frac{1}{2}}\Big(\text{ax}^2+\text{bx}+\text{c}\Big)\text{dx}$
$=\int\Big(\text{ax}^{2+\frac{1}{2}}+\text{bx}^{\frac{1}{2}+1}+\text{cx}^{\frac{1}{2}}\Big)\text{dx}$
$=\text{a}\int\text{x}^{\frac{5}{2}}\text{dx}+\text{b}\int\text{x}^{\frac{3}{2}}\text{dx}+\text{c}\int\text{x}^{\frac{1}{2}}\text{dx}$
$=\text{a}\begin{bmatrix}\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}\end{bmatrix}+\text{b}\begin{bmatrix}\frac{\text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\end{bmatrix}+\text{c}\begin{bmatrix}\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\end{bmatrix}+\text{C}$
$=\frac{2\text{a}}{7}\text{x}^{\frac{7}{2}}+\frac{2\text{b}}{5}\text{x}^{\frac{3}{2}}+\frac{2\text{c}}{3}\text{x}^{\frac{3}{2}}+\text{C}$
View full question & answer→Question 1623 Marks
Evaluate the following integrals:
$\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
Answer$\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
$\text{Let, }\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\text{Now,}\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
$=\int\text{t}^\frac{3}{2}\text{dt}$
$=\Bigg[\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]+\text{C}$
$=\frac{2}{5}\text{t}^\frac{5}{2}+\text{C}$
$=\frac{2}{5}\tan^\frac{5}{2}\text{x}+\text{C}$
View full question & answer→Question 1633 Marks
Evaluate the following integrals:$\int\frac{\cos2\text{x}}{\sqrt{\sin^22\text{x}+8}}\text{ dx}$
Answer$\int\frac{\cos(2\text{x}).\text{dx}}{\sqrt{\sin^22\text{x}+8}}$
Let $\sin(2\text{x})=\text{t}$
$\Rightarrow\cos(2\text{x})\times2.\text{dx}=\text{dt}$
$\Rightarrow\cos(2\text{x}).\text{dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\cos(2\text{x}).\text{dx}}{\sqrt{\sin^22\text{x}+8}}$
$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}^2+\big(2\sqrt2\big)^2}}$
$=\frac{1}{2}\log\Big|\text{t}+\sqrt{\text{t}^2+8}\Big|+\text{C}$
$=\frac{1}{2}\log\Big|\sin(2\text{x})+\sqrt{\sin^2(2\text{x})+8}\Big|+\text{C}$
View full question & answer→Question 1643 Marks
Evaluate the following integrals:
$\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{dx}$
Answer$\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}=\frac{\cos\text{x}-\sin\text{x}}{(\sin^2\text{x}+\cos^2\text{x})+2\sin\text{x}\cos\text{x}}$
$[\sin^2\text{x}+\cos^2\text{x}=1;\ \sin2\text{x}=2\sin\text{x}\cos\text{x}]$
$=\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}$
$\sin\text{x}+\cos\text{x}=\text{t}$
$\therefore(\sin\text{x}+\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{dx}=\int\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}\text{dx}$
$=\int\frac{\text{dt}}{\text{t}^2} $
$=\int\text{t}^{-2}\text{dt}$
$=-\text{t}^{-1}+\text{C}$
$=-\frac{1}{\text{t}}+\text{C}$
$=\frac{-1}{\sin\text{x}+\cos\text{x}}+\text{C}$
View full question & answer→Question 1653 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{-\frac{\pi}{2}}{\sqrt{\cos\text{x}\sin^2\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{-\frac{\pi}{2}}{\sqrt{\cos\text{x}\sin^2\text{x}}}\text{ dx}$
$=-\frac{\pi}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{1}{\sqrt{\cos\text{x}\sin^2\text{x}}}\text{ dx}$
$=-\frac{\pi}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{1}{\sqrt{\cos\text{x}}\ |\sin\text{x}|}\text{ dx}$
$=-\frac{\pi}{2}\times2\int\limits^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\cos\text{x}}\ |\sin\text{x}|}\text{ dx}$ $\Big[\text{f}(-\text{x})=\sqrt{\cos(-\text{x})}\ |\sin(-\text{x})|=\sqrt{\cos\text{x}}\ |-\sin\text{x}|=\sqrt{\cos\text{x}}\ |\sin\text{x}|=\text{f(x)}\Big]$
$=-\pi\int\limits^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\cos\text{x }}\sin\text{x}}\text{ dx}$ $\Big(|\sin\text{x}|=\sin\text{x},0\leq\text{x}\leq\frac{\pi}{2}\Big)$
$=-\pi\int\limits^{\frac{\pi}{2}}_{0}\frac{\sin\text{x}}{\sqrt{\cos\text{x }}(1-\cos^2\text{x})}\text{ dx}$
Put $\cos\text{x}=\text{z}^2$
$\therefore\ -\sin\text{x dx}=2\text{zdz}$
When $\text{x}\rightarrow0,\text{z}\rightarrow1$
When $\text{x}\rightarrow\frac{\pi}{2},\text{z}\rightarrow0$
$\therefore\ \text{I}=2\pi\int\limits^0_1\frac{\text{zdz}}{\text{z}(1-\text{z}^4)}$
$=2\pi\int\limits^0_1\frac{\text{dz}}{(1-\text{z}^4)}$
$=2\pi\int\limits^0_1\frac{\text{dz}}{(1-\text{z})(1+\text{z})(1+\text{z}^2)}$
Now,
$\frac{1}{(1-\text{z})(1+\text{z})(1+\text{z}^2)}=\frac{\text{A}}{1-\text{z}}+\frac{\text{B}}{1+\text{z}}+\frac{\text{Cz}+\text{D}}{1+\text{z}^2}$
$1=\text{A}(1+\text{z})(1+\text{z}^2)+\text{B}(1-\text{z})(1+\text{z}^2)+(\text{Cz}+\text{D})(1-\text{z})(1+\text{z})$
Putting z = 1, we get
$\text{A}=\frac{1}{4}$
Putting z = -1
$\text{B}=\frac{1}{4}$
Putting z = 0
$1=\text{A}+\text{B}+\text{D}$
$\text{D}=1-\frac{1}{4}-\frac{1}{4}=\frac{1}{2}$
$\text{D}=\frac{1}{2}$
Equating coefficient of $z^3$ on both side, we get
$\text{A}-\text{B}+\text{C}=0$
$\frac{1}{4}-\frac{1}{4}+\text{C}=0$
$\text{C}=0$
$\therefore\ \text{I}=2\pi\int\limits^0_1\frac{\text{dz}}{(1-\text{z})(1+\text{z})(1+\text{z}^2)}$
$=2\pi\int\limits^0_1\frac{\frac{1}{4}}{1-\text{z}}\text{dz}+2\pi\int\limits^0_1\frac{\frac{1}{4}}{1+\text{z}}\text{dz}+2\pi\int\limits^0_1\frac{\frac{1}{2}}{1+\text{z}^2}\text{dz}$
$=\frac{2\pi}{4}\times\bigg[\frac{\log(1-\text{z})}{-1}\bigg]^0_1+\frac{2\pi}{4}\times\big[\log(1+\text{z})\big]^0_1+\frac{2\pi}{2}\times\big[\tan^{-1}\text{z}\big]^0_1$
$=-\frac{\pi}{2}\big(\log1-\log0\big)+\frac{\pi}{2}\big(\log1-\log2\big)+\pi\big(\tan^{-1}0-\tan^{-1}1\big)$
$=-\frac{\pi}{2}\big[0-(-\infty)\big]+\frac{\pi}{2}(0-\log2)+\pi\Big(0-\frac{\pi}{4}\Big)$
$=-\infty-\frac{\pi}{2}\log2-\frac{\pi^2}{4}$
$=-\infty$
View full question & answer→Question 1663 Marks
$\int\frac{1}{1-\sin\frac{\text{x}}{2}}\text{dx}$
Answer$\int\frac{\text{dx}}{1-\sin\big(\frac{\text{x}}{2}\big)}$
$=\int\frac{\big(1+\sin\frac{\text{x}}{2}\big)}{\big(1-\sin\frac{\text{x}}{2}\big)\big(1+\sin\frac{\text{x}}{2}\big)}\text{dx}$
$=\int\bigg(\frac{1+\sin\frac{\text{x}}{2}}{1-\sin^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\bigg(\frac{1+\sin\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\big(\sec^2\frac{\text{x}}{2}+\sec\frac{\text{x}}{2}\tan\frac{\text{x}}{2}\big)\text{dx}$
$=\frac{\tan\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}+\frac{\sec\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}+\text{c}$
$=2\big(\tan\frac{\text{x}}{2}+\sec\frac{\text{x}}{2}\big)+\text{c}$
View full question & answer→Question 1673 Marks
Evaluate the integral in Exercise:
$\int\limits^{\frac{\pi}{2}}_{0}\sqrt{\sin\phi}\cos^{5}\phi\text{ d }\phi$
Answer$\text{Let}\ \text{I}=\int\limits^\frac{\pi}{2}_{0}\sqrt{\sin\phi}\cos^{5}\phi\ \text{d}\ \phi$
$\text{putting}\ \sin\phi=\text{t}\ \Rightarrow\ \cos\phi=\frac{\text{dt}}{\text{d}\phi}\ \Rightarrow\ \cos\phi\ \text{d}\ \phi=\text{dt}$
To change the limits of integration from $\phi \text{to}\ \text{t}$
$\text{when}\phi=0,\text{t}=\sin\phi=\sin0^{\text{o}}=0$
$\text{when}\phi=\frac{\pi}{2},\text{t}=\sin\phi=\sin\frac{\pi}{2}=1$
$\therefore$ From eq.(i),$\ \text{I}=\int\limits_{0}^\frac{\pi}{2}\sqrt{\sin}\phi\cos^{5}\phi\cos\phi\ \text{d}\ \phi=\int^\frac{\pi}{2}_{0}\sqrt{\sin\phi}(\cos^{2}\phi)^{2}\cos\phi\ \text{d}\ \phi$
$=\int\limits_{0}^\frac{\pi}{2}\sqrt{\sin}\phi(1-\sin^{2}\phi)^{2}\cos\phi\ \text{d}\ \phi=\int\limits_{0}^{1}\sqrt{\text{t}}(1-\text{t}^{2})^{2}\text{dt}=\int\limits_{0}^{1}\text{t}^\frac{1}{2}(1+\text{t}^{4}-2\text{t}^{2})\text{dt}$
$=\int\limits_{0}^{1}\bigg(\text{t}^\frac{1}{2}+\text{t}^{\frac{1}{2}+4}-2\text{t}^{\frac{1}{2}+2}\bigg)\text{dt}=\int\limits_{0}^{1}\bigg(\text{t}^\frac{1}{2}+\text{t}^\frac{9}{2}-2\text{t}^\frac{5}{2}\bigg)\text{dt}=\int\limits_{0}^{1}\text{t}^\frac{1}{2}\text{dt}+\int\limits_{0}^{1}\text{t}^\frac{9}{2}\text{dt}-2\int\limits_{0}^{1}\text{t}^\frac{5}{2}\text{dt}$
$=\frac{\bigg(\text{t}^\frac{3}{2}\bigg)^{1}_{0}}{\frac{3}{2}}+\frac{\bigg(\text{t}^\frac{11}{2}\bigg)^{1}_{0}}{\frac{11}{2}}+\frac{\bigg(\text{t}\frac{7}{2}\bigg)^{1}_{0}}{\frac{7}{2}}=\frac{2}{3}(1-0)+\frac{2}{11}(1-0)-\frac{4}{7}(1-0)=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$
$=\frac{154+42-132}{231}=\frac{64}{231}$
View full question & answer→Question 1683 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_{\frac{-\pi}{2}}\big\{\sin|\text{x}|+\cos|\text{x}|\big\}\text{dx}$
AnswerWe know,
$\text{I}=\int^\limits{\frac{\pi}{2}}_{\frac{-\pi}{2}}\big\{\sin|\text{x}|+\cos|\text{x}|\big\}\text{dx}$
Let $\text{f(x)}=\sin|\text{x}|+\cos|\text{x}|$
Then, $\text{f(x)}=\text{f(-x)}$
$\therefore\ \text{f(x)}$ is an even function.
So,
$\text{I}=\int^\limits{\frac{\pi}{2}}_{\frac{-\pi}{2}}\big\{\sin|\text{x}|+\cos|\text{x}|\big\}\text{dx}$
$=2\int\limits^{\frac{\pi}{2}}_0(\sin\text{x}+\cos\text{x})\text{dx}$
$\\=2\big[-\cos\text{x}+\sin\text{x}\big]^{\frac{\pi}{2}}_0$
$=4$
View full question & answer→Question 1693 Marks
Evalute the following integrals:
$\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{dx}$
AnswerNote: Here we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{dx}$
Putting $\log\sin\text{x}=\text{t}$
$\Rightarrow\cot\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cot\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\log\sin\text{x}|+\text{C}\ \big[\because\text{t}=\log\sin\text{x}\big]$
View full question & answer→Question 1703 Marks
Integrate the function in Exercise:
$\frac{1}{\text{x}\sqrt{\text{ax}-\text{x}^{2}}}$
$\big[\text{Hint:putx}=\frac{\text{a}}{\text{t}}\big]$
Answer$\frac{1}{\text{x}\sqrt{\text{ax}-\text{x}^{2}}}$
$\text{Let}\ \text{x}=\frac{\text{a}}{\text{t}}\Rightarrow\text{dx}=-\frac{\text{a}}{\text{t}^{2}}\text{dt}$
$\Rightarrow\int\frac{1}{\text{x}\sqrt{\text{ax}-\text{x}^{2}}}\text{dx}=\int\frac{1}{\frac{\text{a}}{\text{t}}\sqrt{\text{a}.\frac{\text{a}}{\text{t}}-\big(\frac{\text{a}}{\text{t}}}\big)^{2}}\Big(-\frac{\text{a}}{\text{t}^{2}}\text{dt}\Big)$
$=-\int\frac{1}{\text{at}}.\frac{1}{\sqrt{\frac{1}{\text{t}}-\frac{1}{\text{t}^{2}}}}\text{dt}$
$=-\frac{1}{\text{a}}\int\frac{1}{\sqrt{\frac{\text{t}^{2}}{\text{t}}-\frac{\text{t}^{2}}{\text{t}^{2}}}}\text{dt}$
$=-\frac{1}{\text{a}}\int\frac{1}{\sqrt{\text{t}-1}}\text{dt}$
$=-\frac{1}{\text{a}}\big[2\sqrt{\text{t}-1}\big]+\text{C}$
$=-\frac{1}{\text{a}}\bigg[2\sqrt{\frac{\text{a}}{\text{x}}-1}\bigg]+\text{C}$
$=-\frac{2}{\text{a}}\bigg(\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{x}}}\bigg)+\text{C}$
$=-\frac{2}{\text{a}}\bigg(\sqrt{\frac{\text{a}-\text{x}}{\text{x}}}\bigg)+\text{C}$
View full question & answer→Question 1713 Marks
Evalute the following integrals:
$\int\frac{2\cos2\text{x}+\sec^2\text{x}}{\sin2\text{x}+\tan\text{x}-5}\text{dx}$
AnswerLet $\text{I}=\int\frac{2\cos2\text{x}+\sec^2\text{x}}{\sin2\text{x}+\tan\text{x}-5}\text{dx}$
Putting $\sin2\text{x}+\tan\text{x}-5=\text{t}$
$\Rightarrow2\cos2\text{x}+\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(2\cos2\text{x}+\sec^2\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}|\sin2\text{x}+\tan\text{x}-5|+\text{C} $ $\big[\because\text{t}=\sin2\text{x}+\tan\text{x}-5\big]$
View full question & answer→Question 1723 Marks
Evaluate the following integrals:
$\int\frac{\text{dx}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
Answer$\int\frac{\text{dx}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
$=\int\frac{\text{dx}}{\text{e}^{\text{x}}+\frac{1}{\text{e}^{\text{x}}}}$
$=\int\frac{\text{e}^{\text{x}}\text{ dx}}{\text{e}^{2\text{x}}+1}$
Let $\text{e}^{\text{x}}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\text{ dx = dt}$
Now, $\int\frac{\text{e}^{\text{x}}\text{ dx}}{\text{e}^{2\text{x}}+1}$
$=\int\frac{\text{dt}}{1+\text{t}^2}$
$=\tan^{-1}(\text{t})+\text{C}$
$=\tan^{-1}(\text{e}^{\text{x}})+\text{C}$
View full question & answer→Question 1733 Marks
Evaluate the following integrals:
$\int\sqrt{2\text{ax}-\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{2\text{ax}-\text{x}^2}\text{dx}$
$=\int\sqrt{\text{a}^2+2\text{ax}-\text{x}^2-\text{a}^2}\text{dx}$
$=\int\sqrt{\text{a}^2-(\text{x}^2-2\text{ax}+\text{a}^2)}\text{dx}$
$=\int\sqrt{\text{a}^2-(\text{x}-\text{a})^2}\text{dx}$
$=\Big(\frac{\text{x}-\text{a}}{2}\Big)\sqrt{2\text{ax}-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\Big(\frac{\text{x}-\text{a}}{\text{a}}\Big)+\text{C}$
View full question & answer→Question 1743 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^{\frac{3}{2}}}\text{dx}$
Answer$\int\frac{1}{\text{x}^\frac{3}{2}}\text{dx}=\int\text{x}\frac{-3}{2}\text{dx}$
$=\int\text{x}^\frac{-3}{2}\text{dx}$
$=\frac{\text{x}^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+\text{c}$
$=\frac{\text{x}^\frac{-1}{2}}{\frac{-1}{2}}+\text{c}$
$=-2\text{x}\frac{1}{\sqrt{\text{x}}}+\text{c}$
$=\frac{-2}{\sqrt{\text{x}}}+\text{c}$
View full question & answer→Question 1753 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\log\tan\text{x dx}$
AnswerLet, $\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\tan\text{x dx}\ ....(\text{i})$ $=\int\limits^{\frac{\pi}{2}}_0\log\tan\Big(\frac{\pi}{2}-\text{x}\Big)\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$ $=\int\limits^{\frac{\pi}{2}}_0\log\cot\text{x dx}\ ...(\text{ii})$ Adding (i) and (ii) we get $2\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\tan\text{x dx}+\int\limits^{\frac{\pi}{2}}_0\log\cot\text{x dx}$ $=\int\limits^{\frac{\pi}{2}}_0\log(\tan\text{x}\cdot\cot\text{x})\text{ dx}$ $=\int\limits^{\frac{\pi}{2}}_0\log1\text{ dx}=0$Hence, $\text{I}=0$
View full question & answer→Question 1763 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{(1-\text{x}^2)\big\{9+\big(\sin^{-1}\text{x}\big)^2\big\}}}\text{ dx}.$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{(1-\text{x}^2)\Big[9+\big(\sin^{-1}\text{x}\big)^2\Big]}}\text{ dx}$
Let $\sin^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$
$\Rightarrow\text{I}=\int\frac{\text{dt}}{\sqrt{(3)^2+\text{t}^2}}$
$\Rightarrow\text{I}=\log\Big|\text{t}+\sqrt{9+\text{t}^2}\Big|+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\log\Big|\text{x}+\sqrt{\text{a}^2+\text{x}^2}\Big|+\text{C}\Big]$
$\text{I}=\log\Big|\sin^{-1}\text{x}+\sqrt{9+\big(\sin^{-1}\text{x}\big)^2}\Big|+\text{C}$
View full question & answer→Question 1773 Marks
If $\int\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}=\text{f(x)}\text{e}^{\text{x}}+\text{C},$ then write the value of f(x).
Answer$\int\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}=\int\Big(\frac{\text{x}}{\text{x}^2}-\frac{1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{dx}$
$=\int\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$
Consider, $\text{f(x)}=\frac{1}{\text{x}},$ then $\text{f}'(\text{x})=-\frac{1}{\text{x}^2}$
Thus, the given integrand is of the from $\text{e}^{\text{x}}\big|\text{f}(\text{x})+\text{f}'(\text{x})\big|$
Therefore, $\int\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}=\frac{1}{\text{x}}\text{e}^{\text{x}}+\text{C}$
Hence, $\text{f(x)}=\frac{1}{\text{x}}$
View full question & answer→Question 1783 Marks
Evaluate the following integrals:$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
Answer$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
Let $\text{t}=\cos^2\text{x}\rightarrow-\text{dt}=2\cos\text{x}\sin\text{x}\text{ dx}$
$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
$=\int\frac{-1}{\sqrt{\text{t}^2-(1-\text{t})+2}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+1}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+\frac{1}{4}+\frac{3}{4}}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\big(\text{t}+\frac{1}{2}\big)^2+\frac{3}{4}}}\text{ dt}$
$=-\log\Big|\Big(\text{t}+\frac{1}{2}\Big)+\sqrt{\text{t}^2+\text{t}+1}\Big|$
$=-\log\Big|\Big(\cos^2\text{x}+\frac{1}{2}\Big)+\sqrt{\cos^4\text{x}+\cos^2\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 1793 Marks
Write a value of $\int\frac{\sin2\text{x}}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
Let $\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}=\text{t}$
$\big(2\text{a}^2\sin\text{x}\cos\text{x}-2\text{b}^2\cos\text{x}\sin\text{x}\big)\text{dx}=\text{dt}$
$2(\text{a}^2-\text{b}^2)\sin\text{x}\cos\text{x dx}=\text{dt}$
$(\text{a}^2-\text{b}^2)\sin2\text{x}\text{ dx}=\text{dt}$
$\text{I}=\frac{1}{\text{a}^2-\text{b}^2}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{\text{a}^2-\text{b}^2}\log\text{t}+\text{C}$
$\text{I}=\frac{1}{\text{a}^2-\text{b}^2}\log\big(\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}\big)+\text{C}$
View full question & answer→Question 1803 Marks
Evalute the following integrals:
$\int\frac{1}{\text{x}(3+\log\text{x})}\text{dx}$
AnswerHere, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.Let $\text{I}=\int\frac{1}{\text{x}(3+\log\text{x})}\text{dx}$
Putting $\log\tan\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{dx}}{\text{x}}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{3+\text{t}}$
$=\log|3+\text{t}|+\text{C}$
$=\log|3+\log\text{x}|+\text{C}\ \big[\because\text{t}=\log\text{x}\big]$
View full question & answer→Question 1813 Marks
Evaluate the following integrals:
$\int\tan\text{x }\sec^4\text{x}\text{dx}$
AnswerLet $\text{I}=\int\tan\text{x }\sec^4\text{x}\text{dx}$ Then
$\text{I}=\int\tan\text{x }\sec^2\text{x}\sec^2\text{x}\text{dx}$
$=\int\tan\text{x}(1+\tan^2\text{x})\sec^2\text{x}\text{dx}$
$\text{I}=\int\big(\tan\text{x}+\tan^3\text{x}\big)\sec^2\text{x}\text{dx}$
Substituting $\tan\text{x}=\text{t}$ and $\sec^2\text{xdx}=\text{dt},$ we get
$\text{I}=\int(\text{t}+\text{t}^3)\text{dt}$
$=\frac{\text{t}^2}{2}+\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\tan^2\text{x}}{2}+\frac{\tan^4}{4}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\times\tan^2\text{x}+\frac{1}{4}\times\tan^4\text{x}+\text{C}$
View full question & answer→Question 1823 Marks
Integrate the function in Exercise:
$\frac{1}{\sqrt{8+3\text{x}-\text{x}^2}}$
Answer$\int\frac{1}{\sqrt{8+3\text{x}-\text{x}^2}}\text{ dx}$
$=\int\frac{1}{\sqrt{-\text{x}^2+3\text{x}+8}}\text{ dx}$
$=\int\frac{1}{\sqrt{-\big(\text{x}^2-3\text{x}-8\big)}}\text{ dx}$
$=\int\frac{1}{\sqrt{-\bigg\{\text{x}^2-3\text{x}+\bigg(\frac{3}{2}\bigg)^2-\bigg(\frac{3}{2}\bigg)^2-8\bigg\}}}\text{ dx}$
$=\int\frac{1}{\sqrt{-\bigg\{\bigg(\text{x}-\frac{3}{2}\bigg)^2-\frac{41}{4}\bigg\}}}\text{ dx}$
$=\int\frac{1}{\sqrt{\bigg(\frac{\sqrt{41}}{2}\bigg)^2-\bigg(\text{x}-\frac{3}{2}\bigg)^2}}\text{ dx}$
$=\sin^{-1}\frac{\text{x}-\frac{3}{2}}{\frac{\sqrt{41}}{2}}+\text{c}=\sin^{-1}\bigg(\frac{2\text{x}-3}{\sqrt{41}}\bigg)+\text{c}$
View full question & answer→Question 1833 Marks
Integrate the function in Exercise:$\frac{1}{\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}}}$
Answer$\frac{1}{\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}}}=\frac{1}{\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}}}\times\frac{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}$
$=\frac{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}{(\text{x}+\text{a)}-(\text{x}+\text{b)}}$
$=\frac{\big(\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}\big)}{\text{(a}-\text{b)}}$
$\Rightarrow\int\frac{1}{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}\text{dx}=\frac{1}{\text{a}-\text{b}}\int\Big(\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}\Big)\text{dx}$
$=\frac{1}{\text{(a}-\text{b)}}\left[\frac{\text{(x}+\text{a)}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{\text{(x}+\text{b)}^{\frac{3}{2}}}{\frac{3}{2}}\right]$
$=\frac{2}{3\text{(a}-\text{b)}}\left[\text{(x}+\text{a)}^{\frac{3}{2}}-\text{(x}+\text{b)}^{\frac{3}{2}}\right]+\text{C}$
View full question & answer→Question 1843 Marks
Evaluate the following integrals:
$\int\cos^5\text{x}\text{ dx}$
Answer$\int\cos^5\text{x}\text{ dx}$
$=\int\cos^4\text{x}\cdot\cos\text{x}\text{ dx}$
$=\int(1-\sin^2\text{x})^2\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int(1-\sin^2\text{x})^2\cos\text{x}\text{ dx}$
$=\int(1-\text{t}^2)^2\text{ dt}$
$=\int(1+\text{t}^4-2\text{t}^2)\text{dt}$
$=\int\text{dt}+\int\text{t}^4\text{ dt}-2\int\text{t}^2\text{ dt}$
$=\text{t}+\frac{\text{t}^5}{5}-\frac{2\text{t}^3}{3}+\text{C}$
$=\sin\text{x}+\frac{\sin^5\text{x}}{5}-\frac{2}{3}\sin^3\text{x}+\text{C}$
View full question & answer→Question 1853 Marks
Evaluate the following:
$\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{dx}$ $=\int\frac{\sqrt{1+\text{x}^2}}{\text{x}}\cdot\frac{1}{\text{x}^3}\text{dx}$
$=\int\sqrt{\frac{1+\text{x}^2}{\text{x}^2}}\cdot\frac{1}{\text{x}^3}\text{dx}$ $=\int\sqrt{\frac{1}{\text{x}^2}+1}\cdot\frac{1}{\text{x}^3}\text{dx}$
Put $1+\frac{1}{\text{x}^2}=\text{t}^2\Rightarrow\frac{-2}{\text{x}^3}\text{dx}=2\text{tdt}$
$\Rightarrow-\frac{1}{\text{x}^3}=\text{tdt}$
$\therefore\ \text{I}=-\int\text{t}^2\text{dt}=-\frac{\text{t}^3}{3}+\text{C}$ $=-\frac{1}{3}\Big(1+\frac{1}{\text{x}^2}\Big)^{\frac{3}{2}}+\text{c}$
View full question & answer→Question 1863 Marks
Evaluate the following:
$\int\limits^2_1\frac{\text{dx}}{\sqrt{(\text{x}-1)(2-\text{x})}}$
AnswerLet $\text{I}=\int\limits^2_1\frac{\text{dx}}{\sqrt{(\text{x}-1)(2-\text{x})}}$ $=\int\limits^2_1\frac{\text{dx}}{\sqrt{2\text{x}-\text{x}^2-2+\text{x}}}$
$=\int\limits^2_1\frac{\text{dx}}{\sqrt{-(\text{x}^2-3\text{x}+2)}}$
$\int\limits^2_1\frac{\text{dx}}{\sqrt{-\bigg[\text{x}^2-2\cdot\frac{3}{2}\text{x}+\Big(\frac{3}{2}\Big)^2+2-\frac{9}{4}\bigg]}}$
$=\int\limits^2_1\frac{\text{dx}}{\sqrt{-\bigg\{\Big(\text{x}-\frac{3}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2\bigg\}}}$
$=\int\limits^2_1\frac{\text{dx}}{\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{3}{2}\Big)^2}}$ $=\Bigg[\sin^{-1}\bigg(\frac{\text{x}-\frac{3}{2}}{\frac{1}{2}}\bigg)\Bigg]^2_1$
$=\big[\sin^{-1}(2\text{x}-3)\big]^2_1=\sin^{-1}1-\sin^{-1}(-1)$
$=\frac{\pi}{2}+\frac{\pi}{2}$ $\Big[\because\sin\frac{\pi}{2}=1\text{ and }\sin(-\theta)=-\sin\theta\Big]$
$=\pi$
View full question & answer→Question 1873 Marks
Evaluate the following integrals:
$\int\text{x}^3\sin\text{x}^4\text{dx}$
Answer$\int\text{x}^3.\sin\text{x}^4\text{dx}$
Let $\text{x}^4=\text{t}$
$\Rightarrow4\text{x}^3=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{x}^3\text{dx}=\frac{\text{dt}}{4}$
Now, $\int\text{x}^3.\sin\text{x}^4\text{dx}$
$=\frac{1}{4}\int\sin\text{t}\text{ dt}$
$=\frac{1}{4}[-\cos(\text{t})]+\text{C}$
$=\frac{1}{4}\big[-\cos\text{x}^4\big]+\text{C}$
View full question & answer→Question 1883 Marks
Evaluate the following integrals:$\int\text{x}^3\cos\text{x}^2\text{dx}$
AnswerLet $\text{I}=\int\text{x}^3\cos\text{x}^2\text{dx}$
Let $\text{x}^2=\text{t}$
$2\text{x dx = dt}$
$\text{I}=\frac{1}{2}\int\text{t}\cos\text{t dt}$
Using integration by parts,
$=\frac{1}{2}[\text{t}\int\cos\text{t dt}-\int(1\times\int\cos\text{t dt})\text{dt}]$
$=\frac{1}{2}[\text{t}\times\sin\text{t}-\int\sin\text{t dt}]$
$=\frac{1}{2}[\text{t}\sin\text{t}+\cos\text{t}]+\text{C}$
$\text{I}=\frac{1}{2}[\text{x}^2\sin\text{x}^2+\cos\text{x}^2]+\text{C}$
View full question & answer→Question 1893 Marks
Evaluate the following:
$\int\tan^2\text{x}\sec^4\text{x dx}$
AnswerLet $\text{I}=\int\tan^2\text{x}\sec^4\text{xdx}$
$=\int\tan^2\text{x}\sec^2\text{x}\sec^2\text{xdx}$
$=\int\tan^2\text{x}(1+\tan^2\text{x})\sec^2\text{xdx}$
Put $\tan\text{x}=\text{t}\Rightarrow\sec^2\text{x}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\text{t}^2(1+\text{t}^2)\text{dt}$
$=\int(\text{t}^2+\text{t}^4)\text{dt}=\frac{\text{t}^3}{3}+\frac{\text{t}^5}{5}+\text{C}$ $=\frac{\tan^5\text{x}}{5}+\frac{\tan^3\text{x}}{3}+\text{C}$
View full question & answer→Question 1903 Marks
Write a value of $\int\frac{\cos\text{x}}{\sin\text{x}\log\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{\sin\text{x}\log\sin\text{x}}\text{ dx}$
$\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{ dx}$
Let $\log\sin\text{x}=\text{t}$
$\cot\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log\text{t}+\text{C}$
$=\log(\log\sin\text{x})+\text{C}$
View full question & answer→Question 1913 Marks
$\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\times\text{dx}.$ Then,
$\text{I}=\int\frac{2\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\times\text{dx}$
$=\int\frac{\sin^2\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}\times\text{dx}$
$=\int\tan^2\frac{\text{x}}{2}\text{dx}$
$=\int\Big(\sec^2\frac{\text{x}}{2}-1\Big)\text{dx}$
$=\frac{\tan\frac{\text{x}}{2}}{\frac{1}2{}}-\text{x+c}$
$=2\tan\frac{\text{x}}{2}-\text{x+c}$
View full question & answer→Question 1923 Marks
Write a value of $\int(\text{e}^{\text{x}\log_\text{e}\text{a}}+\text{e}^{\text{a}\log_\text{e}\text{x}})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}\log_\text{e}\text{a}}+\text{e}^{\text{a}\log_\text{e}\text{x}}\text{ dx}$
$=\int(\text{e}\log\text{a}^{\text{x}}+\text{e}\log\text{x}^{\text{a}})\text{dx}$
$=\int(\text{a}^{\text{x}}+\text{x}^{\text{a}})\text{dx}$
$=\int\text{a}^{\text{x}}\text{dx}+\int\text{x}^{\text{a}}\text{dx}$
$=\frac{\text{a}^{\text{x}}}{\log_\text{e}\text{a}}+\frac{\text{x}^{\text{a}+1}}{\text{a}+1}+\text{C}$
$\therefore\ \text{I}=\frac{\text{a}^{\text{x}}}{\log_\text{e}\text{a}}+\frac{\text{x}^{\text{a}+1}}{\text{a}+1}+\text{C}$
View full question & answer→Question 1933 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int^{\frac{\pi}{2}}_{0}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{dx}$ $\Rightarrow\ \ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin\bigg(\frac{\pi}{2}-\text{x}}\bigg)}{\sqrt{\sin\bigg(\frac{\pi}{2}-\text{x}}\bigg)+\sqrt{\cos\bigg(\frac{\pi}{2}-\text{x}\bigg)}}\text{dx}\ \ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\ \text{dx}=\int\limits_{0}^{\text{a}}\text{f}(\text{a}-\text{x})\text{dx}=\bigg]$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}+\sqrt{\sin\text{x}}}}\text{dx}$Adding eq. (i) and (ii),
$21=\int\limits_{0}^{\frac{\pi}{2}}\bigg(\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}+\sqrt{\cos\text{x}}}}+\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}+\sqrt{\sin\text{x}}}}\bigg)\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\bigg(\frac{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\bigg)\text{dx}$ $\Rightarrow\ \ \ \ 21=\int\limits_{0}^{\frac{\pi}{2}}1\ \text{dx}=\bigg(\text{x}^{\frac{\pi}{2}}_{0}\bigg)\ \ \ \Rightarrow\ \ \ \ 21=\frac{\pi}{2}\ \ \Rightarrow\ \text{I}=\frac{\pi}{4}$
View full question & answer→Question 1943 Marks
Evaluate the following intregals:
$\int\frac{\sin2\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{2\tan\text{x}\sec^2\text{x}}{\tan^4\text{x}+1}\ \text{dx}$
Let $\tan^2\text{x}=\text{t}$
$2\tan\text{x}\sec^2\text{x}\text{ dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2+1}$
$=\tan^{-1}\text{t}+\text{C}$
$\text{I}=\tan^{-1}(\tan^2\text{x})+\text{C}$
View full question & answer→Question 1953 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}\text{dx}$
Answer$\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}$
$\text{Let,}\sin^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}=\text{dt}$
$\text{Now,}\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}$
$=\int\frac{\text{dt}}{\text{t}^2}$
$=\int\text{t}^{-2}\text{dt}$
$=\frac{\text{t}^{-2+1}}{-2+1}+\text{C}$
$=\frac{-1}{\text{t}}+\text{C}$
$=-\frac{1}{\sin^{-1}\text{x}}+\text{C}$
View full question & answer→Question 1963 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int^{2}_{0}\text{x}\sqrt{2-\text{x}}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{2}\text{x}\sqrt{2-\text{x}}\ \text{dx}=\int\limits_{0}^{2}(2-\text{x})\sqrt{2-(2-\text{x)}}\ \text{dx}\ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\text{dx}=\int\limits_{0}^{\text{a}}(\text{a}-\text{x})\text{dx}=\bigg]$
$\Rightarrow\ \text{I}=\int\limits_{0}^{2}(2-\text{x})\sqrt{\text{x}}\ \text{dx}=\int\limits_{0}^{2}\bigg({2\text{x}^{\frac{1}{2}}-\text{x}^{\frac{3}{2}}}\bigg)\text{dx} =\Bigg[2.\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{\text{x}^{\frac{5}{2}}}{\frac{5}{2}}\Bigg]^{2}_{0}=\bigg(\frac{4}{3}.2^{\frac{3}{2}}-\frac{2}{5}.2^{\frac{5}{2}}\bigg)-(0-0)$
$\Rightarrow\ \text{I}=\frac{4}{3}\times2\sqrt{2}-\frac{2}{5}\times4\sqrt{2}=\bigg(\frac{8}{3}-\frac{8}{5}\bigg)\sqrt{2}={\frac{16\sqrt{2}}{15}}$
View full question & answer→Question 1973 Marks
Evaluate the following as limit of sums:
$\int\limits^2_0\text{e}^\text{x}\text{dx}$
AnswerWe know that $\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}=\lim\limits_{\text{r}=0}\text{h}\sum\limits^{\text{n}-1}_{\text{r}=0}\text{f}(\text{a}+\text{rh})$
For $\text{I}=\int\limits^2_0\text{e}^\text{x}\text{dx},$ we have a = 0 and b = 2
$\therefore\ \text{h}=\frac{\text{b}-\text{a}}{\text{n}}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
$\therefore\ \text{I}=\int\limits^2_0\text{e}^\text{x}\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[1+\text{e}^\text{h}+\text{e}^{2\text{h}}+\dots+\text{e}^{(\text{n}-1)^\text{h}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\frac{1\cdot(\text{e}^\text{h})^\text{n}-1}{\text{e}^\text{h}-1}\Big]=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big(\frac{\text{e}^{\text{eh}}-1}{\text{e}^\text{h}-1}\Big)$ $=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big(\frac{\text{e}^2-1}{\text{e}^\text{h}-1}\Big)$
$=\text{e}^2=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{e}^\text{h}-1}=\text{e}^2-1$
View full question & answer→Question 1983 Marks
Evaluate the following integrals:
$\int\frac{\cos\text{x}}{\sin^2\text{x}+4\sin\text{x}+5}\text{dx}$
Answer$\int\frac{\cos\text{x dx}}{\sin^2\text{x}+4\sin\text{x}+5}$
Let $\sin\text{x = t}$
$\Rightarrow\cos\text{x dx = dt}$
Now, $\int\frac{\cos\text{x dx}}{\sin^2\text{x}+4\sin\text{x}+5}$
$=\int\frac{\text{dt}}{\text{t}^2+4\text{t}+5}$
$=\int\frac{\text{dt}}{\text{t}^2+2\times\text{t}\times2+4+1}$
$=\int\frac{\text{dt}}{(\text{t}+2)^2+1^2}$
$=\frac{1}{1}\tan^{-1}\Big(\frac{\text{t}+2}{1}\Big)+\text{C}$
$=\tan^{-1}(\sin\text{x}+2)+\text{C}$
View full question & answer→Question 1993 Marks
Evaluate the following integrals:
$\int^\limits\frac{\pi}{4}_{0}\sin^32\text{t}\cos2\text{t}\text{ dt}$
AnswerLet $\text{I}\int^\limits\frac{\pi}{4}_{0}\sin^32\text{t}\cos2\text{t}\text{ dt}$ Then,Let $\sin2\text{t}=\text{u}$ Then, $2\cos2\text{t dt} =\text{du}$
When $\text{t}=0,\text{u}=0$ and $\text{t}=\frac{\pi}{4},\text{u}=1$
$\therefore\ \text{I}=\frac{1}{2}\int^\limits{1}_{0}\text{u}^3\text{ du}$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\frac{\text{u}^4}{4}\Big]^1_0$
$\Rightarrow\text{I}=\frac{1}{2}\Big(\frac{1}{4}-0\Big)$
$\Rightarrow\text{I}=\frac{1}{8}$
View full question & answer→Question 2003 Marks
By Using properties of definite integral, evaluate the following integral in Exercise:
$\int^{4}_{0}|\text{x}-1|\text{dx}$
Answer$\text{Let}\ \text{I}\int^{4}\limits_{0}|\text{x}-1|\text{dx}$
$\text{Here}\ \ \text{x}-1=0\ \ \Rightarrow\ \ \text{x}=1\in(0,4)$
$\therefore\ \ \text{from eq. (i)},\text{I}=\int^{1}\limits_{0}|\text{x}-1|\text{dx}+\int^{4}\limits_{1}|\text{x}-1|\text{dx}=-\int^{1}\limits_{0}(\text{x}-1)\text{dx}+\int^{4}\limits_{1}(\text{x}-1)\text{dx}$
$\Rightarrow\ \ \text{I}=-\bigg(\frac{\text{x}^{2}}{2}-\text{x}\bigg)^1_0+\bigg(\frac{\text{x}^{2}}{2}-\text{x}\bigg)^{4}_{1}$
$=-\Big\{\frac{1}{2}-1-0\Big\}+\Big\{\frac{16}{2}-4-\frac{1}{2}+1\Big\}$
$\Rightarrow\ \ \text{I}=\frac{-1}{2}+1+8-4-\frac{1}{2}+1=6-\frac{2}{2}=6-1=5$
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