Question 2015 Marks
Integrate the function in Exercise:$\frac{1}{\sqrt{\sin^{3}\text{x}\sin(\text{x}+\text{a)}}}$
Answer$\frac{1}{\sqrt{\sin^{3}\text{x}\sin(\text{x}+\text{a)}}}=\frac{1}{\sqrt{\sin^{3}\text{x}(\sin\text{x}\cos\text{a}+\cos\text{x}\sin\text{a)}}}$
$=\frac{1}{\sqrt{\sin^{4}\text{x}\cos\text{a}+\sin^{3}\text{x}\cos\text{x}\sin\text{a}}}$
$=\frac{1}{\sin^{2}\text{x}\sqrt{\cos\text{a}+\cot\text{x}\sin\text{a}}}$
$=\frac{\text{cosec}^{2}\text{x}}{\sqrt{\cos\text{a}+\cot\text{x}\sin\text{a}}}$
$\text{Let}\cos\text{a}+\cot\text{x}\sin\text{a}=\text{t}\Rightarrow-\text{cosec}^{2}\text{x}\sin\text{a}\ \text{dx}=\text{dt}$
$\therefore\int\frac{1}{\sin^{3}\text{x}\sin(\text{x}+\text{a)}}\text{dx}-\frac{\text{cosec}^{2}\text{x}}{\sqrt{\cos\text{a}+\cot\text{x}\sin\text{a}}}\text{dx}$
$=\frac{-1}{\sin\text{a}}\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\frac{-1}{\sin\text{a}}\big[2\sqrt{\text{t}}\big]+\text{C}$
$=\frac{-1}{\sin\text{a}}\big[2\sqrt{\cos\text{a}+\cot\text{x}\sin\text{a}}\big]+\text{C}$
$=\frac{-2}{\sin\text{a}}\sqrt{\cos\text{a}+\frac{\cos\text{x}\sin\text{a}}{\sin\text{x}}}+\text{C}$
$=\frac{-2}{\sin\text{a}}\sqrt{\frac{\sin\text{x}\cos\text{a}+\cos\text{x}\sin\text{a}}{\sin\text{x}}}+\text{C}$
$=-\frac{2}{\sin\text{a}}\sqrt{\frac{\sin(\text{x}+\text{a)}}{\sin\text{x}}}+\text{C}$
View full question & answer→Question 2025 Marks
Evaluate the following definite integrals:
$\int_{\text{e}}^\limits{\text{e}^2}\Big\{\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big\}\text{dx}$
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 2035 Marks
By using properties of definite integral, evaluate the following integral in Exercise:
$\int^{\pi}_{0}\frac{\text{x}\ \text{dx}}{1+\sin\text{x}}$
Answer$\text{Let}\ \text{I}=\int^{\pi}\limits_{0}\frac{\text{x}\ \text{dx}}{1+\sin\text{x}}\text{dx}$ $ \Rightarrow\ \ \text{I}=\int^{\pi}\limits_{0}\frac{\pi-\text{x}}{1+\sin(\pi-\text{x)}}\text{dx}=\int^{\pi}\limits_{0}\frac{\pi-\text{x}}{1+\sin\text{x}}\text{dx}$Adding eq.(i) and (ii),
$21=\int^{\pi}\limits_{0}\bigg(\frac{\text{x}}{1+\sin\text{x}}+\frac{\pi-\text{x}}{1+\sin\text{x}}\bigg)\text{dx}=\int^{\pi}\limits_{0}\bigg(\frac{\text{x}+\pi-\text{x}}{1+\sin\text{x}}\bigg)\text{dx}=\int^{\pi}\limits\limits_{0}\bigg(\frac{\pi}{1+\sin\text{x}}\bigg)\text{dx}=\pi\int^{\pi}\limits_{0}\bigg(\frac{1}{1+\sin\text{x}}\bigg)\text{dx}$
$\Rightarrow\ \ 21=2\pi\int^{\frac{\pi}{2}}\limits_{0}\frac{\text{dx}}{1+\sin\text{x}}\ \ \bigg[\because\int^{2\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx}=2\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx},\text{if}\ \text{f}(2\text{a}-\text{x})=\text{f}\text{(x)}\bigg]$
$\Rightarrow\ \ 21=2\pi\int^{\frac{\pi}{2}}\limits_{0}\frac{\text{dx}}{1+\sin\bigg(\frac{\pi}{2}-\text{x}\bigg)} \ \ \ \bigg[\because\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx}=\int^{\text{a}}\limits_{0}\text{f}(\text{a}-\text{x})\text{dx}=\bigg]$
$\Rightarrow\ \ 21=2\pi\int^{\frac{\pi}{2}}\limits_{0}\frac{\text{dx}}{1+\cos\text{x}}$
$\Rightarrow\ \ \text{I}=\pi\int^{\frac{\pi}{2}}\limits_{0}\frac{\text{dx}}{2\cos^{2}\frac{\text{x}}{2}}=\frac{\pi}{2}\int^{\frac{\pi}{2}}_\limits{0}\sec^{2}\frac{\pi}{2}\text{dx}=\frac{\pi}{2}\Bigg[\frac{\tan\frac{\text{x}}{2}}{\frac{1}{2}}\Bigg]^{\frac{\pi}{2}}_{0}$
$=\pi\bigg(\tan\frac{\pi}{4}-\tan0^{\text{o}}\bigg)=\pi(1-0)=\pi$
View full question & answer→Question 2045 Marks
Evalute the following integrals:
$\int\frac{\sin(\text{x}-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin(\text{x}-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$ then,
$\text{I}=\int\frac{\sin(\text{x}-\alpha+\alpha-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\frac{\sin(\text{x}+\alpha-2\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\frac{\sin(\text{x}+\alpha)\cos2\alpha-\cos(\text{x}+\alpha)\sin2\alpha}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\Big[\frac{\sin(\text{x}+\alpha)\cos2\alpha}{\sin(\text{x}+\alpha)}-\frac{\cos(\text{x}+\alpha)\sin2\alpha}{\sin(\text{x}+\alpha)}\Big]\text{dx}$
$=\int\big(\cos2\alpha-\cot(\text{x}+\alpha)\sin2\alpha\big)\text{dx}$
$=\cos2\alpha\int\text{dx}-\sin2\alpha\int\cot(\text{x}+\alpha)\text{dx}$
$=\text{x}\cos2\alpha-\sin2\alpha\log|\sin(\text{x}+\alpha)|+\text{C}$
$\therefore\text{I}=\text{x}\cos2\alpha-\sin2\alpha\log|\sin(\text{x}+\alpha)|+\text{C}$
View full question & answer→Question 2055 Marks
Evaluate the following intregals:
$\int\frac{\sin2\text{x}}{(1+\sin\text{x})(2+\sin\text{x})}\text{ dx}$
AnswerLet $\int\frac{\sin2\text{x}}{(1+\sin\text{x})(2+\sin\text{x})}\text{ dx}=\frac{\text{A}}{1+\sin\text{x}}+\frac{\text{B}}{2+\sin\text{x}}$
$\Rightarrow\sin2\text{x}=\text{A}(2+\sin\text{B})+\text{B}(1+\sin\text{B})$
$\Rightarrow2\sin\text{x}\cos\text{x}=(2\text{A}+\text{B})+(\text{A}+\text{B})\sin\text{x}$
Equating similar terms, we get,
$2\text{A}+\text{B}=0\Rightarrow\text{B}=-2\text{A}\text{ and}$
$\text{A}+\text{B}=2\cos\Rightarrow-\text{A}=2\cos\text{x}$
$\Rightarrow\text{A}=-2\cos\text{x}$
Thus,
$\text{I}=\int-\frac{2\cos\text{x}}{1+\sin\text{x}}\text{ dx}+\int\frac{4\cos\text{x}}{1+\sin\text{x}}\text{ dx}$
$=-2\log|1+\sin\text{x}|+4\log|2+\sin\text{x}|+\text{C}$
$\text{I}=\log\Big|\frac{(2+\sin\text{x})^4}{(1+\sin\text{x})^2}\Big|+\text{C}$
View full question & answer→Question 2065 Marks
Evaluate the following definite integrals:
$\int_{-\frac{\pi}{4}}^\limits{\frac{\pi}{4}}\frac{1}{1+\sin\text{x}}\text{ dx}$
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 2075 Marks
Evaluate the following integrals:
$\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}}{\sqrt[4]{\text{x}+4}+\sqrt[9]{9-\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}}{\sqrt[4]{\text{x}+4}-\sqrt[9]{9-\text{x}}}\text{ dx}\ ....(\text{i})$
$\text{I}=\int\limits^{5}_0\frac{\sqrt[4]{9-\text{x}}}{\sqrt[4]{9-\text{x}}-\sqrt[4]{\text{x}+4}}\text{ dx}$
$\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$\text{I}=\int\limits^{5}_0\frac{\sqrt[4]{9-\text{x}}}{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}}{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}-\frac{\sqrt[4]{9-\text{x}}}{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}\text{ dx}$
$=\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}\text{ dx}$
$=\int\limits^{5}_0\text{dx}$
$=\big[\text{x}\big]^5_0$
$=5$
Hence, $\text{I}=\frac{5}{2}$
View full question & answer→Question 2085 Marks
Find the integrals of the functions in Exercises:
$\frac{\cos2\text{x}-\cos2\alpha}{\cos\text{x}-\cos\alpha}$
Answer$\frac{\cos2\text{x}-\cos2\alpha}{\cos\text{x}-\cos\alpha}=\frac{-2\sin\frac{2\text{x}+2\alpha}{2}\sin\frac{2\text{x}-2\alpha}{2}}{-2\sin\frac{\text{x}+\alpha}{2}\sin\frac{\text{x}-\alpha}{2}}$$\ \ \ \ \ \ \ \bigg[\cos\text{C}-\cos\text{D}=-2\sin\frac{\text{C}+\text{D}}{2}\sin\frac{\text{C}-\text{D}}{2}\bigg]$
$=\frac{\sin(\text{x}+\alpha)\sin(\text{x}-\alpha)}{\sin\bigg(\frac{\text{x}+\alpha}{2}\bigg)\sin\bigg(\frac{\text{x}-\alpha}{2}\bigg)}$
$=\frac{\bigg[2\sin\bigg(\frac{\text{x}+\alpha}{2}\bigg)\cos\bigg(\frac{\text{x}+\alpha}{2}\bigg)\bigg]\bigg[2\sin\bigg(\frac{\text{x}-\alpha}{2}\bigg)\cos\bigg(\frac{\text{x}-\alpha}{2}\bigg)\bigg]}{\sin\bigg(\frac{\text{x}+\alpha}{2}\bigg)\sin\bigg(\frac{\text{x}-\alpha}{2}\bigg)}$
$=4\cos\bigg(\frac{\text{x}+\alpha}{2}\bigg)\cos\bigg(\frac{\text{x}-\alpha}{2}\bigg)$
$=2\bigg[\cos\bigg(\frac{\text{x}+\alpha}{2}+\frac{\text{x}-\alpha}{2}\bigg)+\cos\frac{\text{x}+\alpha}{2}-\frac{\text{x}-\alpha}{2}\bigg]$
$=2\big[\cos\text{(x)}+\cos\alpha\big]$
$=2\cos\text{x}+2\cos\alpha$
$\therefore\int\frac{\cos2\text{x}-\cos2\alpha}{\cos\text{x}-\cos\alpha}\text{dx}=\int2\cos\text{x}+2\cos\alpha$
$=2\big[\sin\text{x}+\text{x}\cos\alpha\big]+\text{C}$
View full question & answer→Question 2095 Marks
Integrate the following integrals:
$\int\sin2\text{x}\sin4\text{x}\sin6\text{x dx}$
Answer$\int\sin2\text{x}\sin4\text{x}\sin6\text{x dx}$
$=\frac{1}{2}\int(2\sin2\text{x}\sin4\text{x})\sin6\text{x dx}$
$=\frac{1}{2}\int\big[\cos(2\text{x}-4\text{x})-\cos(2\text{x}+4\text{x})\big]\sin6\text{x dx}$
$=\frac{1}{2}\int\big[\cos(2\text{x})-\cos(6\text{x})\big]\sin6\text{x dx}$
$=\frac{1}{2}\big[\int\cos(2\text{x})\sin(6\text{x})\text{dx}-\int\cos(6\text{x})\sin(6\text{x})\text{dx}\big]$
$=\frac{1}{4}\big[\int2\cos(2\text{x})\sin(6\text{x})\text{dx}-\int2\cos(6\text{x})\sin(6\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big\{\int\big[\sin(2\text{x}+6\text{x})-\sin(2\text{x}-6\text{x})\big]\text{dx}-\int\sin(12\text{x})\text{dx}\Big\}$
$=\frac{1}{4}\Big[\int\sin(8\text{x})\text{dx}+\int\sin(4\text{x})\text{dx}-\int\sin(12\text{x})\text{dx}\Big]$
$=\frac{1}{4}\Big[\frac{-\cos(8\text{x})}{8}+\frac{-\cos(4\text{x})}{4}+\frac{\cos(12\text{x})}{12}\Big]+\text{C}$
$=-\frac{\cos(8\text{x})}{32}-\frac{\cos(4\text{x})}{16}+\frac{\cos(12\text{x})}{48}+\text{C}$
View full question & answer→Question 2105 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2\sin^{-1}\text{x}}{(1-\text{x}^2)^{\frac{3}{2}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2\sin^{-1}\text{x dx}}{(1-\text{x}^2)^{\frac{3}{2}}}$
Putting $\text{x}=\sin\theta$
$\Rightarrow\text{dx}=\cos\theta\text{d}\theta$
$\&\ \theta=\sin^{-1}\text{x}$
$\therefore\text{I}=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{(1-\sin^2\theta)^{\frac{3}{2}}}$
$=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{(\cos^2\theta)^{\frac{3}{2}}}$
$=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{\cos^3\theta}$
$=\int\tan^2\theta.\theta.\text{d}\theta$
$=\int(\sec^2\theta-1)\theta.\text{d}\theta$.
$=\int\theta.\sec^2\theta\text{d}\theta-\int\theta.\text{d}\theta$
$=\theta\int\sec^2\theta\text{d}\theta-\int\Big\{\frac{\text{d}}{\text{d}\theta}(\theta)\int\sec^2\theta\text{d}\theta\Big\}\text{d}\theta-\int\theta.\text{d}\theta$
$=\theta\tan\theta-\int1.\tan\theta\text{d}\theta-\frac{\theta^{\ 2}}{2}$
$=\theta.\tan\theta-\ln\big|\sec\theta\big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\theta.\frac{\sin\theta}{\cos\theta}+\ln\big|\cos\theta\big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\theta.\frac{\sin\theta}{\cos\theta}+\ln\Big|\sqrt{1-\sin^2\theta}\Big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\frac{\theta.\sin\theta}{\sqrt{1-\sin^2\theta}}+\frac{1}2\ln\Big|1-\sin^2\theta\Big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}+\frac{1}{2}\ln\big(1-\text{x}^2\big)-\frac{1}{2}\big(\sin^{-1}\text{x}\big)^2+\text{C}$$\Big[\because\theta=\sin^{-1}\text{x}\Big]$
View full question & answer→Question 2115 Marks
Evaluate the following integrals:
$\int_{1}^\limits{2}\frac{1}{\text{x}\big(1+\log\text{x}\big)^2}\text{ dx}$
AnswerLet $1+\log\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$\frac{1}{\text{x}}\text{ dx}=\text{dt}$
Now, $\text{x}=1\Rightarrow\text{t}=1$
$=\text{x}=2\Rightarrow\text{t}=1+\log2$
$\therefore\ \int_{1}^\limits{2}\frac{1}{\text{x}\big(1+\log\text{x}\big)^2}\text{ dx}=\int^\limits{1+\log2}_1\frac{\text{dt}}{\text{t}^2}$
$=\Big[\frac{-1}{\text{t}}\Big]^{1+\log2}_1$
$=\bigg[\frac{-1}{1+\log2}+1\bigg]$
$=\bigg[\frac{-1+1+\log2}{1+\log2}\bigg]$
$=\bigg[\frac{\log2}{1+\log2}\bigg]$ $\big[\because\log\text{e}=1\big]$
$=\frac{\log2}{\log\text{e}+\log2}$ $\big[\log\text{a}+\log\text{b}=\log\text{ab}\big]$
$=\frac{\log2}{\log2\text{e}}$
View full question & answer→Question 2125 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{1}{1+2\text{x}+2\text{x}^2+2\text{x}^3+\text{x}^4}\text{ dx}$
Answer$\int_{0}^\limits{1}\frac{1}{1+2\text{x}+2\text{x}^2+2\text{x}^3+\text{x}^4}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{\big(\text{x}^2+1\big)^2+2\text{x}\big(\text{x}^2+1\big)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{\big(\text{x}^2+1\big)\big(\text{x}^2+1+2\text{x}\big)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{(\text{x}^2+1)(\text{x}+1)}\text{ dx}$
Let $\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{Cx}+\text{D}}{\text{x}^2+1}$
$\Rightarrow1=\text{A}(\text{x}+1)(\text{x}^2+1)+\text{B}(\text{x}^2+1)+(\text{Cx}+\text{D})(\text{x}+1)^2$
Putting x = -1, we have
$1=2\text{B}$
$\Rightarrow\text{B}=\frac{1}{2}\ ...(\text{i})$
Putting x = 0, we have
$\text{A}+\text{B}+\text{C}=1\ ...(\text{ii})$
Equating co-efficient of $x^3$ on both sides, we have
$\text{A}+\text{C}=0\ ...(\text{iii}) $
Equating co-efficient of $x^2$ on both sides, we have
$\text{A}+\text{B}+2\text{C}+\text{D}=0\ ...(\text{iv})$
$\Rightarrow2\text{C}=-1$ [Using (i)]
$\Rightarrow\text{C}=-\frac{1}{2}$
$\therefore\ \text{A}=\frac{1}{2}$ [Using (iii)]
Putting $\text{A}=\frac{1}{2},\text{ B}=\frac{1}{2}$ and $\text{C}=-\frac{1}{2}$ in (iv), we have
$\text{D}=0$
$\therefore\ \int_{0}^\limits{1}\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{\frac{1}{2}}{\text{x}+1}\text{ dx}+\int_{0}^\limits{1}\frac{\frac{1}{2}}{(\text{x}+1)^2}\text{ dx}+\int_{0}^\limits{1}\frac{-\frac{1}{2}\text{x}}{\text{x}^2+1}$
$=\Big[\frac{1}{2}\log(\text{x}+1)\Big]^1_0+\Big[\frac{1}{2}\times\Big(-\frac{1}{\text{x}+1}\Big)\Big]^1_0-\frac{1}{4}\int_{0}^\limits{1}\frac{2\text{x}}{\text{x}^2+1}\text{ dx}$
$=\frac{1}{2}\big(\log2-\log1\big)-\frac{1}{2}\Big(\frac{1}{2}-1\Big)-\Big[\frac{1}{4}\log(\text{x}^2+1)\Big]^1_0$
$=\frac{1}{2}\log2+\frac{1}{4}-\frac{1}{4}\big(\log2-\log1\big)$ $(\log1=0)$
$=\frac{1}{2}\log2+\frac{1}{4}\log\text{e}-\frac{1}{4}\log2$
$=\frac{1}{4}\log2+\frac{1}{4}\log_\text{e}$
$=\frac{1}{4}\big(\log2+\log_\text{e}\big)$
$=\frac{1}{4}\log(2\text{e})$
View full question & answer→Question 2135 Marks
Evaluate the following integrals:
$\int\text{x}\sin^3\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\sin^3\text{x dx}$
$\sin(3\text{A})=3\sin\text{A}-4\sin^3\text{A}$
$\sin^3\text{A}=\frac{1}4{}\big[3\sin\text{A}-\sin3\text{A}\big]$
$\therefore\text{I}=\frac{1}{4}\int\text{x}(3\sin\text{x}-\sin3\text{x})\text{dx}$
$=\frac{3}{4}\int\text{x}\sin\text{x dx}-\frac{1}{4}\int\text{x}\sin(3\text{x})\text{dx}$
$=\frac{3}{4}\big[\text{x}(-\cos\text{x})-\int1.(-\cos\text{x})\text{dx}\big]\\-\frac{1}{4}\Big[\text{x}\Big(-\frac{\cos3\text{x}}{3}\Big)-\int1.\Big(-\frac{\cos3\text{x}}{3}\Big)\text{dx}\Big]$
$=-\frac{3\text{x}\cos\text{x}}{4}+\frac{3}{4}\sin\text{x}+\frac{\text{x}\cos3\text{x}}{12}-\frac{1}{36}\sin3\text{x}+\text{C}$
View full question & answer→Question 2145 Marks
Evaluate the following integrals:
$\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{8}_2\frac{\sqrt{10-(2+8-\text{x}})}{\sqrt{2+8-\text{x}}+\sqrt{10-(2+8-\text{x}})}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{8}_2\frac{\sqrt{\text{x}}}{\sqrt{10-\text{x}}+\sqrt{\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{8}_2\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^8_2$
$\Rightarrow2\text{I}=8-2=6$
$\Rightarrow\text{I}=3$
View full question & answer→Question 2155 Marks
Evaluate the following integrals:
$\int\limits^0_{-5}\text{f(x)}\text{dx,}$ Where $\text{f(x)}=|\text{x}|+|\text{x}+2|+|\text{x}+5|$
AnswerWe have,
$\text{I}=\int\limits^0_{-5}\big(|\text{x}|+|\text{x}+2|+|\text{x}+5|\big)\text{dx}\\=\int\limits^0_{-5}|\text{x}|\text{dx}+\int\limits^0_{-5}|\text{x}+2|\text{dx}+\int\limits^0_{-5}|\text{x}+5|\text{dx}$
$\Rightarrow\text{I}=\int\limits^0_{-5}|\text{x}|\text{dx}+\int\limits^0_{-5}|\text{x}+2|\text{dx}+\int\limits^0_{-5}|\text{x}+5|\text{dx}$
$=\Big[\frac{-\text{x}^2}{2}\Big]^0_{-5}+\Big[\frac{-\text{x}^2}{2}-2\text{x}\Big]^{-2}_{-5}+\Big[\frac{\text{x}^2}{2}+2\text{x}\Big]^0_{-2}+\Big[\frac{\text{x}^2}{2}+5\text{x}\Big]^0_{-5}$
$=\Big[\frac{25}{2}\Big]-\big[\frac{4}{2}-4-\frac{25}{2}+10\Big]+\Big[0+0-\frac{4}{2}+4\Big]+\Big[0+0-\frac{25}{2}+25\big]$
$=\frac{25}{2}-\Big[8-\frac{25}{2}\Big]+\big[2\big]+\Big[25-\frac{25}{2}\Big]$
$=\frac{25}{2}-8+\frac{25}{2}+2+25-\frac{25}{2}$
$=19+\frac{25}{2}$
$=31\frac{1}{2}$
$=\frac{36}{2}$
View full question & answer→Question 2165 Marks
Evaluate the following integrals:
$\int\text{cosec x}\log({\text{cosec x}-\cot\text{x})}\text{dx}$
AnswerLet $\text{I}=\int\text{cosec x}\log(\text{cosec x}-\cot\text{x})\text{dx}\ ....(1)$
Let $\log(\text{cosec x}-\cot\text{x})=\text{t}$ then,
$\text{dx}[\log(\text{cosec x}-\cot\text{x})]=\text{dt}$
$\Rightarrow\text{cosec x}\text{ dx}=\text{dt}$ $\Big[\because\ \frac{\text{d}}{\text{dx}}(\log(\text{cosec x}-\cot\text{x}))=\text{cosec x}\Big]$
Putting $\log(\text{cosec x}-\cot\text{x})=\text{t}$ and $\text{cosec x}\text{ dx}=\text{dt}$ in equation (1), we get
$\text{I}=\int\text{t dt}$
$=\frac{\text{t}^2}{2}+\text{C}$
$\text{I}=\frac{1}{2}[\log(\text{cosec x}-\cot\text{x})]^2+\text{C}$
View full question & answer→Question 2175 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^3_0(\text{x}+4)\text{dx}$
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, $\text{a}=0,\text{ b}=3,\text{ f(x)}=\text{x}+4,\text{ h}=\frac{3-0}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_0(\text{x}+4)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}(0+(\text{n}-1)\text{h})\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+4)+(\text{h}+4)+\ ....\ +((\text{n}-1)\text{h}+4)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}(1+2+\ ....\ +(\text{n}-1))\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{3}{\text{n}}\Big[4\text{n}+\frac{3}{\text{n}}\times\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\Big[12+\frac{9}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big]$
$=12+\frac{9}{2}$
$=\frac{33}{2}$
View full question & answer→Question 2185 Marks
Evaluate the following integrals:
$\int\limits^{\text{b}}_{\text{a}}\frac{\text{x}^{\frac{1}{\text{n}}}}{\text{x}^\frac{1}{\text{n}}+\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}\text{ dx},\text{ n}\in\text{N},\text{n}\leq2$
AnswerLet $\text{I}=\int\limits^{\text{b}}_{\text{a}}\frac{\text{x}^{\frac{1}{\text{n}}}}{\text{x}^\frac{1}{\text{n}}+\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\text{b}}_{\text{a}}\frac{\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}{\big(\text{a}+\text{b}-\text{x}\big)^\frac{1}{\text{n}}+\big[\text{a}+\text{b}-\big(\text{a}+\text{b}-\text{x}\big)\big]^{\frac{1}{\text{n}}}}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_{\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\text{b}}_{\text{a}}\frac{\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}{\big(\text{a}+\text{b}-\text{x}\big)^\frac{1}{\text{n}}+\text{x}^{\frac{1}{\text{n}}}}\text{ dx}$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\text{b}}_{\text{a}}\frac{\text{x}^{^{\frac{1}{\text{n}}}}+\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}{\text{x}^\frac{1}{\text{n}}+\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\text{b}}_{\text{a}}\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{\text{b}}_{\text{a}}=(\text{b}-\text{a})$
$\Rightarrow\text{I}=\frac{\text{b}-\text{a}}{2}$
View full question & answer→Question 2195 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2+9}{\text{x}^4+81}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+9}{\text{x}^4+81}\ \text{dx}$Dividing numerator and denominator by $x^2$
$\text{I}=\int\frac{1+\frac{9}{\text{x}^2}}{\text{x}^2+\frac{81}{\text{x}^2}}\ \text{dx}$
$=\int\frac{1+\frac{9}{\text{x}^2}}{\Big(\text{x}-\frac{9}{\text{x}}\Big)^2+18}$
Let $\Big(\text{x}-\frac{9}{\text{x}}\Big)=\text{t}\Rightarrow\Big(1+\frac{9}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+18}$
$\text{I}=\frac{1}{3\sqrt{2}}\tan^{-1}\Big(\frac{\text{t}}{3\sqrt{2}}\Big)+\text{C}$
Thus,
$\text{I}=\frac{1}{3\sqrt{2}}\tan^{-1}\Big(\frac{\text{x}^2-9}{3\sqrt{2}}\Big)+\text{C}$
View full question & answer→Question 2205 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\sin\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\sin\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\big(\frac{\pi}{2}-\text{x}\big)+\text{b}\cos\big(\frac{\pi}{2}-\text{x}\big)}{\sin\big(\frac{\pi}{2}-\text{x}\big)+\cos\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\cos\text{x}+\text{b}\sin\text{x}}{\cos\text{x}+\sin\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big(\frac{\text{a}\sin\text{x}+\text{b}\cos\text{x}}{\cos\text{x}+\sin\text{x}}+\frac{\text{a}\cos\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\Big)\text{dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big(\frac{\text{a}\sin\text{x}+\text{b}\cos\text{x}+\text{a}\cos\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\Big)\text{dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{(\text{a}+\text{b})\sin\text{x}+(\text{a}+\text{b})\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{(\text{a}+\text{b})(\sin\text{x}+\cos\text{x})}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0(\text{a}+\text{b})\text{dx}$
$\Rightarrow2\text{I}=(\text{a}+\text{b})\times\big[\text{x}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow2\text{I}=(\text{a}+\text{b})\times\Big(\frac{\pi}{2}-0\Big)$
$\Rightarrow2\text{I}=\frac{\pi}{2}(\text{a}+\text{b})$
$\Rightarrow\text{I}=\frac{\pi}{4}(\text{a}+\text{b})$
View full question & answer→Question 2215 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^3_{1}(2\text{x}+3)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=3,\text{ f(x)}=2\text{x}+3,\text{ h}=\frac{3-1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_{1}(2\text{x}+3)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(2+3)+(2+2\text{h}+3)+\\\ ....\ +\{2+2(\text{n}-1)\text{h}+3\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[5\text{n}+2\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[5\text{n}+2\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[5\text{n}+2\text{n}-2\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big({7}-\frac{2}{\text{n}}\Big)$
$=14$
View full question & answer→Question 2225 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2-10\text{x}+34}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{x}^2-10\text{x}+34}\text{dx}$
$=\int\frac{1}{\text{x}^2-2\text{x}\times5+(5)^2-(5)^2+34}\text{dx}$
$=\int\frac{1}{(\text{x}-5)^2+9}\text{dx}$
Let $(\text{x}-1)=\text{t} \dots(1)$
$\Rightarrow\text{dx = dt}$
so,
$\text{I}=\int\frac{1}{\text{t}^2+(3)^2}\text{dt}$
$\text{I}=\frac{1}{3}\tan^{-1}\big(\frac{\text{t}}{3}\big)+\text{C}$ $\Big[\text{since,}\int\frac{1}{\text{x}^2+\text{a}^2}\text{dx}=\frac{1}{\text{a}}\tan^{-1}\big(\frac{\text{x}}{2}\big)+\text{C}\Big]$
$\text{I}=\frac{1}{3}\tan^{-1}\Big(\frac{\text{x}-5}{3}\Big)+\text{C}$ [using (1)]
View full question & answer→Question 2235 Marks
If $\text{f}(\text{a}+\text{b}-\text{x})=\text{f(x)},$ then prove that, $\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$
Answer$\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})(\text{a}+\text{b}-\text{x}){}\text{dx}$ $\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\int\limits^{\text{b}}_{\text{a}}(\text{a}+\text{b}-\text{x})\text{f}(\text{x})\text{dx}$ $\big[\text{f}(\text{a}+\text{b}-\text{x})=\text{f}(\text{x})\big]$
$\Rightarrow\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\int\limits^{\text{b}}_{\text{a}}(\text{a}+\text{b})\text{f}(\text{x})\text{dx}-\int\limits^{\text{b}}_{\text{a}}\text{x}\text{f}(\text{x})\text{dx}$
$\Rightarrow2\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=(\text{a}+\text{b})\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$
$\Rightarrow\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$
View full question & answer→Question 2245 Marks
If f'(x) = x + b, f'(1) = 5, f'(2) = 13, find f'(x).
Answer$\text{f}'\text{(x)}=\text{x + b},\text{f}'(1)=5,\text{f}'(2)=13$
$\text{f}'\text{(x)}=\text{x + b}$
$\int\text{f}'\text{(x) dx}=\int(\text{x + b})\text{dx}$
$\text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\text{bx}+\text{C}\ \dots(1)$
$\text{f}'(1)=5,\text{f}'(2)=13$ (Given)
putting x = 1 in (1)
$\text{f}'(1)=\frac{1^2}{2}+\text{b}_1+\text{C}$
$5=\frac{1}{2}+\text{b + C}\ \dots(2)$
Putting x = 2 in (1)
$\text{f}'(2)=\frac{2^2}{2}+\text{b}_2+\text{C}$
$13=\frac{4}{2}+2\text{b + C}$
$13=2+2\text{b + C}\ \dots(3)$
Solving (2) and (3) we get,
$\text{b}=\frac{13}{2}\text{ and }\text{C}=-2$
Thus, $\text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\frac{13}{2}\text{x}-2$
View full question & answer→Question 2255 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 2265 Marks
Evaluate the following integrals:
$\int^\limits2_{0}\big|\text{x}^2-3\text{x}+2\big|\text{dx}$
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 2275 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 2285 Marks
Evaluate the following integrals:
$\int\limits^{\frac{3}{2}}_0\big|\text{x}\cos\pi\text{x}\big|\text{dx}$
Answer$\int\limits^\frac{3}{2}_{0}|\text{x} \cos\pi \text{ x}| \text{dx} = \int\limits^{1/2}_{0}\text{x}\cos\pi \text{x dx}{-}\int\limits^{3/2}_{1/2}\text{x}\cos\pi \text{x dx}$
$= \left\{\frac{\text{x}\sin\pi \text{x}}{\pi} + \frac{\cos\pi\text{x}}{\pi^{2}}\right\}^{1/2}_{0}= \left\{\frac{\text{x}\sin\pi \text{x}}{\pi} + \frac{\cos\pi\text{x}}{\pi^{2}}\right\}^{3/2}_{1/2}$
$=\frac{1}{2\pi} - \frac{1}{\pi^{2}} - \bigg(-\frac{3}{2\pi}-\frac{1}{2\pi}\bigg) = \frac{5}{2\pi}-\frac{1}{\pi^{2}}$
View full question & answer→Question 2295 Marks
Evaluate the following integrals:
$\int\cos(\log\text{x})\text{dx}$
AnswerLet $\text{I}=\int\cos(\log\text{x})\text{dx}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\text{x}=\text{e}^\text{t}$
$\Rightarrow\text{dx}=\text{e}^\text{t}\text{dt}$
$\text{I}=\int\text{e}^\text{t}\cos\text{(t)dt}$
Considering cos (t) as first function and $e^t$ as second function
$\text{I}=\cos\text{t}\text{e}^\text{t}-\int(-\sin\text{t})\text{e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\cos\text{t}\text{e}^\text{t}+\int\sin\text{t}\text{e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\cos\text{te}^\text{t}+\text{I}_1\ \dots(1)$
where $\text{I}_1=\int\text{e}^\text{t}\sin\text{t dt}$
$\text{I}_1=\int\text{e}^\text{t}\sin\text{t dt}$
Considering sin t as first function and $e^t$ as second function
$\text{I}_1=\sin\text{te}^\text{t}-\int\cos\text{te}^\text{t}\text{dt}$
$\Rightarrow\text{I}_1=\sin\text{te}^\text{t}-\text{I}\ \dots(2)$
From (1) & (2)
$\text{I}=\cos\text{te}^\text{t}+\sin\text{te}^\text{t}-\text{I}$
$\Rightarrow2\text{I}=\text{e}^\text{t}(\sin\text{t}+\cos\text{t})$
$\Rightarrow\text{I}=\frac{\text{e}^\text{t}(\sin\text{t}+\cos\text{t})}{2}+\text{C}$
$\Rightarrow\text{I}=\frac{\text{e}^{\log\text{x}}\big[\sin(\log\text{x})+\cos(\log\text{x})\big]}{2}+\text{C}$
$\Rightarrow\text{I}=\frac{\text{x}}{2}\big[\sin(\log\text{x})+\cos(\log\text{x})\big]+\text{C}$
View full question & answer→Question 2305 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\text{n}}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx},\text{ n}\in\text{N}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\text{n}}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\text{n}}\big(\frac{\pi}{2}-\text{x}\big)}{\sin^{\text{n}}\big(\frac{\pi}{2}-\text{x}\big)+\cos^{\text{n}}\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\cos^\text{n}\text{x}+\sin^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\bigg[\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}+\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\bigg]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\big[\text{x}\big]^{\frac{\pi}{2}}_0$
$=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→Question 2315 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_0\cos^5\text{x dx}$
Answer$\int^\limits{\frac{\pi}{2}}_0\cos^5\text{x dx}=\int^\limits{\frac{\pi}{2}}_0\big(1-\sin^2\text{x}\big)^2\cos\text{x dx}$
Let $\sin\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$\cos\text{x dx}=\text{dt}$
When $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=1$
$=\int^\limits{\frac{\pi}{2}}_0\big(1-\sin^2\text{x}\big)^2\cos\text{x dx}$
$=\int^\limits{1}_0\big(1-\text{t}^2\big)^2\text{ dt}$
$=\int^\limits{1}_0\big(1-2\text{t}^2+\text{t}^4\big)\text{dt}$
$=\Big[\text{t}-\frac{2}{3}\text{t}^3+\frac{\text{t}^5}{5}\Big]^1_0$
$=1-\frac{2}{3}+\frac{1}{5}$
$=\frac{8}{15}$
View full question & answer→Question 2325 Marks
Evaluvate the following intregals:
$\int\frac{2\sin\text{x}+3\cos\text{x}}{3\sin\text{x}+4\cos\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{2\sin\text{x}+3\cos\text{x}}{3\sin\text{x}+4\cos\text{x}}\ \text{dx}$
Let $(2\sin\text{x}+3\cos\text{x})=\lambda\frac{\text{d}}{\text{dx}}(3\sin\text{x}+4\cos\text{x})+\mu(3\sin\text{x}+4\cos\text{x})+\text{v}$
$(2\sin\text{x}+3\cos\text{x})=\lambda(3\cos\text{x}-4\sin\text{x})+\mu(3\sin\text{x}+4\cos\text{x})+\text{v}$
$(2\sin\text{x}+3\cos\text{x})=(3\lambda+4\mu)\cos\text{x}+(-4\lambda+3\mu)\sin\text{x}+\text{v}$
Compairing the coefficient of $\sin\text{x},\cos\text{x}$ on both the sides,
$3\lambda+4\mu=3\dots\dots(1)$
$-4\lambda+3\mu=2\dots\dots(2)$
$\text{v}=0\dots\dots(3)$
Solving the equation (1), (2) and (3)
$\lambda=\frac{1}{25}$
$\mu=\frac{18}{25}$
$\text{v}=0$
$\text{I}=\frac{1}{25}\int\frac{(3\cos\text{x}-4\sin\text{x})}{(3\sin\text{x}+4\cos\text{x})}\text{dx}+\frac{18}{25}\int\text{dx}$
$\text{I}=\frac{1}{25}\log|3\sin\text{x}+4\cos\text{x}|+\frac{18}{25}\text{x}+\text{C}$
View full question & answer→Question 2335 Marks
Evaluate the following definite integrals:
$\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}$ Then,
$\text{I}=\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\bigg(\frac{1-2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\sin^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\Big(\frac{1}{2}\text{cosec}^2\frac{\text{x}}{2}-\cot\frac{\text{x}}{2}\Big)\text{dx}$
$\Rightarrow\text{I}=\int_{\frac{\pi}{2}}^\limits{\pi}\frac{1}{2}\text{e}^{\text{x}}\text{cosec}^2\frac{\text{x}}{2}\text{ dx}-\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\cot\frac{\text{x}}{2}\text{ dx}$
Integrating second term by parts,
$\text{I}=\bigg\{-\Big[\text{e}^{\text{x}}\cot\frac{\text{x}}{2}\Big]^{\pi}_\frac{\pi}{2}-\int_{\frac{\pi}{2}}^\limits{\pi}\frac{1}{2}\text{ e}^{\text{x}}\text{cosec}^2\frac{\text{x}}{2}\text{ dx}\bigg\}+\int_{\frac{\pi}{2}}^\limits{\pi}\frac{1}{2}\text{ e}^{\text{x}}\text{cosec}^2\frac{\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=-\Big[0-\text{e}^{\frac{\pi}{2}}\Big]$
$\Rightarrow\text{I}=\text{e}^{\frac{\pi}{2}}$
View full question & answer→Question 2345 Marks
Evaluate the following integrals:
$\int\limits^{\frac{3}{2}}_0\big|\text{x}\sin\pi\text{x}\big|\text{dx}$
AnswerFor $0<\text{x}<1,\text{ x}>0$ and $\sin\pi\text{x}>0\Rightarrow\text{x}\sin\pi\text{x}>0$
For $1<\text{x}<\frac{3}{2},\text{ x}>0$ and $\sin\pi\text{x}>0\Rightarrow\text{x}\sin\pi\text{x}>0$
$\therefore\ \int\limits^{\frac{3}{2}}_0\big|\text{x}\sin\pi\text{x}\big|\text{dx}=\int\limits^1_0\text{x}\sin\pi\text{x dx}-\int\limits^{\frac{3}{2}}_1\text{x}\sin\pi\text{x dx}$
Let $\text{I}=\int\text{x}\sin\pi\text{ dx}$
$=\text{x}\int\sin\pi\text{x}-\int\Big(\frac{\text{d}}{\text{dx}}\text{x}\int\sin\pi\text{x dx}\Big)\text{dx}$
$=\text{x}\Big(\frac{-\cos\pi\text{x}}{\pi}\Big)-\int\Big(\frac{-\cos\pi\text{x}}{\pi}\Big)\text{dx}$
$=\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}$
Applying the limits, we get
$\int\limits^{\frac{3}{2}}_0|\text{x}\sin\pi\text{x}|\text{dx}=\Big[\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}\Big]^1_0-\Big[\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}\Big]^{\frac{3}{2}}_1$
$=\Big[\Big(\frac{\cos\text{x}}{\pi}+\frac{\sin\pi}{\pi^2}\Big)-(0+0)\Big]-\Bigg[\bigg(\frac{-\frac{3}{2}\cos\frac{3\pi}{2}}{\pi}+\frac{\sin\frac{3\pi}{2}}{\pi^2}\bigg)-\Big(\frac{-\cos\pi}{\pi}+\frac{\sin\pi}{\pi^2}\Big)\Bigg]$
$=\bigg[\Big(\frac{1}{\pi}+0\Big)\bigg]-\bigg[\Big(0-\frac{1}{\pi^2}\Big)-\Big(\frac{1}{\pi}+0\Big)\bigg]$
$=\frac{1}{\pi}+\frac{1}{\pi^2}+\frac{1}{\pi}$
$=\frac{2\pi+1}{\pi^2}$
View full question & answer→Question 2355 Marks
Evaluate the following integrals:
$\int\limits^4_{0}\big(|\text{x}|+|\text{x}+2|+|\text{x}+4|\big)\text{dx}$
Answer$\text{I}=\int\limits^4_{0}\big\{|\text{x}|+|\text{x}+2|+|\text{x}+4|\big\}\text{dx}$
$\Rightarrow\text{I}=\int\limits^4_0|\text{x}|\text{dx}+\int\limits^4_0|\text{x}-2|\text{dx}+\int\limits^4_0|\text{x}-4|\text{dx}$
We know that,
$|\text{x}|=\begin{cases}-\text{x},&-5\leq\text{x}\leq0\\\text{x},&\text{x}>0\end{cases}$
$|\text{x}-2|=\begin{cases}-(\text{x}-2),&0\leq\text{x}\leq2\\\text{x}-2,&2<\text{x}\leq4\end{cases}$
$|\text{x}-4|=\begin{cases}-(\text{x}-4),&0\leq\text{x}\leq4\\\text{x}-4,&\text{x}>4\end{cases}$
$\therefore\ \text{I}=\int\limits^4_{0}\text{x dx}-\int\limits^2_{0}(\text{x}-2)\text{dx}+\int\limits^4_{2}(\text{x}-2)\text{dx}-\int\limits^4_{0}(\text{x}-4)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}^2}{2}\Big]^4_0-\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^2_0+\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^4_2-\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^4_0$
$\Rightarrow\text{I}=8-(2-4)+8-8-2+4-(8-16)$
$\Rightarrow\text{I}=20$
View full question & answer→Question 2365 Marks
Evaluate the following integrals:
$\int\text{cosec}^43\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\text{cosec}^43\text{x}\text{ dx}$ Then
$\text{I}=\int\text{cosec}^23\text{x }\text{cosec}^23\text{x}\text{ dx}$
$=\int\big(1+\cot^23\text{x}\big)\text{cosec}^23\text{x}\text{ dx}$
$=\int\big(\text{cosec}^23\text{x}+\cot^23\text{x }\text{cosec}^23\text{x}\big)\text{dx}$
$\text{I}=\int\text{cosec}^23\text{x}\text{ dx}+\int\cot^23\text{x }\text{cosec}^23\text{x}\text{ dx}$
Sunbstituting $\cot3\text{x}=\text{t}$ and $\text{cosec}^23\text{x}\text{ dx}=-\text{dt}$ in $2^{nd}$ integral, we get
$\text{I}=\int\text{cosec}^23\text{x}-\int\text{t}^2\frac{\text{dt}}{3}$
$=\frac{-1}{3}\cot3\text{x}-\frac{\text{t}^3}{9}+\text{C}$
$=\frac{-1}{3}\cot3\text{x}-\frac{\cot^33\text{x}}{9}+\text{C}$
$\therefore\ \text{I}=\frac{-1}{3}\cot3\text{x}-\frac{1}{9}\infty\text{t}^3\text{3x}+\text{C}$
View full question & answer→Question 2375 Marks
Evaluate the following integrals:
$\int\frac{\tan\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
Answer$\int\frac{\tan\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
$\Rightarrow\int\frac{\sin\text{x}}{\cos\text{x}\sqrt{\cos\text{x}}}\text{dx}$
$\Rightarrow\int\frac{\sin\text{x}}{\cos^\frac{3}{2}\text{x}}\text{dx}$
$\text{Let }\cos\text{x}=\text{t}$
$\Rightarrow-\sin\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\sin\text{x}=-\frac{\text{dt}}{\text{dx}}$
$\text{Now,}\int\frac{\sin\text{x}}{\cos^\frac{3}{2}\text{x}}\text{dx}$
$=\int-\frac{1}{\text{t}^\frac{3}{2}}\text{dt}$
$=-\int\text{t}^{-\frac{3}{2}}\text{dt}$
$=-\bigg[\frac{\text{t}^{-\frac{3}{2}+1}}{\frac{-3}{2}+1}\bigg]+\text{C}$
$=\frac{2}{\sqrt{\text{t}}}+\text{C}$
$=\frac{2}{\sqrt{\cos\text{x}}}+\text{C}$
View full question & answer→Question 2385 Marks
If f'(x) = a sin x + b cos x and f'(0) = 4, f(0) = 3, $\text{f}\Big(\frac{\pi}{2}\Big)=5$, find f(x).
Answer$\text{f'(x)}=\text{a}\sin\text{x}+\text{b}\cos\text{x}$
$\text{f'}(0)=4,\text{f}(0)=3$
$\text{f}\Big(\frac{\pi}{2}\Big)=5$
$\text{f'(x)}=\text{a}\sin\text{x + b}\cos\text{x}$
$\int\text{f'(x)}\text{dx}=\int(\text{a}\sin\text{x + b}\cos\text{x})\text{dx}$
$\text{f(x)}=-\text{a}\cos\text{x}+\text{b}\sin\text{x}+\text{C}\ \dots(1)$
Now putting x = 0 in eq. (1)
$\text{f}(0)=-\text{a}\cos0+\text{b}\sin0+\text{C}$
$3=-\text{a}\times1+\text{b}\times0+\text{C}$
$3=-\text{a + C}\ \dots(2)$
Now putting $\text{x}=\frac{\pi}{2}$ in eq. (1)
$\text{f}\Big(\frac{\pi}{2}\Big)=-\text{a}\cos\frac{\pi}{2}+\text{b}\sin\frac{\pi}{2}+\text{C}$
$5=-\text{a}\cos\frac{\pi}{2}+\text{b}\sin\Big(\frac{\pi}{2}\Big)+\text{C}$
$5=-\text{a}\times0+\text{b}\times1+\text{C}$
$5=\text{b + C}\ \dots(3)$
We also have f'(0) = 4
$\text{f'(x)}=\text{a}\sin\text{x + b}\cos\text{x}$
$\text{f'(0)}=\text{a}\sin0+\text{b}\cos0$
$4=\text{a}\times0+\text{b}\times1$
$4=\text{b}\ \dots(4)$
Solving (2), (3) and (4) we get,
$\text{b}=4$
$\text{C}=1$
$\text{a}=-2$
$\therefore\ \text{f(x)}=2\cos\text{x}+4\sin\text{x}+1$
View full question & answer→Question 2395 Marks
Evaluate the following integrals:
$\int\limits^1_0\log\Big(\frac{1}{\text{x}}-1\Big)\text{dx}$
AnswerLet $\text{I}=\int\limits^1_0\log\Big(\frac{1}{\text{x}}-1\Big)\text{dx}\ ....(\text{i})$
$=\int\limits^1_0\log\Big(\frac{1}{1-\text{x}}-1\Big)\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$\text{I}=\int\limits^1_0\log\Big(\frac{\text{x}}{\text{x}-1}\Big)\text{dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^1_0\log\Big(\frac{1-\text{x}}{\text{x}}\Big)\log\Big(\frac{1-\text{x}}{\text{x}}\Big)\text{dx}$
$=\int\limits^1_0\log\Big(\frac{1-\text{x}}{\text{x}}\times\frac{\text{x}}{1-\text{x}}\Big)\text{dx}$
$=\int\limits^1_0\log1\text{ dx}$
$=0$
Hence, $\text{I}=0$
View full question & answer→Question 2405 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cot\text{x}}}{\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cot\text{x}}}{\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cot\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\cot\big(\frac{\pi}{2}-\text{x}\big)}+\sqrt{\tan\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\bigg(\frac{\sqrt{\cot\text{x}}}{\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}+\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\bigg)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→Question 2415 Marks
Evaluate the following integrals:
$\int\text{x}\sin\text{x}\cos2\text{x dx}$
Answer$\int\text{x}.\cos2\text{x}\sin\text{x dx}$
$=\frac{1}{2}\int\text{x}(2\cos2\text{x}\sin\text{x})\text{dx}$ $\big[\therefore2\cos\text{A}\sin\text{B}=\sin(\text{A+B})-\sin(\text{A}-\text{B})\big]$
$=\frac{1}{2}\int\text{x}(\sin3\text{x}-\sin\text{x})\text{dx}$
$=\frac{1}{2}\int\text{x}\sin3\text{x dx}-\frac{1}{2}\int\text{x}\sin\text{x dx}$
$=\frac{1}{2}\int\text{x}\sin3\text{x dx}-\frac{1}{2}\int\text{x}\sin\text{x dx}$
$=\frac{1}{2}\Big[\text{x}\int\sin3\text{x dx}-\int\Big\{\frac{\text{x}}{\text{dx}}(\text{x})\int\sin3\text{x dx}\Big\}\text{dx}\Big]\\-\frac{1}{2}\Big[\text{x}\int\sin\text{x dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{x dx}\Big\}\text{dx}\Big]$
$=\frac{1}{2}\Big[\text{x}\Big(\frac{-\cos3\text{x}}{3}\Big)-\int1\Big(\frac{-\cos3\text{x}}{3}\Big)\text{dx}\Big]\\-\frac{1}{2}\big[\text{x}(-\cos\text{x})-\int1(-\cos\text{x})\text{dx}\big]$
$=\frac{1}{2}\Big[\text{x}\Big(\frac{-\cos3\text{x}}{3}\Big)+\frac{1}{9}\sin3\text{x}\big]-\frac{1}{2}\big[\text{x}(-\cos\text{x})+\sin\text{x}\big]$
$=-\frac{\text{x}\cos3\text{x}}{6}+\frac{\sin3\text{x}}{18}+\frac{\text{x}\cos\text{x}}{2}-\frac{\sin\text{x}}{2}+\text{C}$
View full question & answer→Question 2425 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{6}}_{0}\cos^{-3}2\theta\sin2\theta\text{ d}\theta$
AnswerWe have,
$\int^\limits{\frac{\pi}{6}}_{0}\cos^{-3}2\theta\sin2\theta\text{ d}\theta$
$=\int^\limits{\frac{\pi}{6}}_{0}\frac{\sin2\theta}{\cos^32\theta}\text{ d}\theta$
$=\int^\limits{\frac{\pi}{6}}_{0}\tan2\theta\cdot\sec^22\theta\text{ d}\theta$
Let $\tan2\theta=\text{t}$
Differentiating w.r.t. x, we get
$2\sec^22\theta\text{d}\theta=\text{dt}$
Now, $\theta=0\Rightarrow\text{t}=0$
$\theta=\frac{\pi}{6}\Rightarrow\text{t}=\sqrt{3}$
$\therefore\ \int^\limits{\frac{\pi}{6}}_{0}\tan2\theta\cdot\sec^22\theta\text{ d}\theta=\frac{1}{2}\int^\limits{\sqrt{3}}_0\text{t dt}=\frac{1}{2}\Big[\frac{\text{t}^2}{2}\Big]^{\sqrt{\text{3}}}_0$
$=\frac{3}{4}$
View full question & answer→Question 2435 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^1_{-1}(\text{x}+3)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=-1,\text{ b}=1,\text{ f(x)}=\text{x}+3,\text{ h}=\frac{1+1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^1_{-1}(\text{x}+3)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(-1)+\text{f}(-1+\text{h})+\\\ ....\ +\text{f}\big\{-1+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(-1+3)+(-1+\text{h}+3)+\\ ....\ +\{-1+(\text{n}-1)\text{h}+3\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[\text{n}+\text{n}-1\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big(3-\frac{1}{\text{n}}\Big)$
$=6$
View full question & answer→Question 2445 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-1}\text{ dx}\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-1}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^3+\text{x}+1}{\text{x}^2-1}\ \text{dx}$
$=\int\Big(\text{x}+\frac{2\text{x}+1}{\text{x}^2-1}\Big)\text{dx}$
Now,
$\frac{2\text{x}+1}{\text{x}^2-1}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{\text{x}-1}$
$\Rightarrow2\text{x}+1=\text{A}(\text{x}-1)+\text{B}(\text{x}+1)$
put x = 1
$\Rightarrow3=2\text{B}\Rightarrow\text{B}=\frac{3}{2}$
put x = -1
$\Rightarrow-1=-2\text{A}\Rightarrow\text{A}=\frac{1}{2}$
$\therefore\text{I}=\int\text{xdx}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}+\frac{3}{2}\int\frac{\text{dx}}{\text{x}-1}$
$\text{I}=\frac{\text{x}^2}{2}+\frac{1}{2}\log|\text{x}+1|+\frac{3}{2}\log|\text{x}-1|+\text{C}$
View full question & answer→Question 2455 Marks
Find the integrals of the functions in Exercises:
$\sin^4\text{x}$
Answer$\sin^4\text{x}=\sin^2\text{x}\sin^2\text{x}$
$=\bigg(\frac{1-\cos2\text{x}}{2}\bigg)\bigg(\frac{1-\cos2\text{x}}{2}\bigg)$
$=\frac{1}{4}(1-\cos2\text{x})^2$
$=\frac{1}{4}\Big[1+\cos^22\text{x}-2\cos2\text{x}\Big]$
$=\frac{1}{4}\Bigg[1+\bigg(\frac{1+\cos4\text{x}}{2}\bigg)-2\cos2\text{x}\Bigg]$
$=\frac{1}{4}\Bigg[1+\frac{1}{2}+\frac{1}{2}\cos4\text{x}-2\cos2\text{x}\Bigg]$
$=\frac{1}{4}\Bigg[\frac{3}{2}+\frac{1}{2}\cos4\text{x}-2\cos2\text{x}\Bigg]$
$\therefore\int\sin^4\text{x}\text{ dx}=\frac{1}{4}\int\bigg[\frac{3}{2}+\frac{1}{2}\cos4\text{x}-2\cos2\text{x}\bigg]\text{ dx}$
$=\frac{1}{4}\bigg[\frac{3}{2}\text{x}+\frac{1}{2}\bigg(\frac{\sin4\text{x}}{4}\bigg)-\frac{2\sin2\text{x}}{2}\bigg]+\text{C}$
$=\frac{1}{8}\bigg[3\text{x}+\frac{\sin4\text{x}}{4}-2\sin2\text{x}\bigg]+\text{C}$
$=\frac{3\text{x}}{8}-\frac{1}{4}\sin2\text{x}+\frac{1}{32}\sin4\text{x}+\text{C}$
View full question & answer→Question 2465 Marks
$\int\frac{1}{\text{x}^{\frac{1}{3}}\big(\text{x}^{\frac{1}{3}}-1\big)}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{x}^{\frac{1}{3}}\big(\text{x}^{\frac{1}{3}}-1\big)}\text{dx}$
$=\int\frac{1}{\text{x}^{\frac{2}{3}}-\text{x}^{\frac{1}{3}}}\text{dx}$
Let $\text{x}=\text{t}^{3}$
On differentiating both sides, we get
$\text{dx}=3\text{t}^{2}\text{dt}$
$\therefore\ \text{I}\int\frac{3\text{t}^{2}}{(\text{t})^{\frac{2}{3}}-(\text{t}^{3})^{\frac{1}{3}}}\text{dt}$
$=\int\frac{3\text{t}^{2}}{\text{t}^2-\text{t}}\text{dt}$
$=3\int\frac{\text{t}}{\text{t}-1}\text{dt}$
$=3\int\frac{(\text{t}-1)+1}{\text{t}-1}\text{dt}$
$=3\int\Big[(1)+\frac{1}{\text{t}-1}\Big]\text{dt}$
$=\big[1+\log(\text{t}-1)\big]+\text{C}$
$=3\text{x}^\frac{1}{3}+3\log\big({\text{x}^\frac{1}{3}-1\big)}+\text{C}$
Hence, $\int\frac{1}{\text{x}^{\frac{1}{3}}\big(\text{x}^{\frac{1}{3}}-1\big)}\text{dx}=3\text{x}^\frac{1}{3}+3\log\big({\text{x}^\frac{1}{3}-1\big)}+\text{C}$
View full question & answer→Question 2475 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\cos^2\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\cos^2\text{x dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0(\pi-\text{x})\cos^2(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\cos^2\text{x}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\cos^2\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\cos^2\text{x}\text{ dx}$
$=\pi\int\limits^{\pi}_0\frac{1+\cos2\text{x}}{2}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\big(1+\cos2\text{x}\big)\text{dx}$
$=\frac{\pi}{2}\Big[\text{x}+\frac{\sin2\text{x}}{2}\Big]^{\pi}_0$
$=\frac{\pi}{2}(\pi-0)$
Hence, $\text{I}=\frac{\pi^2}{4}$
View full question & answer→Question 2485 Marks
Evaluate the following definite integral as limit of sum:
$\int\limits_{1}^{4}(\text{x}^{2}-\text{x)}\ \text{dx}$
Answer$\text{we}\ \text{know}\ \text{thet}\ \int\limits_{\text{a}}^{\text{b}}\ \text{f}\ \text{(x)}\ \text{dx}=\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[\text{f} \ \text{(a)}+\text{f}\ \text{(a}+\text{h)}+\text{f}(\text{a}+2\text{h)}+.......+\text{f}\text{(a}+\text{(n}-1)\text{h)}\big]$
$\text{where}\ \text{nh}=\text{b}-\text{a}$
$\text{Here},\ \ \text{a}=1,\text{b}=4,\text{nh}=3\ \text{and}\ \text{f}\ \text{(x)}=\text{x}^{2}-\text{x}$
$\therefore \ \ \int\limits_{1}^{4}\text{(x}^{2}-\text{x)}\ \text{dx}=\lim\limits_{\text{h}\rightarrow0}\ \text{h}\bigg[0+\text{h}+\text{h}^{2}+2\text{h}+4\text{h}^{2}+....+\text{(n}-1)\text{h}+\text{(n}-1)^{2}\text{h}^ {2}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0} \ \text{h}\bigg[\text{h}(1+2+3+....+\text{(n}-1)+\text{h}^{2}(1^{2}+2^{2}+.....\text{(n}-1)^{2}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{nh}\frac{\text{(nh}-\text{h)}}{2}+\frac{\text{nh}\text{(nh}-\text{h)}(2\text{nh}-\text{h)}}{6}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[3\frac{(3-\text{h)}}{2}+\frac{3(3-\text{h)}(2.3-\text{h)}}{6}\bigg]$
$=\bigg[3\frac{(3-0)}{2}+\frac{3(3-0)(6-0)}{6}\bigg]=\bigg[\frac{9}{2}+9\bigg]=\frac{27}{9}$
View full question & answer→Question 2495 Marks
Evaluate the following integrals:
$\int\cos^3\sqrt{\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\cos^3\sqrt{\text{x}}\text{dx}$
Let $\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt }$
$=2\int\text{t}\cos^3\text{t dt}$
$=2\int\text{t}\Big(\frac{3\cos\text{t}+\cos3\text{t}}{4}\Big)\text{dt}$
$=\frac{1}{2}\int\text{t}(3\cos\text{t}+\cos3\text{t})\text{dt}$
Using integral\tion by parts,
$\text{I}=\frac{1}{2}\Big[\text{t}\Big(3\sin\text{t}+\frac{1}{3}\sin3\text{t}\Big)+\int\Big(1\times3\sin\text{t}+\frac{\sin3\text{t}}{3}\Big)\text{dt}\Big]$
$=\frac{1}{2}\Big[\text{t}\Big(\frac{9\sin\text{t}+\sin3\text{t}}{3}\Big)+3\cos\text{t}+\frac{\cos3\text{t}}{9}\Big]+\text{C}$
$=\frac{1}{18}\big[27\text{t}\sin\text{t}+3\text{t}\sin3\text{t}+9\cos\text{t}+\cos3\text{t}\big]+\text{C}$
$\text{I}=\frac{1}{18}\big[27\sqrt{\text{x}}\sin\sqrt{\text{x}}+3\sqrt{\text{x}}\sin3\sqrt{\text{x}}+9\cos\sqrt{\text{x}}+\cos3\sqrt{\text{x}}\big]+\text{C}$
View full question & answer→Question 2505 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}\ \text{dx}$
AnswerConsider the integrals
$\text{I}=\int\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}\ \text{dx}$
Let $y = x^2$
Thus,
$\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}=\frac{\text{y}+1}{(\text{y}+4)(\text{y}+25)}$
$\Rightarrow\frac{\text{y}+1}{(\text{y}+4)(\text{y}+25)}=\frac{\text{A}}{\text{y}+4}+\frac{\text{B}}{\text{y}+25}$
$\Rightarrow\frac{\text{y}+1}{(\text{y}+4)(\text{y}+25)}=\frac{\text{A}(\text{y}+25)+\text{B}(\text{y}+4)}{(\text{y}+4)(\text{y}+25)}$
$\Rightarrow\text{y}+1=\text{Ay}+25\text{A}+\text{By}+4\text{B}$
Compairing the coefficient, we have,
$A + B = 1$ and $25A + 4B = 1$
Solving the above equations, we have,
$\text{A}=\frac{-1}{7}\text{ and }\text{B}=\frac{8}{7}$
Thus, $\int\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}\ \text{dx}$
$=\int\frac{\frac{1}{7}}{\text{x}^2+4}\ \text{dx}+\int\frac{\frac{8}{7}}{\text{x}^2+25}\ \text{dx}$
$=\frac{-1}{7}\int\frac{1}{\text{x}^2+4}\text{ dx}+\frac{8}{7}\int\frac{1}{\text{x}^2+25}\ \text{dx}$
$=\frac{-1}{7}\times\frac{1}{2}\tan^{-1}\frac{\text{x}}{2}+\frac{8}{7}\times\frac{1}{5}\tan^{-1}\frac{\text{x}}{5}+\text{C}$
$=\frac{-1}{14}\tan^{-1}\frac{\text{x}}{2}+\frac{8}{35}\tan^{-1}\frac{\text{x}}{5}+\text{C}$
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