Question 1013 Marks
Evaluate the following integrals:
$\int\limits^{2\pi}_0\cos^{-1}(\cos\text{x})\text{dx}$
Answer$\int\limits^{2\pi}_0\cos^{-1}(\cos\text{x})\text{dx}$
$=\int\limits^{\pi}_0\cos^{-1}(\cos\text{x})\text{dx}+\int\limits^{2\pi}_\pi\cos^{-1}(\cos\text{x})\text{dx}$
$=\int\limits^{\pi}_0\text{x}\text{ dx}+\int\limits^{2\pi}_\pi(2\pi-\text{x})\text{dx}$ $\big[{\pi}\leq\text{x}\leq2\pi\Rightarrow-2\pi\leq-\text{x}\leq-{\pi}\Rightarrow0\leq2\pi-\text{x}\leq{\pi}\big]$
$=\Big[\frac{\text{x}^2}{2}\Big]+\bigg[\frac{(2\pi-\text{x})}{2\times(-1)}\bigg]^{2\pi}_\pi$
$=\frac{1}{2}(\pi^2-0)-\frac{1}{2}(0-\pi^2)$
$=\frac{\pi^2}{2}+\frac{\pi^2}{2}$
$=\pi^2$
View full question & answer→Question 1023 Marks
Evaluate the definite integral in Exercise:
$\int_{0}^{1}(\text{x}\text{e}^{\text{x}}+\sin\frac{\pi\text{x}}{4})\text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{1}\bigg(\text{xe}^{\text{x}}+\sin\frac{\pi\text{x}}{4}\bigg)\text{dx}$
$\int\bigg(\text{xe}^{\text{x}}+\sin\frac{\pi\text{x}}{4}\bigg)\text{dx}=\text{x}\int\text{e}^\text{x}\text{dx}-\int\left\{\bigg(\frac{\text{d}}{\text{dx}}\text{x}\bigg)\int\text{e}^\text{x}\text{dx}\right\}\text{dx}+\left\{\frac{-\cos\frac{\pi\text{x}}{4}}{\frac{\pi}{4}}\right\}$
$=\text{xe}^\text{x}-\int\text{e}^\text{x}\text{dx}-\frac{4}{\pi}\cos\frac{\pi\text{x}}{4}$
$=\text{xe}^\text{x}-\text{e}^\text{x}-\frac{4}{\pi}\cos\frac{\pi\text{x}}{4}$
$=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(1)-\text{F}(0)$
$=\bigg(1.\text{e}^{1}-\text{e}^{1}-\frac{4}{\pi}\cos\frac{\pi}{4}\bigg)-\bigg(0.\text{e}^{0}-\text{e}^{0}-\frac{4}{\pi}\cos0\bigg)$
$=\text{e}-\text{e}-\frac{4}{\pi}\bigg(\frac{1}{\sqrt{2}}\bigg)+1+\frac{4}{\pi}$
$=1+\frac{4}{\pi}-\frac{2\sqrt{2}}{\pi}$
View full question & answer→Question 1033 Marks
Evaluate the following integrals:
$\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$
Answer$\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$ Let $\text{x}+\tan^{-1}\text{x}=\text{t}$ $\Big(1+\frac{1}{1+\text{x}^2}\Big)=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\Big(\frac{\text{x}^2-1+1}{\text{x}^2+1}\Big)\text{dx}=\text{dt}$ $\Rightarrow\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}=\text{dt}$Now, $\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$
$=\int5^\text{t}\text{dt}$
$=\frac{5^\text{t}}{\log5}+\text{C}$
$=\frac{5^{\text{x}+\tan^{-1}\text{x}}}{\log5}+\text{C}$
View full question & answer→Question 1043 Marks
Evaluate the following integrals:
$\int (3\text{x}\sqrt{\text{x}}+4\sqrt{\text{x}}+5)\text{dx}$
Answer$\int(3\text{x}\sqrt{5}+4\sqrt{\text{x}}+5)\text{dx}$
$=\int3\text{x}\sqrt{5}\text{dx}+\int4\sqrt{\text{x}}\text{dx}+\int5\text{dx}$
$=\int3\text{x}^{\frac{3}{2}}\text{dx}+4\int\text{x}^{\frac{1}{2}}\text{dx}+5\int\text{dx}$
$=\frac{\text{x}\frac{3}{2}+1}{\frac{3}{2}+1}+\frac{4\text{x}^{\frac{1}{2}}}{\frac{1}{2}+1}+5\text{x}+\text{C}$
$=\frac{6}{5}\text{x}^{\frac{5}{2}}+\frac{8}{3}\text{x}^{\frac{3}{2}}+5\text{x}+\text{C}$
View full question & answer→Question 1053 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
Here, $\text{f(x)}=\log\sin\text{x}$ Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{f}'\text{(x)}=\cot\text{x}$
let $\text{e}^{\text{x}}\log\sin\text{x = t}$
Diff. both sides w.r.t x
$\text{e}^{\text{x}}\log(\sin\text{x})+\text{e}^{\text{x}}\times\frac{1}{\sin\text{x}}\times\cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\big[\text{e}^{\text{x}}\log(\sin\text{x})+\text{e}^{\text{x}}\cot\text{x}\big]\text{dx = dt}$
$\Rightarrow\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx = dt}$
$\therefore \int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx} =\int\text{dt}$
$=\text{t + C}$
$=\text{e}^{\text{x}}\log\sin\text{x}+\text{C}$
View full question & answer→Question 1063 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\pi}\frac{1}{1+\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\pi}\frac{1}{1+\sin\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\pi}\frac{1-\sin\text{x}}{(1+\sin\text{x})(1-\sin\text{x})}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\frac{1-\sin\text{x}}{1+\sin^2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{ dx}$ $\big[\because\sin^2\text{x}+\cos^2\text{x}=1\big]$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\sec^2\text{x}-\sec\text{x}\tan\text{x}\text{ dx}$
$\Rightarrow\text{I}=\big[\tan\text{x}-\sec\text{x}\big]^{\pi}_0$
$\Rightarrow\text{I}=(\tan\pi-\sec\pi)-(\tan0-\sec0)$
$\Rightarrow\text{I}=0+1-(0-1)$
$\Rightarrow\text{I}=1+1$
$\Rightarrow\text{I}=2$
View full question & answer→Question 1073 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 1083 Marks
Evaluate the following integrals:
$\int\text{x}\frac{\tan^{-1}\text{x}^2}{1+\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int\text{x}\frac{\tan^{-1}\text{x}^2}{1+\text{x}^4}\text{ dx}\ ....(1)$ Let $\tan^{-1}\text{x}^2=\text{t}$ then, $\text{d}\big(\tan^{-1}\text{x}^2\big)=\text{dt}$ $\Rightarrow\frac{1\times2\text{x}}{1+(\text{x}^2)^2}\text{ dx}=\text{dt}$ $\Rightarrow\frac{1\times\text{x}}{1+\text{x}^4}\text{ dx}=\frac{\text{dt}}{2}$ Putting, $\tan^{-1}\text{x}^2=\text{t}$ and $\frac{\text{x}}{1+\text{x}^4}\text{ dx}=\frac{\text{dt}}{2}$ in equation (1),we get,
$\text{I}=\int\text{t}\frac{\text{dx}}{2}$
$=\frac{1}{2}\int\text{t dt}$
$=\frac{1}{2}\times\frac{\text{t}^2}{2}+\text{C}$
$\text{I}=\frac{\text{t}^2}{4}+\text{C}$
$=\frac{(\tan^{-1}\text{x}^2)^2}{4}+\text{C}$
$\text{I}=\frac{1}{4}\big(\tan^{-1}\text{x}^2\big)^2+\text{C}$
View full question & answer→Question 1093 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\text{x}(1-\text{x})^5\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{1}\text{x}(1-\text{x})^5\text{ dx}$
Expanding $(1 - x)^5$ by Binomial theorem,
$\therefore\ (1-\text{x})^5=1^5+{^5\text{C}_1}(-\text{x})+{^5\text{C}_2}(-\text{x})^2\\+{^5\text{C}_3}(-\text{x})^3+{^5\text{C}_4}(-\text{x})^4+{^5\text{C}_5}(-\text{x})^5$
$=1-5\text{x}+10\text{x}^2-10\text{x}^3+5\text{x}^4-\text{x}^5$
$=\int_{0}^\limits{1}\text{x}(1-5\text{x}+10\text{x}^2-10\text{x}^3+5\text{x}^4-\text{x}^5)\text{dx}$
$=\Big[\frac{\text{x}^2}{2}-\frac{5\text{x}^3}{3}+\frac{10\text{x}^4}{4}-\frac{10\text{x}^5}{5}+\frac{5\text{x}^6}{6}-\frac{\text{x}^7}{7}\Big]^1_0$
$=\frac{1}{2}-\frac{5}{3}+\frac{10}{4}-\frac{10}{5}+\frac{5}{6}-\frac{1}{7}$
$=\frac{1}{42}$
View full question & answer→Question 1103 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 1113 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 1123 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 1133 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 1143 Marks
Evaluate the following:
$\int\sqrt{5-2\text{x}+\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{5-2\text{x}+\text{x}^2}\text{dx}$
$=\int\sqrt{\text{x}^2-2\text{x}+1+4}\text{dx}$
$=\int\sqrt{(\text{x}-1)^2+(2)^2}\text{dx}$
Using $\int\sqrt{\text{x}^2+\text{a}^2}\text{dx}$ $=\frac{1}{2}\text{x}\sqrt{\text{x}^2+\text{a}^2}+\frac{\text{a}^2}{2}\log\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|+\text{C},$ we get
$\int\sqrt{(\text{x}-1)^2+(2)^2}\text{dx}$ $=\frac{\text{x}-1}{2}\sqrt{2^2+(\text{x}+1)^2}+2\log\Big|\text{x}-1+\sqrt{2^2+(\text{x}-1)^2}\Big|+\text{C}$
$\Rightarrow\ \int\sqrt{5-2\text{x}+\text{x}^2}\text{dx}$ $=\frac{\text{x}-1}{2}\sqrt{5-2\text{x}+\text{x}^2}+2\log\Big|\text{x}-1+\sqrt{5-2\text{x}+\text{x}^2}\Big|+\text{C}$
View full question & answer→Question 1153 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 1163 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 1173 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 1183 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 1193 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 1203 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 1213 Marks
Evaluate the following integrals:
$\frac{\text{x}+1}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\frac{\text{x}+1}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{(\text{x}+1)(1+\text{xe}^\text{x}-\text{xe}^\text{x})}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{\int(\text{x}+1)(1+\text{xe}^\text{x})}{\text(1+\text{xe}^\text{x})}\ \text{dx}-\int\frac{(\text{x}+1)(\text{xe}^\text{x})}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{(\text{x}+1)}{\text{x}}\ \text{dx}-\int\frac{\text{e}^\text{x}(\text{x}+1)}{1+\text{xe}^\text{x}}\ \text{dx}$
$=\int\frac{(\text{x}+1)\text{e}^\text{x}}{\text{xe}^\text{x}}\ \text{dx}-\int\frac{\text{e}^\text{x}(\text{x}+1)}{1+\text{xe}^\text{x}}\ \text{dx}$
$=\log|\text{xe}^\text{x}|-\log|1+\text{xe}^\text{x}|+\text{C}$
$\therefore\text{I}=\log\Big|\frac{\text{xe}^\text{x}}{1+\text{xe}^\text{x}}\Big|+\text{C}$
View full question & answer→Question 1223 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 1233 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 1243 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 1253 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 1263 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 1273 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 1283 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 1293 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 1303 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 1313 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 1323 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 1333 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 1343 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 1353 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 1363 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 1373 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 1383 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 1393 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 1403 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 1413 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 1423 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$
View full question & answer→Question 1433 Marks
Find the integrals of the functions in Exercises:
$\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}$
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}}$$\ \ \ \ \ \ \ \big[\sin^2\text{x}+\cos^2\text{x}=1;\ \sin2\text{x}=2\sin\text{x }\cos\text{x}\big]$
$=\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}$
$\text{Let }\sin\text{x}+\cos{\text{x}}=\text{t}$
$\therefore(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\Rightarrow\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{ dx}=\int\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}\text{ dx}$
$=\int\frac{\text{dt}}{\text{t}^2}$
$=\int\text{t}^{-2}\text{dt}$
$=-\text{t}^{-1}+\text{C}$
$=-\frac{1}{\text{t}}+\text{C}$
$=\frac{-1}{\sin\text{x}+\cos\text{x}}+\text{C}$
View full question & answer→Question 1443 Marks
Evaluate the definite integral in Exercise:
$\int^{\frac{\pi}{4}}_{0}(2\sec^{2}\text{x}+\text{x}^{3}+2)\text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{4}}(2\sec^{2}\text{x}+\text{x}^{3}+2)\text{dx}$ $\int\limits(2\sec^{2}\text{x}+\text{x}^{3}+2)\text{dx}=2\tan\text{x}+\frac{\text{x}^{4}}{4}+2\text{x}=\text{F}\text{(x)}$ By second fundamental theorem of calculus, we obtain $\text{I}=\text{F}\bigg(\frac{\pi}{4}\bigg)-\text{F}(0)$ $=\left\{\Bigg(2\tan\frac{\pi}{4}+\frac{1}{4}\bigg(\frac{\pi}{4}\bigg)^{4}+2\bigg(\frac{\pi}{4}\bigg)\Bigg)-(2\tan0+0+0)\right\}$ $=2\tan\frac{\pi}{4}+\frac{\pi^{4}}{4^{5}}+\frac{\pi}{2}$$=2+\frac{\pi}{2}+\frac{\pi^{4}}{1024}$
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If $\int\text{e}^{\text{x}}(\tan\text{x}+1)\sec\text{x dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C},$ then write the value of f(x).
AnswerSince, $\int\text{e}^{\text{x}}\big(\text{f(x)}+\text{f}(\text{x})\big)\text{dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C}$
We can write the expression as
$\int\text{e}^{\text{x}}(\tan\text{x}+1)\sec\text{x dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C}$
$\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{f}(\text{x})\text{e}^{\text{x}}+\text{C}$
Comparing with equation $\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{f}(\text{x})\text{e}^{\text{x}}+\text{C}$ with standard equation we have
$\text{f(x)}=\sec\text{x},\text{f}'(\text{x})=\sec\text{x}\tan\text{x}$
Therefore,
$\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{e}^{\text{x}}\sec\text{x}+\text{C}$
Thus, $\text{f(x)}=\sec\text{x}$
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Find the integrals of the functions in Exercises:
$\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}$
Answer$\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}=\frac{\cos2\text{x}}{\cos^2\text{x}+\sin^2\text{x}+2\sin\text{x}\cos\text{x}}=\frac{\cos2\text{x}}{1+\sin2\text{x}}$
$\therefore\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{ dx}=\int\frac{\cos2\text{x}}{(1+\sin2\text{x})}\text{dx}$
$\text{Let }1+\sin2\text{x}=\text{t}$
$\Rightarrow2\cos2\text{x}\text{ dx}=\text{dt}$
$\therefore\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{2}\log|\text{t}|+\text{C}$
$=\frac{1}{2}\log|1+\sin2\text{x}|+\text{C} $
$=\frac{1}{2}\log|(\sin\text{x}+\cos\text{x})^2|+\text{C} $
$=\log|\sin\text{x}+\cos\text{x}|+\text{C}$
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Find the integrals of the functions in Exercises:
$\sin3\text{x}\cos4\text{x}$
AnswerIt is know that, $\sin\text{A}\cos\text{B}=\frac{1}{2}\big\{\sin(\text{A+B})+\sin(\text{A}-\text{B}) \big\} $
$\therefore\int\sin3\text{x}\cos4\text{x dx}=\frac{1}{2}\int\big\{\sin(3\text{x}+4\text{x})+\sin(\text{3x}-\text{4x}) \big\}\text{ dx} $
$=\frac{1}{2}\int\big\{\sin\text{7x}+\sin(-\text{x}) \big\}\text{ dx} $
$=\frac{1}{2}\int\big\{\sin\text{7x}-\sin\text{x} \big\}\text{ dx} $
$=\frac{1}{2}\int\sin\text{7x dx}-\frac{1}{2}\int\sin\text{x} \text{ dx} $
$=\frac{1}{2}\bigg(\frac{-\cos7\text{x}}{7}\bigg)-\frac{1}{2}(-\cos\text{x}) +\text{ C} $
$=\frac{-\cos7\text{x}}{14}+\frac{\cos\text{x}}{2}+\text{C} $
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Evaluate the following integrals:
$\int\sqrt{4\text{x}^2-5}\text{dx}$
Answer$\text{I}=\int\sqrt{4\text{x}^2-5}\text{dx}$
$=\int\sqrt{4\Big(\text{x}^2-\frac{5}{4}\Big)}\text{dx}$
$=2\int\sqrt{\text{x}^2-\Big(\frac{\sqrt5}{2}}\Big)^2\text{dx}$
$=2\Big[\frac{\text{x}}{2}\sqrt{\text{x}^2-\frac{5}{4}}-\frac{5}{8}\int\Big|\text{x}+\sqrt{\text{x}^2-\frac{5}{4}}\Big|\Big]+\text{C}$
$\Big[\because\ \int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2-\text{a}^2}\frac{1}{2}\text{a}^2\int\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|+\text{C}\Big]$
$=\text{x}\sqrt{\text{x}^2-\frac{5}{4}}-\frac{5}{4}\int\bigg|\text{x}+\sqrt{\text{x}^2-\frac{5}{4}}\bigg|+\text{C}$
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Evaluate the following integrals:
$\int_{0}^\limits{1}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
Answer$\int_{0}^\limits{1}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
$=\int_{0}^\limits{1}2\tan^{-1}\text{x dx}$
$=2\Big[\text{x}\tan^{-1}\text{x}\Big]^1_0-2\int_{0}^\limits{1}\frac{\text{x}}{1+\text{x}^2}\text{ dx}$
$=2\Big[\text{x}\tan^{-1}\text{x}\Big]^1_0-\Big[\log\big(1+\log\text{x}^2\big)\Big]$
$=2\frac{\pi}{4}-0-\log2+0$
$=\frac{\pi}{2}-\log2$
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Evaluate the following integrals:
$\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$
Answer$\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$ Let $2+3\log\text{x}=\text{t}$ $\Rightarrow\frac{3}{\text{x}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\frac{\text{dx}}{\text{x}}=\frac{\text{dt}}{3}$ Now, $\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$$=\frac{1}{3}\int\sin\text{t dt}$
$=\frac{1}{3}[-\cos\text{t}]+\text{C}$
$=-\frac{1}{3}\cos(2+3\log\text{x})+\text{C}$
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