Question 13 Marks
Evaluate the following integrals:
$\int^\limits4_{-4}|\text{x}+2|\text{dx}$
AnswerWe have,
$\int^\limits4_{-4}|\text{x}+2|\text{dx}$
$=\int^\limits{-2}_{-4}-(\text{x}+2)\text{dx}+\int^\limits{4}_{-2}(\text{x}+2)\text{dx}$
$=-\Big[\frac{\text{x}^2}{2}+2\text{x}\Big]^{-2}_{-4}+\Big[\frac{\text{x}^2}{2}+2\text{x}\Big]^4_{-2}$
$=-\bigg[\Big(\frac{4}{2}-4-\Big(\frac{16}{2}-8\Big)\bigg]+\bigg[\Big(\frac{16}{2}+8\Big)-\Big(\frac{4}{2}-4\Big)\bigg]$
$=-\big[(-2)-(0)\big]+\big[(16)-(-2)\big]$
$=-\big[-2\big]+\big[16+2\big]$
$=2-18$
$=20$
View full question & answer→Question 23 Marks
Evaluate the following definite integrals:
$\int_{4}^\limits{9}\frac{1}{\sqrt{\text{x}}}\text{ dx}$
AnswerWe know that,
$\text{x}^{\text{n}}\text{ dx}-\frac{\text{x}^{\text{n}-1}}{\text{n}+1}+\text{C}$
Now,
$\int_{4}^\limits{9}\frac{1}{\sqrt{\text{x}}}\text{ dx}$
$=\Bigg[\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]^9_4$
$=\Bigg[\frac{\sqrt{\text{x}}}{\frac{1}{2}}\Bigg]^9_4$
$=2\big[\sqrt{9}-\sqrt{4}\big]$
$=2\big[3-2\big]$
$=2$
View full question & answer→Question 33 Marks
Evaluate the following integrals:
$\int^\limits1_{-1}5\text{x}^4\sqrt{\text{x}^5+1}\text{ dx}$
AnswerLet $\text{I}=\int^\limits1_{-1}5\text{x}^4\sqrt{\text{x}^5+1}\text{ dx}$ Then,
Let $\text{x}^5+1=\text{t}$ Then, $5\text{x}^4\text{ dx}=\text{dt}$
When $\text{x}=-1,\text{t}=0$ and $\text{x}=1,\text{t}=2$
$\therefore\ \text{I}=\int\limits^2_0\sqrt{\text{t}}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{2}{3}\text{t}^{\frac{3}{2}}\Big]^6_0$
$\Rightarrow\text{I}=\frac{2}{3}\sqrt{8}$
$\Rightarrow\text{I}=\frac{4\sqrt{2}}{3}$
View full question & answer→Question 43 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{1}{2}}\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{1}{2}}\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{ dx}$
Put $\text{x}=\sin\theta$
$\therefore\text{ dx}=\cos\theta\text{ d}\theta$
When $\text{x}\rightarrow0,\theta\rightarrow0$
When $\text{x}\rightarrow\frac{1}{2},\theta\rightarrow\frac{\pi}{6}$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{6}}\frac{\sin\theta\sin^{-1}(\sin\theta)}{\cos\theta}\cos\theta\text{ d}\theta$
$=\int_{0}^\limits{\frac{\pi}{6}}\theta\sin\theta\text{ d}\theta$
Applying integration by parts, we have
$\text{I}=\big[\theta\big(-\cos\theta\big)\big]^{\frac{\pi}{6}}_0-\int_{0}^\limits{\frac{\pi}{6}}1\times\big(-\cos\theta\big)\text{d}\theta$
$=-\Big(\frac{\pi}{6}\cos\frac{\pi}{6}-0\Big)+\int_{0}^\limits{\frac{\pi}{6}}\cos\theta\text{ d}\theta$
$=-\frac{\pi}{6}\times\frac{\sqrt{3}}{2}+\big[\sin\theta\big]^{\frac{\pi}{6}}_0$
$=-\frac{\pi}{4\sqrt{3}}+\Big(\sin\frac{\pi}{6}-\sin0\Big)$
$=-\frac{\pi}{4\sqrt{3}}+\Big(\frac{1}{2}-0\Big)$
$=\frac{1}{2}-\frac{\pi}{4\sqrt{3}}$
View full question & answer→Question 53 Marks
Evaluate the following integrals:
$\int^\limits{\pi}_{0}5\big(5-4\cos\theta\big)^{\frac{1}{4}}\sin\theta\text{ d}\theta$
AnswerLet $\text{I}=\int^\limits{\pi}_{0}5\big(5-4\cos\theta\big)^{\frac{1}{4}}\sin\theta\text{ d}\theta$
Let $\big(5-4\cos\theta\big)=\text{t}$ Then, $4\sin\theta\text{ d}\theta=\text{dt}$
When $\theta=0,\text{t}=1$ and $\theta=\pi,\text{t}=9$
$\therefore\ \text{I}=\frac{5}{4}\int\limits^9_1\text{t}^{\frac{1}{4}}\text{ dt}$
$\Rightarrow\text{I}=\frac{5}{4}\Bigg[\frac{4\text{t}^{\frac{5}{4}}}{5}\Bigg]^9_1$
$\Rightarrow\text{I}=\big(9\sqrt{3}-1\big)$
View full question & answer→Question 63 Marks
Evaluate the following integrals:
$\int\limits^1_{-1}\text{x|x|}\text{dx}$
Answer$|\text{x}|=\begin{cases}-\text{x},&-1<\text{x}<0\\\text{x},&0<\text{x}<1\end{cases}$
$\therefore\ \text{x}|\text{x}|=\begin{cases}-\text{x}^2,&-1<\text{x}<0\\\text{x}^2,&0<\text{x}<1\end{cases}$
Now, $\int\limits^1_{-1}\text{x|x|}\text{dx}$
$=\int\limits^0_{-1}-\text{x}^2\text{ dx}+\int\limits^1_{0}\text{x}^2\text{ dx}$
$=-\int\limits^0_{-1}\text{x}^2\text{ dx}+\int\limits^1_{0}\text{x}^2\text{ dx}$
$=-\Big[\frac{\text{x}^3}{3}\Big]^0_{-1}+\Big[\frac{\text{x}^3}{3}\Big]^1_0$
$=-\Big(0+\frac{1}{3}\Big)+\Big(\frac{1}{3}-0\Big)$
$=0-\frac{1}{3}+\frac{1}{3}-0$
$=0$
View full question & answer→Question 73 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\cos\text{x}}{1+\sin^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\cos\text{x}}{1+\sin^2\text{x}}\text{ dx}$
Let $\sin\text{x}=\text{t}$ Then, $\cos\text{x dx}=\text{dt}$
When, $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2},\text{t}=1$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\cos\text{x}}{1+\sin^2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{1}\frac{1}{1+\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=\big[\tan^{-1}\text{t}\big]^1_0$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 83 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ d}\theta$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ d}\theta$
Let $\cos\theta=\text{t}$ Then, $-\sin\theta\text{ d}\theta=\text{dt}$
When $\theta=0,\text{t}=1$ and $\theta=\frac{\pi}{2},\text{t}=0$
$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\theta}{\sqrt{1+\cos\theta}}\text{ d}\theta$
$=\int_{1}^\limits{0}\frac{-\text{dt}}{\sqrt{1+\text{t}}}$
$=\int_{0}^\limits{1}\frac{\text{dt}}{\sqrt{1+\text{t}}}$
$=2\big[\sqrt{1+\text{t}}\big]^1_0$
$=2\big(\sqrt{2}-1\big)$
View full question & answer→Question 93 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{\frac{-\pi}{2}}\log\Big(\frac{2-\sin\text{x}}{2+\sin\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_{\frac{-\pi}{2}}\log\Big(\frac{2-\sin\text{x}}{2+\sin\text{x}}\Big)\text{dx}$
Here, $\text{f(x)}=\log\Big(\frac{2-\sin\text{x}}{2+\sin\text{x}}\Big)$
$\text{f}(-\text{x})=\log\Big(\frac{2-\sin\text{x}}{2+\sin\text{x}}\Big)$
$=\log\Big(\frac{2-\sin\text{x}}{2+\sin\text{x}}\Big)=-\log\Big(\frac{2-\sin\text{x}}{2+\sin\text{x}}\Big)=-\text{f(x)}$
Hence f(x) is an odd function
$\therefore\ \text{I}=0$
View full question & answer→Question 103 Marks
Evaluate the following definite integrals:
$\int_{1}^\limits{2}\log\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{2}\log\text{x}\text{ dx}$ Then,
$\text{I}=\int_{1}^\limits{2}1\log\text{x}\text{ dx}$
Integrating by parts.
$\Rightarrow\text{I}=\big[\text{x }\log\text{ x}\big]^2_1-\int_{1}^\limits{2}\frac{1}{\text{x}}\text{x}\text{ dx}$
$\Rightarrow\text{I}=\big[\text{x }\log\text{ x}\big]^2_1-\int_{1}^\limits{2}\text{dx}$
$\Rightarrow\text{I}=\big[\text{x }\log\text{ x}\big]^2_1-\big[\text{x}\big]^2_1$
$\Rightarrow\text{I}=2\log2-2+1$
$\Rightarrow\text{I}=2\log2-1$
View full question & answer→Question 113 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\frac{2\text{x}+3}{5\text{x}^2+1}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{2\text{x}+3}{5\text{x}^2+1}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{1}\frac{2\text{x}+3}{5\text{x}^2+1}\text{ dx}+\int_{0}^\limits{1}\frac{3}{5\text{x}^2+1}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{5}\int_{0}^\limits{1}\frac{10\text{x}}{5\text{x}^2+1}\text{ dx}+3\int_{0}^\limits{1}\frac{1}{(\sqrt{5}\text{x})^2+1}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{5}\Big[\log\big(5\text{x}^2+1\big)\Big]^1_0+\frac{3}{\sqrt{5}}\Big[\tan^{-1}\big(\sqrt{5}\text{x}\big)\Big]^1_0$
$\Rightarrow\text{I}=\frac{1}{5}\log6+\frac{3}{\sqrt{5}}\tan^{-1}\sqrt{5}$
View full question & answer→Question 123 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\text{xe}^{\text{x}^2}\text{ dx}$
AnswerLet $\text{x}^2=\text{t}$
Differentiating w.r.t. x, we get
$2\text{xdx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=1\Rightarrow\text{t}=1$
$\therefore\ \int_{0}^\limits{1}\text{xe}^{\text{x}^2}\text{ dx}=\int_{0}^\limits{1}\frac{\text{e}^{\text{t}}\text{ dt}}{2}$
$=\frac{1}{2}\int_{0}^\limits{1}\text{e}^{\text{t}}\text{ dt}$
$=\frac{1}{2}\big[\text{e}^{\text{t}}\big]^1_0$
$=\frac{1}{2}\big[\text{e}^1-\text{e}_0\big]$ $\big[\because\text{e}^0=1\big]$
$=\frac{1}{2}\big(\text{e}-1\big)$
View full question & answer→Question 133 Marks
Evaluate the following integrals:
$\int_{\pi}^\limits{\frac{3\pi}{2}}\sqrt{1-\cos2\text{x}}\text{ dx}$
Answer$\int_{\pi}^\limits{\frac{3\pi}{2}}\sqrt{1-\cos2\text{x}}\text{ dx}$
$=\int_{\pi}^\limits{\frac{3\pi}{2}}\sqrt{2\sin^2\text{x}}\text{ dx}$
$=\sqrt{2}\int_{\pi}^\limits{\frac{3\pi}{2}}\big[\sin\text{x}\big]\text{dx}$
$=\sqrt{2}\int_{\pi}^\limits{\frac{3\pi}{2}}\sin\text{x}\text{ dx}$ $(\sin\text{x}<0\text{ for }\pi\leq\text{x}\leq2\pi)$
$=\sqrt{2}\big[(-\cos\text{x})\big]^{\frac{3\pi}{2}}_\pi$
$=\sqrt{2}\Big(\cos\frac{3\pi}{2}-\cos\pi\Big)$
$=\sqrt{2}\big[0-(-1)\big]$
$=\sqrt{2}\times1$
$=\sqrt{2}$
View full question & answer→Question 143 Marks
Evaluate the following definite integrals:
$\int_{\frac{\pi}{4}}^\limits{\frac{\pi}{2}}\cot\text{x}\text{ dx}$
AnswerWe have,
$\int_{\frac{\pi}{4}}^\limits{\frac{\pi}{2}}\cot\text{x}\text{ dx}$
We know that $\int\cot\text{x dx}=\log(\sin\text{x})$
Now, $\int_{\frac{\pi}{4}}^\limits{\frac{\pi}{2}}\cot\text{x}\text{ dx}$
$=\big[\log(\sin\text{x})\big]^{\frac{\pi}{2}}_\frac{\pi}{4}$
$=\Big[\log\Big(\sin\frac{\pi}{2}\Big)-\log\Big(\sin\frac{\pi}{4}\Big)\Big]$
$=\Big[\log1-\log\frac{1}{\sqrt{2}}\Big]$
$=\big[0-\log\text{2}\big]$
$=\log\sqrt{2}$ $[\because\log\text{a}^{\text{n}}-\text{n}\log\text{a}\big]$
$=\frac{1}{2}\log2$
View full question & answer→Question 153 Marks
Evaluate the following integrals:
$\int\limits^{{\pi}}_0\cos^5\text{x dx}$
AnswerLet $\text{I}=\int\limits^{{\pi}}_0\cos^5\text{x dx}$
$=\int\limits^{{\pi}}_0\cos\text{x}\big(\cos^2\text{x}\big)^2\text{dx}$
$=\int\limits^{{\pi}}_0\cos\text{x}\big(1-\sin^2\text{x}\big)^2\text{dx}$
Let $\sin\text{x}=\text{t},$ then $\cos\text{x dx}=\text{dt}$
When, $\text{x}\rightarrow0;\text{ t}\rightarrow0$ and $\text{x}\rightarrow\pi;\text{ t}\rightarrow0$
Therefore,
$\text{I}=\int\limits^0_0\big(1-\text{t}^2\big)^2\text{dt}$
$\text{I}=0$
View full question & answer→Question 163 Marks
Evaluate the following integrals:
$\int\limits^4_0|\text{x}-1|\text{dx}$
Answer$\int\limits^4_0|\text{x}-1|\text{dx}$
We know that,
$|\text{x}-1|=\begin{cases}-(\text{x}-1),&0\leq\text{x}\leq1\\\text{x}-1,&1<\text{x}\leq4\end{cases}$
$\therefore\ \text{I}=\int\limits^4_0|\text{x}-1|\text{dx}$
$\Rightarrow\text{I}=\int\limits^{1}_0-(\text{x}-1)\text{dx}+\text{I}=\int\limits^{4}_1(\text{x}-1)\text{dx}$
$\Rightarrow\text{I}=\Big[-\frac{\text{x}^2}{2}+\text{x}\big]^1_0+\Big[\frac{\text{x}^2}{2}-\text{x}\big]^4_1$
$\Rightarrow\text{I}=\frac{-1}{2}+1-0+8-4-\frac{1}{2}+1$
$\Rightarrow\text{I}=5$
View full question & answer→Question 173 Marks
Evaluate the following integrals:
$\int\limits^8_2|\text{x}-5|\text{dx}$
Answer$\int\limits^8_2|\text{x}-5|\text{dx}$
We know that,
$|\text{x}-5|=\begin{cases}-(\text{x}-5),&2\leq\text{x}\leq5\\\text{x}-5,&5<\text{x}\leq8\end{cases}$
$\therefore\ \text{I}=\int\limits^8_2|\text{x}-5|\text{dx}$
$\Rightarrow\text{I}=\int\limits^5_2-(\text{x}-5)\text{dx}+\int\limits^8_5(\text{x}-5)\text{dx}$
$\Rightarrow\text{I}=-\Big[\frac{\text{x}^2}{2}-5\text{x}\big]^5_2+\Big[\frac{}{}\frac{\text{x}^2}{2}-5\text{x}\Big]^8_5$
$\Rightarrow\text{I}=\frac{-25}{2}+25+2-10+32-40-\frac{25}{2}+25$
$\Rightarrow\text{I}=9$
View full question & answer→Question 183 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{3+5\cos\text{x}}{3+5\sin\text{x}}\Big)\text{dx}$
AnswerLet, $\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{3+5\cos\text{x}}{3+5\sin\text{x}}\Big)\text{dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\log\Bigg[\frac{3+5\cos\big(\frac{\pi}{2}-\text{x}\big)}{3+5\sin\big(\frac{\pi}{2}-\text{x}\big)}\Bigg]$
$=\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{3+5\sin\text{x}}{3+5\cos\text{x}}\Big)\text{dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\bigg[\log\Big(\frac{3+5\cos\text{x}}{3+5\sin\text{x}}\Big)+\log\Big(\frac{3+5\sin\text{x}}{3+5\cos\text{x}}\Big)\bigg]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{3+5\cos\text{x}}{3+5\sin\text{x}}\times\frac{3+5\sin\text{x}}{3+5\cos\text{x}}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log1\text{ dx}=0$
Hence, $\text{I}=0$
View full question & answer→Question 193 Marks
Evaluate the following definite integrals:
$\int_{2}^\limits{3}\frac{\text{x}}{\text{x}^2+1} \text{ dx}$
AnswerLet $\text{I}=\int_{2}^\limits{3}\frac{\text{x}}{\text{x}^2+1} \text{ dx}$ Then,
$\text{I}=\frac{1}{2}\int_{2}^\limits{3}\frac{2\text{x}}{\text{x}^2+1} $
$\Rightarrow\text{I}=\frac{1}{2}\big[\log(\text{x}^2-1)\big]^3_2$
$\Rightarrow\text{I}=\frac{1}{2}\big(\log10-\log5\big)$
$\Rightarrow\text{I}=\frac{1}{2}\log\frac{10}{5}$ $\Big[\because\log\text{a}-\log\text{b}=\log\frac{\text{a}}{\text{b}}\Big]$
$\Rightarrow\text{I}=\frac{1}{2}\log2$
View full question & answer→Question 203 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\pi}\Big(\sin^2\frac{\text{x}}{2}-\cos^2\frac{\text{x}}{2}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\pi}\Big(\sin^2\frac{\text{x}}{2}-\cos^2\frac{\text{x}}{2}\Big)\text{dx}$
$=-\int_{0}^\limits{\pi}\Big(\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}\big)\text{dx}$
$=-\int_{0}^\limits{\pi}\cos\text{x dx}$
$\int\cos\text{x dx}=\sin\text{x}=\text{F(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(\pi)-\text{F}(0)$
$=\sin(\pi)-\sin0$
$=0$
View full question & answer→Question 213 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_{0}|\cos2\text{x}|\text{dx}$
Answer$\int^\limits{\frac{\pi}{2}}_{0}|\cos2\text{x}|\text{dx}$
We know that,
$|\cos2\text{x}|=\begin{cases}-\cos2\text{x},&\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{2}\\\cos2\text{x},&0<\text{x}\leq\frac{\pi}{4}\end{cases}$
$\therefore\ \text{I}=\int^\limits{2}_{-2}|\cos2\text{x}|\text{dx}$
$\Rightarrow\text{I}=\int^\limits{\frac{\pi}{4}}_{0}\cos2\text{x }\text{dx}-\int^\limits{\frac{\pi}{2}}_{\frac{\pi}{4}}\cos2\text{x }\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0-\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_{\frac{\pi}{4}}$
$\Rightarrow\text{I}=\frac{1}{2}-0-0+\frac{1}{2}$
$\Rightarrow\text{I}=1$
View full question & answer→Question 223 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\cos^2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\cos^2\text{x}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\cos^2\text{x}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{2}}(1+2\cos2\text{x})\text{dx}$ $[\because\cos2\text{x}=2\cos^2\text{x}-1\big]$
$\Rightarrow\text{I}=\Big[\frac{\pi}{2}+\frac{\sin2\text{x}}{4}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=\frac{\pi}{4}+0-0$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 233 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\frac{1}{\sqrt{1+\text{x}}-\sqrt{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{1}{\sqrt{1+\text{x}}-\sqrt{\text{x}}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{1}\bigg(\frac{1}{\sqrt{1+\text{x}}-\sqrt{\text{x}}}\times\frac{\sqrt{1+\text{x}}+\sqrt{\text{x}}}{\sqrt{1+\text{x}}+\sqrt{\text{x}}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{1}\frac{\sqrt{1+\text{x}}-\sqrt{\text{x}}}{1+\text{x}-\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{1}\Big({\sqrt{1+\text{x}}+\sqrt{\text{x}}}\Big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{2}{3}(1+\text{x})^{\frac{3}{2}}+\frac{2}{3}\text{x}^{\frac{3}{2}}\Big]^1_0$
$\Rightarrow\text{I}=\frac{2}{3}\times2\sqrt{2}+\frac{2}{3}-\frac{2}{3}$
$\Rightarrow\text{I}=\frac{4\sqrt{2}}{3}$
View full question & answer→Question 243 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x }\sin2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x }\sin2\text{x}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}2\sin^2\text{x }\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}2(1-\cos^2\text{x})\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(2\cos\text{x}-2\cos^3\text{x}\big)\text{dx}$
$\Rightarrow\text{I}=\bigg[2\sin\text{x}-2\Big(\sin\text{x}-\frac{\sin^3\text{x}}{3}\Big)\bigg]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=\Big[2-2\Big(1-\frac{1}{3}\Big)\Big]-0$
$\Rightarrow\text{I}=\frac{2}{3}$
View full question & answer→Question 253 Marks
Evaluate the following integrals:
$\int^\limits4_1\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}7\text{x}+3,&\text{if }\ 1\leq\text{x}\leq3\\8\text{x},&\text{if }\ 3\leq\text{x}\leq9\end{cases}$
AnswerWe have,
$\int^\limits4_1\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}7\text{x}+3,&\text{if }\ 1\leq\text{x}\leq3\\8\text{x},&\text{if }\ 3\leq\text{x}\leq9\end{cases}$
$\text{I}=\int^\limits4_1\text{f(x)}\text{dx}$
$\Rightarrow\text{I}=\int^\limits{3}_1\text{f(x)}\text{dx}+\int^\limits4_3\text{f(x)}\text{dx}$ [Additive property]
$\Rightarrow\text{I}=\int^\limits{3}_1(7\text{x}+3)\text{dx}+\int^\limits{4}_38\text{x dx}$
$\Rightarrow\text{I}=\Big[\frac{7\text{x}^2}{2}+3\text{x}\big]^3_1+\big[4\text{x}^2\big]^4_2$
$\Rightarrow\text{I}=\frac{63}{2}+9-\frac{7}{2}-3+64-36$
$\Rightarrow\text{I}=\frac{56}{2}+34$
$\Rightarrow\text{I}=62$
View full question & answer→Question 263 Marks
Evaluate the following definite integrals:
$\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{4}}\text{cosec}\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{4}}\text{cosec}\text{x}\text{ dx}$
$\int\text{cosec}\text{x dx}=\log|\text{cosecx}-\cot\text{x}|=\text{F}(\text{x})$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}\Big(\frac{\pi}{4}\Big)-\text{F}\Big(\frac{\pi}{6}\Big)$
$=\log\Big|\text{cosec}\frac{\pi}{4}-\cot\frac{\pi}{4}\Big|-\log\Big|\text{cosec}\frac{\pi}{6}-\cot\frac{\pi}{6}\Big|$
$=\log\big|\sqrt{2}-1\big|-\log\big|2-\sqrt{3}\big|$
$=\log\bigg(\frac{\sqrt{2}-1}{2-\sqrt{3}}\bigg)$
View full question & answer→Question 273 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\frac{1}{2\text{x}^2+\text{x}+1}\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{1}\frac{1}{2\text{x}^2+\text{x}+1}\text{ dx}$
$=\frac{1}{2}\int_{0}^\limits{1}\frac{1\text{ dx}}{\big(\text{x}^2+\frac{1}{2}\text{x}+\frac{1}{2}\big)}$
$=\frac{1}{2}\int_{0}^\limits{1}\frac{\text{dx}}{\big(\text{x}+\frac{1}{4}\big)^2+\frac{1}{2}-\frac{1}{16}}$ $\Big[\text{Adding }\frac{1}{16}\&\text{ substracting }\frac{1}{16}\text{ in numerator}\Big]$
$=\frac{1}{2}\int_{0}^\limits{1}\frac{\text{dx}}{\big(\text{x}+\frac{1}{4}\big)^2+\frac{7}{16}}$
$=\frac{1}{2}\int_{0}^\limits{1}\frac{\text{dx}}{\big(\text{x}+\frac{1}{4}\big)^2+\big(\frac{\sqrt{7}}{4}\big)^2}$
$=\frac{1}{2}\cdot\frac{4}{\sqrt{7}}\Bigg[\tan^{-1}\Bigg(\frac{\text{x}+\frac{1}{4}}{\frac{\sqrt{7}}4{}}\Bigg)\Bigg]^1_0$
$=\frac{2}{\sqrt{7}}\bigg\{\tan^{-1}\frac{5}{\sqrt{7}}-\tan^{-1}\Big(\frac{1}{\sqrt{7}}\Big)\bigg\}$
View full question & answer→Question 283 Marks
Evaluate the following definite integrals:
$\int_{1}^\limits{2}\text{e}^{2\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{2\text{x}^2}\big)\text{dx}$
AnswerLet $\text{I}=\int_{1}^\limits{2}\text{e}^{2\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{2\text{x}^2}\big)\text{dx}$ Then,
$\text{I}=\int_{1}^\limits{2}\text{e}^{2\text{x}} \frac{1}{\text{x}}-\int_{1}^\limits{2}\text{e}^{2\text{x}} \frac{1}{2\text{x}^2}\text{ dx}$
Integrating first term by parts,
$\Rightarrow\text{I}=\bigg\{\Big[\frac{\text{e}^{2\text{x}}}{2\text{x}}\Big]^2_1-\int_{1}^\limits{2}-\text{e}^{2\text{x}} \frac{1}{2\text{x}^2}\bigg\}-\int_{1}^\limits{2}\text{e}^{2\text{x}} \frac{1}{2\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{e}^{2\text{x}}}{2\text{x}}\Big]^2_1$
$\Rightarrow\text{I}=\frac{\text{e}^4}{4}-\frac{\text{e}^2}{2}$
$\Rightarrow\text{I}=\frac{\text{e}^4-2\text{e}^2}{4}$
View full question & answer→Question 293 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{4}}\text{x}^2\sin\text{x}\text{ dx}$
AnswerLet $\int_{0}^\limits{\frac{\pi}{4}}\text{x}^2\sin\text{x}\text{ dx}$ Then,
Integrating by parts.
$\text{I}=\big[-\text{x}^2\cos\text{x}\big]^{\frac{\pi}{4}}_0-\int_{0}^\limits{\frac{\pi}{4}}-2\text{x}\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\big[-\text{x}^2\cos\text{x}\big]^{\frac{\pi}{4}}_0+\big[2\text{x }\sin\text{x}\big]^{\frac{\pi}{4}}_0-\int_{0}^\limits{\frac{\pi}{4}}2\sin\text{x dx}$
$\Rightarrow\text{I}=\big[-\text{x}^2\cos\text{x}\big]^{\frac{\pi}{4}}_0+\big[2\text{x }\sin\text{x}\big]^{\frac{\pi}{4}}_0+\big[2\cos\text{x}\big]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\frac{-\pi^2}{16\sqrt{2}}+\frac{\pi}{2\sqrt{2}}+\frac{2}{\sqrt{1}}-2$
$\Rightarrow\text{I}=\sqrt{2}+\frac{\pi}{2\sqrt{2}}-\frac{\pi^2}{16\sqrt{2}}-2$
View full question & answer→Question 303 Marks
If $\int\limits^{\text{a}}_0\frac{1}{4+\text{x}^2}\text{ dx}=\frac{\pi}{8},$ find the value of a.
Answer$\int\limits^{\text{a}}_0\frac{1}{4+\text{x}^2}\text{ dx}=\frac{\pi}{8}$
$\Rightarrow\frac{1}{2}\tan^{-1}\Big[\frac{\text{x}}{2}\Big]^{\text{a}}_0=\frac{\pi}{8}$ $\bigg[\int\frac{1}{\text{a}^2+\text{x}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}+\text{C}\bigg]$
$\Rightarrow\frac{1}{2}\Big(\tan^{-1}\frac{\text{a}}{2}-\tan^{-1}0\Big)=\frac{\pi}{8}$
$\Rightarrow\tan^{-1}\frac{\text{a}}{2}-0=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\frac{\text{a}}{2}=\frac{\pi}{4}$
$\Rightarrow\frac{\text{a}}{2}=\tan\frac{\pi}{4}=1$
$\Rightarrow\text{a}=2$
Thus, the value of a is 2
View full question & answer→Question 313 Marks
Evaluate the following definite integrals:
$\int_{1}^\limits{\text{e}}\frac{\log\text{x}}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{\text{e}}\frac{\log\text{x}}{\text{x}}\text{ dx}$
Let $\log\text{x}=\text{u}$
$\Rightarrow\frac{1}{\text{x}}=\text{dx}=\text{du}$
$\therefore\text{ I}=\int\text{u}\text{ du}$
$\Rightarrow\text{I}=\Big[\frac{\text{u}^2}{2}\Big]$
$\Rightarrow\text{I}=\Big[\frac{(\log\text{x})}{2}\Big]^{\text{e}}_1$
$\Rightarrow\text{I}=\frac{1}{2}-0$
$\Rightarrow\text{I}=\frac{1}{2}$
View full question & answer→Question 323 Marks
Evaluate the following integrals:
$\int\limits^{\sqrt{2}}_0\big[\text{x}^2\big]\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{\sqrt{2}}_0\big[\text{x}^2\big]\text{dx}$
$=\int\limits^{1}_0\big[\text{x}^2\big]\text{dx}+\int\limits^{\sqrt{2}}_1\big[\text{x}^2\big]\text{dx}$
$=\int\limits^{1}_0(0)\text{dx}+\int\limits^{\sqrt{2}}_1(1)\text{dx}$ $\begin{pmatrix}\because\big[\text{x}\big]^2=\begin{cases}0,&0<\text{x}<1\\1,&1<\text{x}<\sqrt{2}\end{cases}\end{pmatrix}$
$=0+\big[\text{x}\big]^{\sqrt{2}}_1$
$=\sqrt{2}-1$
View full question & answer→Question 333 Marks
Evaluate the following integrals:
$\int_{1}^\limits{2}\frac{3\text{x}}{9\text{x}^2-1}\text{ dx}$
AnswerLet $\text{x}^2=\text{t}$ Then, $2\text{x dx}=\text{dt}$
When $\text{x}=1,\text{t}=1$ and $\text{x}=2,\text{t}=4$
$\therefore\ \text{I}=\int_{1}^\limits{2}\frac{3\text{x}}{9\text{x}^2-1}\text{ dx}$
$\Rightarrow\text{I}=\frac{3}{2}\int_{1}^\limits{4}\frac{\text{dt}}{9\text{t}-1}$
$\Rightarrow\text{I}=\frac{3}{18}\big[\log(9\text{t}-1)\big]^4_1$
$\Rightarrow\text{I}=\frac{3}{18}\big(\log35-\log8\big)$
$\Rightarrow\text{I}=\frac{(\log35-\log8)}{6}$
View full question & answer→Question 343 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
Let $\text{f}(\text{x})=\text{x}\cos^2\text{x}$
$\Rightarrow \text{f}(-\text{x})= (-\text{x})\cos^2(-\text{x})$
$= - \text{x}\cos^2\text{x}$
$\therefore \text{f}(-\text{x})=-\text{f}(\text{x})$
i.e., f(x) is odd function.
We know that $\int\limits^\text{a}_{-\text{a}}\text{f}(\text{x})\text{dx} = 0, $ if f(x) is odd function.
$\therefore\ \text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}=0$
View full question & answer→Question 353 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{ dx}$
AnswerLet $\text{e}^\text{x}=\text{t}$ Then, $\text{e}^\text{x}\text{ dx}=\text{dt}$
When $\text{e}^\text{x}=0,\text{t}=1$ and $\text{x}=1,\text{t}=\text{e}$
$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{1}^\limits{\text{e}}\frac{\text{dt}}{1+\text{t}^{2}}$
$\Rightarrow\text{I}=\big[\tan^{-1}\text{x}\big]^\text{e}_1$
$\Rightarrow\text{I}=\tan^{-1}\text{e}-\tan^{-1}1$
$\Rightarrow\text{I}=\tan^{-1}\text{e}-\frac{\pi}{4}$
View full question & answer→Question 363 Marks
Evaluate the following integrals:
$\int_{4}^\limits{12}\text{x}(\text{x}-4)^{\frac{1}{3}}\text{dx}$
AnswerLet $\text{I}=\int_{4}^\limits{12}\text{x}(\text{x}-4)^{\frac{1}{3}}\text{dx}$
Let $\text{x}-4=\text{t}$ Then, $\text{dx}=\text{dt}$
When $\text{x}=4,\text{t}=0$ and $\text{x}=12,\text{t}=8$
$\therefore\ \text{I}=\int\limits^8_0(\text{t}+4)\text{t}^{\frac{1}{3}}\text{dt}$
$\Rightarrow\text{I}=\int\limits^8_0\Big(\text{t}^{\frac{4}{3}}+4\text{t}^{\frac{1}{3}}\Big)\text{dt}$
$\Rightarrow\text{I}=\Big[\frac{3}{7}\text{t}^{\frac{7}{3}}+\frac{3}{1}\text{t}^{\frac{4}{3}}\Big]^8_0$
$\Rightarrow\text{I}=\frac{384}{7}+48$
$\Rightarrow\text{I}=\frac{720}{7}$
View full question & answer→Question 373 Marks
Evaluate the following integrals:
$\int\limits^2_0\text{x}\sqrt{2-\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^2_0\text{x}\sqrt{2-\text{x}}\text{ dx}$
$=\int\limits^2_0(2-\text{x})\sqrt{2-2+\text{x}}\text{ dx}$
$=\int\limits^2_0(2-\text{x})\sqrt{\text{x}}\text{ dx}$
$=\int\limits^2_0\big(2\sqrt{\text{x}}-\text{x}\sqrt{\text{x}}\big)\text{dx}$
$=\int\limits^2_0\Big(2\text{x}^{\frac{1}{2}}-\text{x}^{\frac{3}{2}}\Big)\text{dx}$
$=\Bigg[2\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{\text{x}^{\frac{5}{2}}}{\frac{5}{2}}\Bigg]$
$=\bigg[\frac{4}{3}\text{x}^{\frac{3}{2}}-\frac{2}{5}\text{x}^{\frac{5}{2}}\bigg]^2_0$
$=\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5}$
$=\frac{16\sqrt{2}}{15}$
View full question & answer→Question 383 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$Let $\tan^{-1}\text{x}=\text{t}$ Then, $\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=1,\text{t}=\frac{\pi}{4}$
$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\sqrt{\text{t}}\text{ dt}$
$\Rightarrow\text{I}=\Bigg[\frac{2\text{t}^{\frac{3}{2}}}{3}\Bigg]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\frac{2}{3}\Big(\frac{\pi}{4}\Big)^{\frac{3}{2}}$
$\Rightarrow\text{I}=\frac{1}{12}\pi^{\frac{3}{2}}$
View full question & answer→Question 393 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 403 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 413 Marks
Evaluate the following integrals:
$\int^\limits1_0\frac{1-\text{x}^2}{(1+\text{x}^2)^2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits1_0\frac{1-\text{x}^2}{(1+\text{x}^2)^2}\text{ dx}$ Then,
$\text{I}=\int^\limits1_0\frac{\big(\frac{1}{\text{x}^2}-1\big)}{\big(\text{x}+\frac{1}{\text{x}}\big)^2}\text{ dx}$
Let $\text{x}+\frac{1}{\text{x}}=\text{t}$ Then, $1-\frac{1}{\text{x}^2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=\infty$ and $\text{x}=1,\text{t}=2$
$\therefore\ \text{I}=\int^\limits2_\infty\frac{-\text{dt}}{\text{t}^2}$
$\Rightarrow\text{I}=\Big[\frac{1}{\text{t}}\Big]^2_\infty$
$\Rightarrow\text{I}=\frac{1}{2}-0$
$\Rightarrow\text{I}=\frac{1}{2}$
View full question & answer→Question 423 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{4}\frac{1}{\sqrt{4\text{x}-\text{x}^2}}\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{4\text{x}-\text{x}^2}}$
$=\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{4-4+4\text{x}-\text{x}^2}}$ [Add and subtract 4 in denominator]
$=\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{4-(\text{x}^2-4\text{x}+4)}}$
$=\int_{0}^\limits{4}\frac{\text{dx}}{\sqrt{(2)^2-(\text{x}-2)^2}}$
$=\Big[\sin^{-1}\Big(\frac{\text{x}-2}{2}\Big)\Big]^4_0$ $\bigg[\because\int\frac{\text{dx}}{\sqrt{\text{a}^2-\text{x}^2}}=\sin^{-1}\frac{\text{x}}{\text{a}}\bigg]$
$=\sin^{-1}(1)-\sin^{-1}(-1)$
$=\frac{\pi}{2}-\Big(-\frac{\pi}{2}\Big)$
$=\frac{2\pi}{2}=\pi$
View full question & answer→Question 433 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 443 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 453 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 463 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 473 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 483 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 493 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin^3\text{x}\text{ dx}$
Answer$\text{Let}\text{ I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin^3\text{x}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin\text{x}\sin^2\text{x}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin\text{x}(1-\cos^2\text{x})\text{dx}$
$\text{Let }\cos\text{x} = \text{t}, \text{then}-\sin\text{x}\text{ dx}=\text{dt}$
$\text{when}, \text{x}\rightarrow-\frac{\pi}{2};\text{t}\rightarrow0\text{ and }\text{x}\rightarrow\frac{\pi}{2};\text{t}\rightarrow0$
$\text{I}=\int\limits^0_0(-1+\text{t}^2)\text{dt}$
$= 0$
View full question & answer→Question 503 Marks
Evaluate the following integrals:
$\int_{2}^\limits{4}\frac{\text{x}}{\text{x}^2+1}\text{ dx}$
AnswerLet $\text{x}^2=\text{t}$ Then, $2\text{x dx}=\text{dt}$
When $\text{x}=2,\text{ t}=4$ and $\text{x}=4,\text{ t}=16$
$\therefore\ \text{I}=\int_{2}^\limits{4}\frac{\text{x}}{\text{x}^2+1}\text{ dx}$
$\Rightarrow\text{I}=\int_{4}^\limits{16}\frac{1}{2}\frac{\text{dt}}{\text{t}+1}$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\log(\text{t}+1)\Big]^{16}_4$
$\Rightarrow\text{I}=\frac{1}{2}\log17-\frac{1}{2}\log5$
$\Rightarrow\text{I}=\frac{1}{2}\log\frac{17}{5}$
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