Question 15 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 25 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{\text{dx}}{\text{a}\cos\text{x}+\text{b}\sin\text{x}}\text{ a},\text{b}>0$
Answer$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}\cos\text{x}+\text{b}\sin\text{x}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}\Bigg(\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)+\text{b}\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\big(1+\tan^{2}\frac{\text{x}}{2}\big)}{\text{a}-\text{a}\tan^{2}\frac{\text{x}}{2}+2\text{b}\tan\frac{\text{x}}{2}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sec^2\frac{\text{x}}{2}}{\text{a}-\text{a}\tan^{2}\frac{\text{x}}{2}+2\text{b}\tan\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Then, $\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi}{2},\text{t}=1$
Therefore the integral becomes
$\text{I}=\int^\limits1_0\frac{2\text{dt}}{\text{a}-\text{a}\text{t}^2+2\text{bt}}$
$=\int^\limits1_0\frac{2\text{dt}}{-\text{a}\big[\text{t}^2-\frac{2\text{bt}}{\text{a}-1}\big]}$
$=\frac{2}{\text{a}}\int^\limits1_0\frac{\text{dt}}{-\Big[\big(\text{t}-\frac{\text{b}}{\text{a}}\big)^2-1-\frac{\text{b}^2}{\text{a}^2}\Big]}$
$=\frac{2}{\text{a}}\int^\limits1_0\frac{\text{dt}}{\Big(\frac{\text{b}^2}{\text{a}^2}+1\Big)-\big(\text{t}-\frac{\text{b}}{\text{a}}\big)^2}$
$=\frac{2}{\text{a}}\begin{bmatrix}\frac{1}{2\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}}}\begin{pmatrix}\log\begin{vmatrix}\frac{2\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}}+\big(\text{t}-\frac{\text{b}}{\text{a}}\big)}{\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}-\big(\text{t}-\frac{\text{b}}{\text{a}}\big)}}\end{vmatrix} \end{pmatrix}^1_0\end{bmatrix}$
$=\frac{1}{\sqrt{\text{a}^2+\text{b}^2}}\log\bigg(\frac{\text{a}+\text{b}+\sqrt{\text{a}^2+\text{b}^2}}{\text{a}+\text{b}-\sqrt{\text{a}^2+\text{b}^2}}\bigg)$
View full question & answer→Question 35 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{4}}\Big(\sqrt{\tan\text{x}}+\sqrt{\cot}\text{x}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\Big(\sqrt{\tan\text{x}}+\sqrt{\cot}\text{x}\Big)\text{dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\bigg(\sqrt{\frac{\sin\text{x}}{\cos\text{x}}}+\sqrt{\frac{\cos\text{x}}{\sin\text{x}}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{\sin\text{x}\cos\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\sqrt{2}\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2\sin\text{x}\cos\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\sqrt{2}\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1-(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
Let $\sin\text{x}-\cos\text{x}=\text{t}$ Then, $\cos\text{x}+\sin\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=1$ and $\text{x}=\frac{\pi}{4},\text{t}=0$
$\therefore\ \text{I}=\sqrt{2}\int^\limits0_{-1}\frac{\text{dt}}{\sqrt{1-\text{t}^2}}$
$\Rightarrow\text{I}=\sqrt{2}\Big[\sin^{-1}\text{t}\Big]^0_{-1}$
$\Rightarrow\text{I}=\frac{\pi}{\sqrt{2}}$
View full question & answer→Question 45 Marks
Evaluate the following integrals:
$\int\limits_{0}^{1}\tan^{-1}\text{x dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\tan^{-1}\text{x dx}$ Then,
$\text{I}=\int_{0}^\limits{1}\tan^{-1}\text{x dx}$
Integrating by parts,
$\text{I}=\big[\text{x}\tan^{-1}\text{x}\big]^1_0-\int_{0}^\limits{1}\frac{\text{x}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\big[\text{x}\tan^{-1}\text{x}\big]^1_0-\frac{1}{2}\big[\log\big(\text{x}^2+1\big)\big]^1_0$
$\Rightarrow\text{I}=\frac{\pi}{4}-0-\frac{1}{2}\log2+0$
$\Rightarrow\text{I}=\frac{\pi}{4}-\frac{1}{2}\log2$
View full question & answer→Question 55 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}$
AnswerWe have,
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}=\frac{1}{\text{b}^2}\int_{0}^\limits{\infty}\frac{1}{\big(\frac{\text{a}}{\text{b}}\big)^2+\text{x}^2}\text{ dx}$
We have that,
$\int\frac{1}{\text{a}^2+\text{x}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}$
$\therefore\ \frac{1}{\text{b}^2}\int_{0}^\limits{\infty}\frac{1}{\big(\frac{\text{a}}{\text{b}}\big)^2+\text{x}^2}\text{ dx}=\frac{1}{\text{b}^2}\Big[\frac{\text{b}}{\text{a}}\tan^{-1}\Big(\frac{\text{bx}}{\text{a}}\Big)\Big]^{\infty}_0$
$=\frac{1}{\text{ab}}\Big[\tan^{-1}\Big(\frac{\text{bx}}{\text{a}}\Big)\Big]^{\infty}_0$
$=\frac{1}{\text{ab}}\big[\tan^{-1}\infty-\tan^{-1}0\big]$
$=\frac{1}{\text{ab}}\Big[\frac{\pi}{2}-0\Big]$
$=\frac{\pi}{2\text{ab}}$
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}=\frac{\pi}{2\text{ab}}$
View full question & answer→Question 65 Marks
Evaluate the following integrals:
$\int\limits_{0}^{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}$
AnswerLet $1+\text{x}^2=\text{t}$
Differentiating w.r.t. x, we get
$2\text{xdx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=1$
$\text{x}=1\Rightarrow\text{t}=2$
$\int_{0}^\limits{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}=\int_{1}^\limits{2}\frac{12(\text{t}-1)}{\text{t}^4}\text{ dt}$
$=12\int_{1}^\limits{2}\Big(\frac{1}{\text{t}^3}-\frac{1}{\text{t}^4}\Big)\text{dt}$
$=12\Big[-\frac{1}{2\text{t}^2}-\frac{1}{3\text{t}^3}\Big]^2_1$
$=12\Big[-\frac{1}{8}+\frac{1}{24}+\frac{1}{2}-\frac{1}{3}\Big]$
$=12\Big[\frac{-3+1+12-8}{24}\Big]$
$=\frac{12\times2}{24}=1$
View full question & answer→Question 75 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}+3\cos\text{x}+2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}+3\cos\text{x}+2}\text{ dx}$ Then,
Let $\cos\text{x}=\text{t},$ Then, $-\sin\text{x dx}=\text{dt}$
When, $\text{x}=0,\text{ t}=1$ and $\text{x}=\frac{\pi}{2},\text{ t}=0$
$\therefore\ \text{I}=-\int^\limits0_1\frac{\text{t dt}}{\text{t}^2+3\text{t}+2}$
$\Rightarrow\text{I}=\int\limits^0_1\frac{-\text{t dt}}{\text{t}^3+3\text{t}+2}$
$\Rightarrow\text{I}=\int\limits^0_1\Big(\frac{1}{(\text{t}+1)}-\frac{2}{(\text{t}+2)}\Big)\text{dt}$
$\Rightarrow\text{I}=\Big[\log(\text{t}+1)-2\log(\text{t}+2)\Big]^0_1$
$\Rightarrow\text{I}=\bigg[\log\frac{(\text{t}+1)}{(\text{t}+2)^2}\bigg]^1_0$
$\Rightarrow\text{I}=\bigg[\log\Big(\frac{1}{4}\Big)-\log\Big(\frac{2}{9}\Big)\bigg]^1_0$
$\Rightarrow\text{I}=\log\frac{9}{8}$
View full question & answer→Question 85 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin\text{x}\cos\text{x}}{1+\sin^4\text{x}}\text{ dx}$
AnswerLet $\sin^2\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$2\sin\text{x}\cos\text{x dx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=1$
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\text{x}\cos\text{x}}{1+\sin^4\text{x}}\text{ dx}$
$=\frac{1}{2}\int\limits^1_0\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{1}{2}\big[\tan^{-1}\text{t}\big]^1_0$
$=\frac{1}{2}\Big[\tan^{-1}(1)-\tan^{-1}(0)\Big]$
$=\frac{1}{2}\Big[\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)-\tan^{-1}(\tan0)\Big]$
$=\frac{1}{2}\times\frac{\pi}{4}$
$=\frac{\pi}{8}$
View full question & answer→Question 95 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{2}\frac{1}{4+\text{x}-\text{x}^2}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{2}\frac{1}{4+\text{x}-\text{x}^2}\text{ dx}$ Then
$\text{I}=\int_{0}^\limits{2}\frac{1}{\text{x}^2-\text{x}-4}\text{ dx}$
$\Rightarrow\text{I}=-\int_{0}^\limits{2}\frac{1}{\Big(\text{x}^2-\text{x}+\frac{1}{4}\Big)-\frac{1}{4}-4}\text{ dx}$
$\Rightarrow\text{I}=-\int_{0}^\limits{2}\frac{1}{\Big(\text{x}-\frac{1}{2}\Big)^2-\frac{17}{4}}\text{ dx}$
$\Rightarrow\text{I}=-\int_{0}^\limits{2}\frac{1}{\Big(\text{x}-\frac{1}{2}\Big)^2-\Big(\frac{\sqrt{17}}{2}\Big)^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{2}\frac{1}{-\big(\frac{2\text{x}-1}{2}\big)^2+\big(\frac{\sqrt{17}}{2}\big)^2}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big[\log\Big(\frac{\sqrt{17}+2\text{x}-1}{\sqrt{17}-2\text{x}+1}\Big)\Big]^2_0$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{\sqrt{17}+3}{\sqrt{17}-3}-\log\frac{\sqrt{17}-1}{\sqrt{17}+1}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{26+6\sqrt{17}}{8}-\log\frac{18-2\sqrt{17}}{16}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{52+12\sqrt{17}}{18-2\sqrt{17}}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{52+12\sqrt{17}}{18-2\sqrt{17}}\times\frac{18+2\sqrt{17}}{18+2\sqrt{17}}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\log\frac{1344+320\sqrt{17}}{256}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\log\frac{21+5\sqrt{17}}{4}$
View full question & answer→Question 105 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{4}}\frac{\tan^{3}\text{x}}{1+\cos2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\tan^{3}\text{x}}{1+\cos2\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\tan^{3}\text{x}}{1+\cos2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{4}}\tan^{3}\text{x}\sec^2\text{x}\text{ dx}$
Let $\tan\text{x}=\text{t}$ Then, $\sec^2\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{4},\text{t}=1$
$\therefore\ \text{I}=\frac{1}{2}\int_{0}^\limits{1}\text{t}^3\text{ d}t$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\frac{\text{t}^4}{4}\Big]^1_0$
$\Rightarrow\text{I}=\frac{1}{2}\Big(\frac{1}{4}-0\Big)$
$\Rightarrow\text{I}=\frac{1}{8}$
View full question & answer→Question 115 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$ Then,
$\text{I}=\int^\limits{\pi}_0\sin\text{x }\sin^2\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
$\Rightarrow\text{I}=\int^\limits{\pi}_0\sin\text{x}(1-2\cos^2\text{x})(1+\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
$\Rightarrow\text{I}=\int^\limits{\pi}_0\sin\text{x}(1-\cos\text{x})(1+2\cos\text{x})(1+\cos\text{x})^3\text{ dx}$
Let $\cos\text{x}=\text{t}$ Then, $-\sin\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=1$ and $\text{x}=\pi,\text{t}=-1$
$\therefore\ \text{I}=-\int^\limits{-1}_1(1-\text{t})(1+2\text{t})(1+\text{t})^3\text{ dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+\text{t}-2\text{t}^2\big)\big(1+\text{t}^3+3\text{t}+3\text{t}^2\big)\text{dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+\text{t}^3+3\text{t}+3\text{t}^3+\text{t}+\text{t}^4\\+3\text{t}^2+3\text{t}^3-2\text{t}^2-2\text{t}^5-6\text{t}^3-6\text{t}^4\big)\text{dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+4\text{t}+4\text{t}^2-2\text{t}^3-5\text{t}^4-2\text{t}^5\big)\text{dt}$
$\Rightarrow\text{I}=\Big[\text{t}+2\text{t}^2+\frac{4\text{t}^3}{3}-\frac{\text{t}^4}{2}-\text{t}^5-\frac{\text{t}^6}{3}\Big]^1_{-1}$
$\Rightarrow\text{I}=1+2+\frac{4}{3}-\frac{1}{2}-1-\frac{1}{3}\\+1-2+\frac{4}{3}+\frac{1}{2}-1+\frac{1}{3}$
$\Rightarrow\text{I}=\frac{8}{3}$
View full question & answer→Question 125 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}$ $\big[\because2\cos\text{C}\cos\text{D}=\cos(\text{C}+\text{D})-\cos(\text{C}-\text{D})\big]$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{6}}2\cos\text{x }\cos2\text{x dx}$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{6}}(\cos3\text{x}+\cos\text{x})\text{dx}$
$=\frac{1}{2}\int\Big[\frac{\sin3\text{x}}{3}+\sin\text{x}\Big]_0^{\frac{\pi}{6}}$
$=\frac{1}{2}\Bigg[\bigg(\frac{\sin3\frac{\pi}{6}}{3}+\sin\frac{\pi}{6}\bigg)-(\sin0-\sin0)\Bigg]$
$=\frac{1}{2}\bigg[\frac{\sin\frac{\pi}{2}}{3}+\sin\frac{\pi}{6}\bigg]$
$=\frac{1}{2}\Big(\frac{1}{3}+\frac{1}{2}\Big)$
$=\frac{1}{2}\Big(\frac{5}{6}\Big)$
$=\frac{5}{12}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}=\frac{5}{12}$
View full question & answer→Question 135 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\tan^{7}\text{x}}{\tan^{7}\text{x}+\cot^7\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan^{7}\text{x}}{\tan^{7}\text{x}+\cot^7\text{x}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan^{7\big(\frac{\pi}{2}-\text{x}\big)}}{\tan^{7}{\big(\frac{\pi}{2}-\text{x}\big)}+\cot^7{\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cot^7\text{x}}{\cot^7\text{x}+\tan^{7}\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan^{7}\text{x}+\cot^7\text{x}}{\tan^{7}\text{x}+\cot^7\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$\Rightarrow2\text{I}=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{2}-0=\frac{\pi}{2}$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 145 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{4}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$
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 155 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 165 Marks
Evaluate the following integrals:
$\int\limits_{0}^{2}\text{x}\sqrt{\text{x}+2}\text{ dx}$
Answer$\int_{0}^\limits{2}\text{x}\sqrt{\text{x}+2}\text{ dx}$
Let $\text{x}+2=\text{t}^2\Rightarrow\text{dx}=2\text{tdt}$
When $\text{x}=0,\text{t}=\sqrt{2}$ and when $\text{x}=2,\text{t}=2$
$\therefore\ \int_{0}^\limits{2}\text{x}\sqrt{\text{x}+2}\text{ dx}=\int_{\sqrt{2}}^\limits{2}\big(\text{t}^2-2\big)\sqrt{\text{t}^2}2\text{t dt}$
$=2\int_{\sqrt{2}}^\limits{2}\big(\text{t}^2-2\big)\text{t}^2\text{ dt}$
$=2\int_{\sqrt{2}}^\limits{2}\big(\text{t}^4-2\text{t}^2\big)\text{dt}$
$=2\Big[\frac{\text{t}^5}{5}-\frac{2\text{t}^3}{3}\Big]^2_\sqrt{2}$
$=2\Big[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}\Big]$
$=2\Big[\frac{96-80-12\sqrt{2}+20\sqrt{2}}{15}\Big]$
$=2\Big[\frac{16+8\sqrt{2}}{15}\Big]$
$=\frac{16\big(2+\sqrt{2}\big)}{15}$
$=\frac{16\sqrt{2}\big(\sqrt{2}+1\big)}{15}$
View full question & answer→Question 175 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^2_{0}\big(\text{x}^2+4\big)\text{dx}$
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, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}^2+4,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{0}\big(\text{x}^2+4\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(0+4)+(\text{h}^2+4)+\ ....\ +\big\{(\text{n}-1)^2\text{h}^2+4\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[4\text{n}+\text{h}^2\big\{1^2+2^2+3^2\ ....\ +(\text{n}-1)^2\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[4\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\bigg[4\text{n}+\frac{2(\text{n}-1)(2\text{n}-1)}{3\text{n}}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{4+\frac{2}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=8+\frac{8}{3}$
$=\frac{32}{3}$
View full question & answer→Question 185 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^2\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^2\text{x dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\pi}_0(\pi-\text{x})\sin(\pi-\text{x})\cos^2(\pi-\text{x})\text{dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin\text{x}\cos^2\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\pi-\text{x}+\text{x})\sin\text{x}\cos^2\text{x dx}$
$\Rightarrow2\text{I}=\pi\int\limits^{\pi}_0\sin\text{x}\cos^2\text{x dx}$
$\Rightarrow2\text{I}=-\pi\int\limits^{\pi}_0\cos^2\text{x}(-\sin\text{x})\text{dx}$
$\Rightarrow2\text{I}=-\pi\Big[\frac{\cos^3\text{x}}{3}\Big]^{\pi}_0$ $\Bigg[\int\big[\text{f(x)}\big]^{\text{n}}\text{f}'(\text{x})\text{dx}=\frac{\big[\text{f(x)}\big]^{\text{n}+1}}{\text{n}+1}+\text{C}\Bigg]$
$\Rightarrow2\text{I}=-\frac{\pi}{3}\big(\cos^3\text{x}-\cos^20\big)$
$\Rightarrow2\text{I}=-\frac{\pi}{3}(-1-1)=\frac{2\pi}{3}$
$\Rightarrow\text{I}=\frac{\pi}{3}$
View full question & answer→Question 195 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 205 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 215 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 225 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 235 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 245 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 255 Marks
Evaluate the following definite integrals:
$\int\limits_{\text{e}}^{\text{e}^2}\Big\{\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big\}\text{dx}$
AnswerLet $\text{I}=\int_{\text{e}}^\limits{\text{e}^2}\Big\{\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big\}\text{dx}$ Then,
$\text{I}=\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{\log\text{x}}\text{ dx}-\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})^2}\text{ dx}$
Integrating by parts
$\Rightarrow\text{I}=\Bigg\{\Big[\frac{\text{x}}{\log\text{x}}\Big]^{\text{e}^2}_\text{e}-\int_{\text{e}}^\limits{\text{e}^2}\frac{-1}{\text{x}(\log\text{x})^2}\text{x dx}\Bigg\}-\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}}{\log\text{x}}\Big]^{\text{e}^2}_\text{e}+\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})^2}\text{ dx}-\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}}{\log\text{x}}\Big]^{\text{e}^2}_\text{e}+0$
$\Rightarrow\text{I}=\frac{\text{e}^2}{\log\text{e}^2}-\frac{\text{e}}{\log\text{e}}$
$\Rightarrow\text{I}=\frac{\text{e}^2}{2\log\text{e}}-\frac{\text{e}}{\log\text{e}}$
$\Rightarrow\text{I}=\frac{\text{e}^2}{2}-\text{e}$
View full question & answer→Question 265 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 275 Marks
Evaluate the following definite integrals:
$\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{1}{1+\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{-\frac{\pi}{4}}^\limits{\frac{\pi}{4}}\frac{1}{1+\sin\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{-\frac{\pi}{4}}^\limits{\frac{\pi}{4}}\frac{1}{1+\sin\text{x}}\times\frac{1-\sin\text{x}}{1-\sin\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{-\frac{\pi}{4}}^\limits{\pi}\frac{1-\sin\text{x}}{1-\sin^2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{-\frac{\pi}{4}}^\limits{\pi}\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{-\frac{\pi}{4}}^\limits{\pi}\Big(\frac{1}{\cos^2\text{x}}-\frac{\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$\Rightarrow\text{I}=\int_{-\frac{\pi}{4}}^\limits{\pi}\big(\sec^2\text{x}-\sec\text{x}\tan\text{x}\big)\text{ dx}$
$\Rightarrow\text{I}=\big[\tan\text{x}-\sec\text{x}\big]^\frac{\pi}{4}_{-\frac{\pi}{4}}$
$\Rightarrow\text{I}=\big(1-\sqrt{2}\big)-\big(-1-\sqrt{2}\big)$
$\Rightarrow\text{I}=2$
View full question & answer→Question 285 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^\limits{\frac{\pi}{2}}\text{x}^2\sin\text{x}\text{ dx}$
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 295 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\cos^3\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\cos^3\text{x}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\cos^3\text{x}\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(1-\sin^2\text{x}\big)\cos\text{x}\text{ dx}$
Let $\text{u}=\sin\text{x},\text{ du}=\cos\text{x dx}$
$\Rightarrow\text{I}=\int(1-\text{u}^2)\text{du}$
$\Rightarrow\text{I}=\Big[\text{u}-\frac{\text{u}^3}{3}\Big]$
$\Rightarrow\text{I}=\Big[\sin\text{x}-\frac{\sin^3\text{x}}{3}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=1-\frac{1}{3}-0$
$\Rightarrow\text{I}=\frac{2}{3}$
View full question & answer→Question 305 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\pi}\frac{1}{3+2\sin\text{x}+\cos\text{x}}\text{ dx}$
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 315 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\pi}\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
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 325 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\sin2\text{x }\tan^{-1}(\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_0\sin2\text{x }\tan^{-1}(\sin\text{x})\text{dx}$
Differentiating w.r.t. x, we get
$\cos\text{x dx}=\text{dt}$
Now, $\text{x}=0,\text{t}=0$
$\text{x}=\frac{\pi}{2},\text{t}=1$
$\therefore\ \int^\limits{\frac{\pi}{2}}_0\sin2\text{x }\tan^{-1}(\sin\text{x})\text{dx}=2\int^\limits{1}_0\text{t }\tan^{-1}\text{t dt}$ $\big[\because\sin2\text{x}=2\sin\text{x }\cos\text{x}\big]$
Using by parts
$=2\Big\{\tan^{-1}\text{t}\int\text{t dt}-\int\big(\int\text{t dt}\big)\frac{\text{d}\tan^{-1}\text{t}}{\text{dt}}\Big\}$
$=2\Big\{\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\frac{1}{2}\int\frac{\text{t}^2}{1+\text{t}^2}\text{ dt}\Big\}$
$=2\bigg\{\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\frac{1}{2}\Big(\int\text{dt}-\int\frac{\text{dt}}{1+\text{t}^2}\text{ dt}\Big)\bigg\}$
$=2\Big[\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\frac{1}{2}\big(\text{t}-\tan^{-1}\text{t}\big)\Big]^1_0$
$=2\bigg\{\frac{1}{2}\frac{\pi}{4}-\frac{1}{2}\Big(1-\frac{\pi}{4}\Big)\bigg\}$
$=2\Big\{\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}\Big\}$
$=2\Big(\frac{\pi}{4}-\frac{1}{2}\Big)$
$=\frac{\pi}{2}-1$
View full question & answer→Question 335 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\cos^4\text{x}\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{\frac{\pi}{2}}\cos^4\text{x}\text{ dx}$
$=\frac{1}{4}\int_{0}^\limits{\frac{\pi}{2}}(1+\cos2\text{x})^2\text{dx}$ $\big[\because2\cos^2\text{x}=1+\cos2\text{x}\big]$
$=\frac{1}{4}\int_{0}^\limits{\frac{\pi}{2}}\big(1+\cos^22\text{x}+2\cos2\text{x}\big)\text{dx}$
$=\frac{1}{4}\int_{0}^\limits{\frac{\pi}{2}}\Big(1+\frac{1+\cos4\text{x}}{2}+2\cos2\text{x}\Big)\text{dx}$
$=\frac{1}{4}\Big[\text{x}+\frac{1}{2}\text{x}+\frac{\sin4\text{x}}{8}+\sin2\text{x}\Big]^{\frac{\pi}{2}}_0$ $\Big[\because\int\cos4\text{x dx}=\frac{\sin4\text{x}}{4}\Big]$
$=\frac{1}{4}\Big[\frac{\pi}{2}+\frac{\pi}{4}+0+0-0-0-0-0\Big]$
$=\frac{1}{4}\times\frac{3\pi}{4}$
$=\frac{3\pi}{16}$
View full question & answer→Question 345 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{(\sin^3\text{x}\cos^3\text{x})}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{(\sin^3\text{x}\cos^3\text{x})}\text{ dx}$
$=\int^\limits{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{\cos^6\text{x}(\tan^3\text{x}+1)^2}\text{ dx}=\int^\limits{\frac{\pi}{4}}_{0}\frac{\tan^2\text{x}\sec^2\text{x}}{(\tan^3\text{x}+1)}\text{ dx}$
Put $\tan^3\text{x}+1=\text{z}$
$\therefore\ 3\tan^{2}\text{x}\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\tan^{2}\text{x}\sec^2\text{x dx}=\frac{\text{dz}}{3}$
When $\text{x}\rightarrow0,\text{z}\rightarrow1$
When $\text{x}\rightarrow\frac{\pi}{4},\text{z}\rightarrow2$
$\therefore\ \text{I}=\frac{1}{3}\int^\limits{2}_1\frac{\text{dz}}{\text{z}^2}$
$=\frac{1}{3}\times-\Big[\frac{1}{\text{z}}\Big]^2_1$
$=-\frac{1}{3}\Big(\frac{1}{2}-1\Big)$
$=-\frac{1}{3}\times\Big(-\frac{1}{2}\Big)$
$=\frac{1}{6}$
View full question & answer→Question 355 Marks
Evaluate the following integrals:
$\int\limits^2_{0}\big|\text{x}^2-3\text{x}+2\big|\text{dx}$
Answer$\int^\limits2_{0}\big|\text{x}^2-3\text{x}+2\big|\text{dx}$
We know that,
$\big|\text{x}^2-3\text{x}+2\big|\text{dx}=\begin{cases}-(\text{x}^2-3\text{x}+2),&(\text{x}-1)(\text{x}-2)\leq0\text{ or },&1\leq\text{x}\leq2\$\text{x}^2-3\text{x}+2),&\text{x}^2-3\text{x}+2\leq0\text{ or },&\text{x}\in(-\infty,1)(2,\infty)\end{cases}$
$\therefore\ \text{I}=\ \int^\limits2_{0}\big(\text{x}^2-3\text{x}+2\big)\text{dx}$
$\Rightarrow\text{I}=\int^\limits1_{0}\big(\text{x}^2-3\text{x}+2\big)\text{dx}-\int^\limits2_1\big(\text{x}^2-3\text{x}+2\big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}^3}{3}-\frac{3\text{x}^2}{2}+2\text{x}\Big]^1_0-\Big[\frac{\text{x}^3}{3}-\frac{3\text{x}^2}{2}+2\text{x}\Big]^2_1$
$\Rightarrow\text{I}=\frac{1}{3}-\frac{3}{2}+2-\Big[\frac{8}{3}-6+4-\frac{1}{3}+\frac{3}{2}-2\Big]$
$\Rightarrow\text{I}=\frac{1}{3}-\frac{3}{2}+2-\frac{8}{3}+6-2+\frac{1}{3}-\frac{3}{2}$
$\Rightarrow\text{I}=1$
View full question & answer→Question 365 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{1}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\int_{0}^\limits{\frac{\pi}{2}}\begin{pmatrix}\frac{\frac{1}{\cos^2\text{x}}}{\text{a}^2\frac{\sin^2\text{x}}{\cos^2\text{x}}+\text{b}^2\frac{\cos^2\text{x}}{\cos^2\text{x}}} \end{pmatrix}\text{dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\Big(\frac{\sec^2\text{x}}{\text{a}^2\tan^{2}\text{x}+\text{b}^2}\Big)\text{dx}$
$=\frac{1}{\text{a}^2}\int_{0}^\limits{\frac{\pi}{2}}\Bigg(\frac{\sec^2\text{x}}{\tan^{2}\text{x}+\big(\frac{\text{b}}{\text{a}}\big)^2}\Bigg)\text{dx}$
Let $\tan\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$\sec^2\text{xdx}=\text{dt}$
When $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=\infty$
$\therefore\ \frac{1}{\text{a}^2}\int_{0}^\limits{\frac{\pi}{2}}\Bigg(\frac{\sec^2\text{x}}{\tan^{2}\text{x}+\big(\frac{\text{b}}{\text{a}}\big)^2}\Bigg)\text{dx}$
$=\frac{1}{\text{a}^2}\int_{0}^\limits{\infty}\frac{\text{dt}}{\big(\frac{\text{b}}{\text{a}}\big)^2+\text{t}^2}$
$=\frac{1}{\text{a}^2}\Big[\frac{\text{a}}{\text{b}}\tan^{-1}\frac{\text{at}}{\text{b}}\Big]^{\infty}_0$
$=\frac{1}{\text{a}^2}\frac{\text{a}}{\text{b}}\big[\tan^{-1}\infty-\tan^{-1}0\big]$
$=\frac{1}{\text{ab}}\Big[\tan^{-1}\tan\frac{\pi}{2}\Big]$
$=\frac{\pi}{2\text{ab}}$
View full question & answer→Question 375 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\text{ dx}$
AnswerLet $\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\times\frac{\sqrt{1-\sin\text{x}}}{\sqrt{1-\sin\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sqrt{1-\sin^2\text{x}}}{\sqrt{1-\sin^2\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\cos\text{x}}{\sqrt{1-\sin\text{x}}}\text{ dx}$
Let $1-\sin\text{x}=\text{u}$
$\Rightarrow-\cos\text{x dx}=\text{du}$
$\therefore\ \text{I}=\int\frac{-\text{du}}{\sqrt{\text{u}}}$
$\Rightarrow\text{I}=\big[-2\sqrt{\text{u}}\big]$
$\Rightarrow\text{I}=\big[-2\sqrt{1-\sin\text{x}}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0+2$
$\Rightarrow\text{I}=2$
View full question & answer→Question 385 Marks
Evaluate the following definite integrals:
$\int\limits_{1}^{2}\frac{\text{x}}{(\text{x}+1)(\text{x}+2)}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{2}\frac{\text{x}}{(\text{x}+1)(\text{x}+2)}\text{ dx}$ Then,
$\text{I}=\int_{1}^\limits{2}\Big(\frac{-1}{(\text{x}+1)}+\frac{2}{(\text{x}+2)}\Big)\text{dx}$
$\Rightarrow\text{I}=-\int_{1}^\limits{2}\frac{1}{(\text{x}+1)}\text{ dx}+2\int_{1}^\limits{2}\frac{1}{(\text{x}+2)}\text{ dx}$
$\Rightarrow\text{I}=\big[-\log(\text{x}+1)+2\log(\text{x}+2)\big]^2_1$
$\Rightarrow\text{I}=-\log3+2\log4+\log2-2\log3$
$\Rightarrow\text{I}=5\log2-3\log3$
$\Rightarrow\text{I}=\log2^5-\log3^3$
$\Rightarrow\text{I}=\log\frac{32}{27}$
View full question & answer→Question 395 Marks
Evaluate the following integrals:
$\int\limits_{0}^{{\frac{\pi}{4}}}(\tan\text{x}+\cot\text{x})^{-2}\text{ dx}$
Answer$\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}(\tan\text{x}+\cot\text{x})^{-2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\frac{1}{(\tan\text{x}+\cot\text{x})^{2}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\frac{1}{\Big(\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin\text{x}\cos\text{x}}\Big)^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}(\sin\text{x}\cos\text{x})^2\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\sin^2\text{x}(1-\sin^2\text{x})\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\sin^2\text{x dx}-\int_{0}^\limits{{\frac{\pi}{4}}}\sin^4\text{x dx}$
We know that by reduction formula,
$\int\sin^{\text{n}}\text{x dx}=\frac{\text{n}-1}{\text{n}}\int\sin^{\text{n}-2}\text{x dx}-\frac{\cos\text{x}\sin^{\text{n}-1}}{\text{n}}$
For n = 2
$\Rightarrow\int\sin^2\text{x dx}=\frac{2-1}{2}\int1\text{ dx}-\frac{\cos\text{x}\sin\text{x}}{2}$
$\Rightarrow\int\sin^2\text{x dx}=\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}$
For n = 4
$\Rightarrow\int\sin^4\text{x dx}=\frac{4-1}{4}\int\sin^2\text{x dx}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
$\Rightarrow\int\sin^4\text{x dx}=\frac{3}{4}\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
Hence,
$\text{I}=\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}^{\frac{\pi}{4}}_0-\Big\{\frac{3}{4}\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}\Big\}^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\Big\{\frac{\pi}{8}-\frac{1}{4}\Big\}-\Big\{\frac{3}{4}\Big(\frac{\pi}{8}-\frac{1}{4}\Big)-\frac{1}{16}\Big\}$
$\Rightarrow\text{I}=\frac{\pi}{32}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}(\sin^{\text{x}}\cos\text{x})^2\text{dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x}(1-\sin^2\text{x})\text{dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x}-\sin^4\text{x dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x dx}-\int_{0}^\limits{\frac{\pi}{2}}\sin^4\text{x dx}$
We know, by reduction formula,
$\int\sin^{\text{n}}\text{x dx}=\frac{\text{n}-1}{\text{n}}\int\sin^{\text{n}-2}\text{x dx}-\frac{\cos\text{x }\sin^{\text{n}-1}\text{x}}{\text{n}}$
For n = 2
$\Rightarrow\int\sin^2\text{x dx}=\frac{2-1}{2}\int1\text{ dx}-\frac{\cos\text{x}\sin\text{x}}{2}$
$\Rightarrow\int\sin^2\text{x dx}=\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}$
For n = 4
$\Rightarrow\int\sin^4\text{x dx}=\frac{4-1}{4}\int\sin^2\text{x dx}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
$\Rightarrow\int\sin^4\text{x dx}=\frac{3}{4}\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
Hence,
$\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}^{\frac{\pi}{2}}_0-\Big\{\frac{3}{4}\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}\Big\}^{\frac{\pi}{2}}_0$
$\Rightarrow\frac{\pi}{4}-\frac{3}{4}\Big\{\frac{\pi}{4}\Big\}$
$\Rightarrow\frac{\pi}{16}$
View full question & answer→Question 405 Marks
Evaluate the following integrals:
$\int\limits^{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}$
AnswerWe have,
$\int^\limits{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}=(\cos^{-1}\text{x}\big)^2\int\limits^1_0\text{dx}-\int\limits^1_0\big(\int\text{dx}\big)\frac{\text{d}\big(\cos^{-1}\text{x}\big)^2}{\text{dx}}\text{ dx}$
$=\Big[\text{x}\big(\cos^{-1}\text{x}\big)^2\Big]^1_0+\int^\limits{1}_0\frac{\text{x}\cdot2\cos^{-1}}{\sqrt{1-\text{x}^2}}\text{ dx}$
Now,
Let $\cos^{-1}\text{x}=\text{t}\Rightarrow-\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$
When $\text{x}=0\Rightarrow\text{t}=\frac{\pi}{2}$
$\text{x}=1\Rightarrow\text{t}=0$
$\therefore\ \int^\limits{1}_0\frac{2\text{ x}\cos^{-1}}{\sqrt{1-\text{x}^2}}\text{ dx}=-2\int^0_\limits{\frac{\pi}{2}}\text{t}\cos\text{t dt}=2\int^\limits{\frac{\pi}{2}}_0\text{t}\cos\text{t dt}$
$=2\Big[\text{t}\int\cos\text{t dt}-\int\big(\cos\text{t dt}\big)\frac{\text{dt}}{\text{dt}}\text{ dt}\Big]^{\frac{\pi}{2}}_0$
$=2\Big[\text{t}\sin\text{t}-\int\sin\text{t dt}\Big]^{\frac{\pi}{2}}_0$
$=2\Big[\text{t}\sin\text{t}+\cos\text{t}\Big]^{\frac{\pi}{2}}_0$
$=2\Big[\frac{\pi}{2}-1\Big]$
$\int^\limits{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}=\Big[\text{x}\big(\cos^{-1}\text{x}\big)^2\Big]^1_0+\int^\limits{1}_0\frac{\text{x}\cdot2\cos^{-1}}{\sqrt{1-\text{x}^2}}\text{ dx}\\=\Big[\text{x}\big(\cos^{-1}\text{x}\big)^2\Big]^1_0+2\Big(\frac{\pi}{2}-1\Big)$
$=0-0+2\Big(\frac{\pi}{2}-1\Big)$
$=(\pi-2)$
View full question & answer→Question 415 Marks
Evaluate the following definite integrals:
$\int\limits_{0}^{1}\frac{1}{\sqrt{(\text{x}-1)(2-\text{x})}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{1}{\sqrt{(\text{x}-1)(2-\text{x})}}\text{ dx}$
Put $\text{x}=\cos^2\theta+2\sin^2\theta$
$\therefore\ \text{dx}=2\cos\theta(-\sin\theta)\text{d}\theta+4\sin\theta\cos\theta\text{ d}\theta$
$=2\sin\theta\cos\theta\text{ d}\theta$
Also, $\text{x}=\cos^2\theta+2\sin^2\theta$
$\Rightarrow\text{x}=1+\sin^2\theta$
$\Rightarrow\sin\theta=\sqrt{\text{x}-1}$
When $\text{x}\rightarrow1,\sin\theta\rightarrow0$ or $\theta\rightarrow0$
When $\text{x}\rightarrow2,\sin\theta\rightarrow1$ or $\theta\rightarrow\frac{\pi}{2}$
$\therefore\ \text{I}=\int\limits^2_1\frac{1}{\sqrt{(\text{x}-1)(2-\text{x})}}\text{ dx}$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{2\sin\theta\cos\theta\text{ d}\theta}{\sqrt{(\cos^2\theta+2\sin^2\theta-1)(2-\cos^2\theta-2\sin^2\theta)}}$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{2\sin\theta\cos\theta\text{ d}\theta}{\sqrt{\sin^2\theta\cos^2\theta}}$ $\big(\sin^2\theta+\cos^2\theta=1\big)$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{2\sin\theta\cos\theta\text{ d}\theta}{\sin\theta\cos\theta}$
$\Rightarrow\text{I}=2\int\limits^\frac{\pi}{2}_0\text{d}\theta$
$\Rightarrow\text{I}=2\big[\theta\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}-0\Big)$
$\Rightarrow\text{I}=\pi$
View full question & answer→Question 425 Marks
Evaluate the following integrals:
$\int\limits^{\text{a}}_{-\text{a}}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}}\text{ dx}$
AnswerLet $\text{x}=\text{a}\cos2\theta$
Differentiating w.r.t. x, we get
$\text{dx}=-2\text{a}\sin2\theta$
Now, $\text{x}=-\text{a}\Rightarrow\theta=\frac{\pi}{2}$
$\text{x}=\text{a}\Rightarrow\theta=0$
$\therefore\ \int^\limits{\text{a}}_{-\text{a}}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}}\text{ dx}=\int^\limits0_\frac{\pi}{2}\sqrt{\frac{\text{a}(1-\cos2\theta)}{\text{a}\big(1+\cos2\theta)}}(-2\sin2\theta\big)\text{d}\theta$
$=2\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\frac{\sin\theta}{\cos\theta}\cdot\sin2\theta\text{ d}\theta$ $\begin{bmatrix}\because1-\cos2\theta=2\sin^2\theta\\1+\cos2\theta=2\cos^2\theta\\-\int^\limits\text{b}_\text{a}\text{f(x)}\text{dx}=\int^\limits\text{a}_\text{b}\text{f}(\text{x})\text{dx} \end{bmatrix}$
$=2\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\frac{\sin\theta\cdot2\sin\theta\cos\theta}{\cos\theta}$
$=4\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\sin^{2}\theta\text{ d}\theta$
$=2\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\big(1-\cos2\theta\big)\text{d}\theta$
$=2\text{a}\Big[\theta-\frac{\sin2\theta}{2}\Big]^{\frac{\pi}{2}}_0$
$=2\text{a}\Big[\frac{\pi}{2}-0-0+0\Big]$
$=\pi\text{a}$
View full question & answer→Question 435 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{1}{5+4\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5+4\sin\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5+4\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1+\tan^2\frac{\text{x}}{2}}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)+8\tan\frac{\text{x}}{2}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sec^2\frac{\text{x}}{2}}{5\tan^2\frac{\text{x}}{2}+8\tan\frac{\text{x}}{2}+5}\text{ dx}$
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 $\text{x}=\frac{\pi}{2},\text{t}=1$
$\therefore\ \text{I}=2\int\limits_0^1\frac{1}{5\text{t}^2+8\text{t}+5}\text{ dt}$
$\Rightarrow\text{I}=2\int\limits_0^1\frac{1}{\big(\sqrt{5\text{t}}\big)^2+8\text{t}+5+\Big(\frac{4}{\sqrt{5}}\Big)^2-\Big(\frac{4}{\sqrt{5}}\Big)^2}\text{ dt}$
$\Rightarrow\text{I}=2\int\limits_0^1\frac{1}{\Big(\sqrt{5}\text{t}+\frac{4}{\sqrt{5}}\Big)^2+\frac{9}{5}}\text{ dt}$
$\Rightarrow\text{I}=\frac{2}{3}\begin{bmatrix}\tan^{-1}\begin{pmatrix}\frac{\sqrt{5}\text{t}+\frac{4}{\sqrt{5}}}{\frac{3}{\sqrt{5}}} \end{pmatrix}\end{bmatrix}^1_0$
$\Rightarrow\text{I}=\frac{2}{3}\Big[\tan^{-1}3-\tan^{-1}\frac{4}{3}\Big]$
$\Rightarrow\text{I}=\frac{2}{3}\Bigg[\tan^{-1}\Bigg(\frac{3-\frac{4}{3}}{1+3\times\frac{4}{3}}\Bigg)\Bigg]$
$\Rightarrow\text{I}=\frac{2}{3}\tan^{-1}\frac{1}{3}$
View full question & answer→Question 445 Marks
Evaluate the following integrals:
$\int\limits_{0}^{{\pi}}\frac{1}{5+3\cos\text{x}}\text{ dx}$
AnswerWe know that,
$\cos\text{x}=\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}$
$\Rightarrow\ \frac{1}{5+3\cos\text{x}}=\frac{1}{5+3\Bigg(\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)}\\=\frac{1+\tan^{2}\frac{\text{x}}{2}}{5\big(1+\tan^{2}\frac{\text{x}}{2}\big)+3\big(1-\tan^{2}\frac{\text{x}}{2}\big)}=\frac{\sec^2\frac{\text{x}}{2}\text{ dx}}{8+2\tan^{2}\frac{\text{x}}{2}}$
$\therefore\ \int_{0}^\limits{{\pi}}\frac{1}{5+3\cos\text{x}}\text{ dx}=\frac{1}{2}\int_{0}^\limits{{\pi}}\frac{\sec^2\frac{\text{x}}{2}}{2^2+2\tan^{2}\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
Differentiating w.r.t. x, we get
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\pi\Rightarrow\text{t}=\infty$
$\therefore\ \frac{1}{2}\int^\limits\pi_0\bigg(\frac{\sec^2\frac{\text{x}}{2}\text{ dx}}{2^2+\tan^{2}\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int^\limits\infty_0\frac{\text{dt}}{2^2+\text{t}^2}$
$=\Big[\frac{1}{2}\tan^{-1}\Big(\frac{\text{t}}{2}\Big)\Big]^{\infty}_0$
$=\frac{1}{2}\Big[\tan^{-1}(\infty)-\tan^{-1}(0)\Big]$
$=\frac{1}{2}\Big[\tan^{-1}\Big(\tan\frac{\pi}{2}\Big)-\tan^{-1}\big(\tan0\big)\Big]$
$=\frac{1}{2}\Big[\frac{\pi}{2}-0\Big]$
$=\frac{\pi}{4}$
View full question & answer→Question 455 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{1}{5\cos\text{x}+3\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5\cos\text{x}+3\sin\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+3\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\text{ dx}$ $\Bigg[\because\sin\text{A}=\bigg(\frac{2\tan\frac{\text{A}}{2}}{1+\tan^2\frac{\text{A}}{2}}\bigg),\cos\text{A}=\bigg(\frac{1-\tan^2\frac{\text{A}}{2}}{1+\tan^2\frac{\text{A}}{2}}\bigg)\Bigg]$
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1+\tan^2\frac{\text{x}}{2}}{5-5\tan^2\frac{\text{x}}{2}+6\tan\frac{\text{x}}{2}}\text{ dx}$
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sec^2\frac{\text{x}}{2}}{5-5\tan^2\frac{\text{x}}{2}+6\tan\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Then, $\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$
Also, $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2}, \text{t}=1$
$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{2\text{ dt}}{5-5\text{t}^2+6\text{t}}$
$\text{I}=\frac{1}{5}\int_{0}^\limits{1}\frac{2\text{dt}}{1-\text{t}^2+\frac{6}{5}\text{t}+\frac{36}{100}-\frac{36}{100}}$
$=\frac{2}{5}\int_{0}^\limits{1}\frac{\text{dt}}{-\big(\text{t}-\frac{6}{10}\big)^2+\frac{136}{100}}$
$=\frac{2}{5}\times\frac{10}{\sqrt{136}}\Bigg[-\log\Bigg(\frac{\text{t}-\frac{6}{10}-\frac{\sqrt{136}}{10}}{\text{t}-\frac{6}{10}+\frac{\sqrt{136}}{10}}\Bigg)\Bigg]^1_0$
$=\frac{1}{\sqrt{34}}\bigg[-\log\bigg(\frac{4-2\sqrt{34}}{4+2\sqrt{34}}\bigg)+\log\bigg(\frac{-6-2\sqrt{34}}{-6+2\sqrt{34}}\bigg)\bigg]$
$=\frac{1}{\sqrt{34}}\log\bigg(\frac{6+2\sqrt{34}}{6-2\sqrt{34}}\times\frac{4+2\sqrt{34}}{4-2\sqrt{34}}\bigg)$
$=\frac{1}{\sqrt{34}}\log\bigg(\frac{160+20\sqrt{34}}{160-20\sqrt{34}}\bigg)$
$=\frac{1}{\sqrt{34}}\log\bigg(\frac{8+\sqrt{34}}{8-\sqrt{34}}\bigg)$
View full question & answer→Question 465 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{0}\frac{\cos\text{x}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\cos\text{x}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^{\text{n}}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\big)}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^{\text{n}-1}}\text{ dx}$
Put $\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}=\text{z}$
$\therefore\ \Big(-\sin\frac{\text{x}}{2}\times\frac{1}{2}+\cos\frac{\text{x}}{2}\times\frac{1}{2}\Big)\text{dx}=\text{dz}$
$\Big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\Big)\text{dx}=2\text{dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow1$
When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow\sqrt{2}$ $\Big(\text{z}=\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\Big)$
$\therefore\ \text{I}=2\int^\limits{\sqrt{2}}_{1}\frac{\text{dz}}{\text{z}^{\text{n}-1}}$
$=2\times\Big[\frac{\text{z}^{\text{n}-1}}{2-\text{n}}\Big]^{\sqrt{2}}_1$
$=\frac{2}{(2-\text{n})}\Big[\big(\sqrt{2}\big)^{2-\text{n}}-1\Big]$
$=\frac{2}{(2-\text{n})}\bigg(2^{\frac{2}{(2-\text{n})}}-1\bigg)$
$=\frac{2}{(2-\text{n})}\Big(2^{1-\frac{\text{n}}{2}}-1\Big)$
View full question & answer→Question 475 Marks
Evaluate the following definite integrals:
$\int\limits_{1}^{4}\frac{\text{x}^2+\text{x}}{\sqrt{2\text{x}+1}}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{4}\frac{\text{x}^2+\text{x}}{\sqrt{2\text{x}+1}}\text{ dx}$
Let $2\text{x}+1=\text{u}$
$\Rightarrow\text{x}=\frac{\text{u}-1}{2}$
$\Rightarrow\text{dx}=\frac{\text{du}}{2}$
$\therefore\ \text{I}=\int\frac{\big(\frac{\text{u}-1}{2}\big)^2+\frac{\text{u}-1}{2}}{\sqrt{\text{u}}}\frac{\text{du}}{2}$
$\Rightarrow\text{I}=\frac{1}{8}\int\frac{\text{u}^2+1-2\text{u}+2\text{u}-2}{\sqrt{\text{u}}}\text{ du}$
$\Rightarrow\text{I}=\frac{1}{8}\int\frac{(\text{u}^2-1)}{\sqrt{\text{u}}}\text{ du}$
$\Rightarrow\text{I}=\frac{1}{8}\int\Big(\text{u}^{\frac{3}{2}}-\text{u}^{-\frac{1}{2}}\Big)\text{du}$
$\Rightarrow\text{I}=\frac{1}{8}\bigg[\frac{2\text{u}^{\frac{5}{2}}}{5}-\frac{2\text{u}^{\frac{1}{2}}}{1}\bigg]$
$\Rightarrow\text{I}=\frac{1}{8}\Big[\frac{2}{5}\times243-6-\frac{2}{5}\times9\sqrt{3}+2\sqrt{3}\Big]$
$\Rightarrow\text{I}=\frac{1}{8}\Big[\frac{456}{5}-\frac{8\sqrt{3}}{5}\Big]$
$\Rightarrow\text{I}=\frac{57-\sqrt{3}}{5}$
View full question & answer→Question 485 Marks
Evaluate the following integrals:
$\int\limits_{0}^{{2\pi}}\sqrt{1-\sin\frac{\text{x}}{2}}\text{ dx}$
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 495 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}-\cos^3\text{x}}(\sec^3\text{x}-1)\cos^2\text{x dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}-\cos^3\text{x}}(\sec^3\text{x}-1)\cos^2\text{x dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}(1-\cos^2\text{x}\big)}(-\tan^2\text{x})\cos^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}(\sin^2\text{x})}\sin^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}}(\sin\text{x})\sin^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}}\big(1-\cos^2\text{x}\big)\sin\text{x dx}$ $\Big(\sin\text{x}=\sin\text{x}\text{ for }0\leq\text{x}\leq\frac{\pi}{2}\Big)$
Put $\cos\text{x}=\text{z}^2$
$\therefore\ -\sin\text{x dx}=2\text{z dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow0$
When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow0$
$\therefore\ \text{I}=-\int^\limits0_1\text{z}(1-\text{z}^4)2\text{z dz}$
$=-2\int^0\limits_1\text{z}^2\text{ dz}+2\int^\limits0_1\text{z}^6\text{ dz}$
$=-2\times\Big[\frac{\text{z}^3}{3}\Big]^0_1+2\times\Big[\frac{\text{z}^7}{7}\Big]^0_1$
$=-\frac{2}{3}\big(0-1\big)+\frac{2}{7}\big(0-1\big)$
$=\frac{2}{3}-\frac{2}{74}$
$=\frac{8}{21}$
View full question & answer→Question 505 Marks
Evaluate the following integrals:
$\int\limits_{\frac{1}{3}}^{1}\frac{\big(\text{x}-\text{x}^3\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int_\limits{\frac{1}{3}}^{1}\frac{\big(\text{x}-\text{x}^3\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\Bigg[\text{x}^3\Big(\frac{\text{x}}{\text{x}^3}-1\Big)\Bigg]^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\text{x}\big(\frac{1}{\text{x}^2}-1\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\text{x}\big(\frac{1}{\text{x}^2}-1\big)^{\frac{1}{3}}}{\text{x}^3}\text{ dx}$
Put $\Big(\frac{1}{\text{x}^2}-1\Big)=\text{Z}$
$\therefore\ -\frac{2}{\text{x}^3}\text{ dx}=\text{dz}$
$\Rightarrow\frac{\text{dx}}{\text{x}^3}=-\frac{\text{dz}}{2}$
When $\text{x}\rightarrow\frac{1}{3},\text{z}\rightarrow8$
When $\text{x}\rightarrow1,\text{z}\rightarrow0$
$\therefore\ \text{I}=-\frac{1}{2}\int^\limits0_8\text{z}^{\frac{1}{3}}\text{ dz}$
$=-\frac{1}{2}\times\Bigg[\frac{\text{z}^{\frac{4}{3}}}{\frac{4}{3}}\Bigg]^0_8$
$=-\frac{3}{8}\Big[0-(8)^{\frac{4}{3}}\Big]$
$=-\frac{3}{8}\times(-16)$
$=6$
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