Question 2013 Marks
Integrate the function in Exercise:$\frac{\sin^{8}-\cos^{8}\text{x}}{1-2\sin^{2}\text{x}\cos^{2}\text{x}}$
Answer$\frac{\sin^{8}\text{x}-\cos^{8}\text{x}}{1-2\sin^{2}\text{x}\cos^{2}\text{x}}=\frac{\Big(\sin^{4}\text{x}+\cos^{4}\text{x)}(\sin^{4}\text{x}-\cos^{4}\text{x}\Big)}{\sin^{2}\text{x}+\cos^{2}\text{x}-\sin^{2}\text{x}\cos^{2}\text{x}-\sin^{2}\text{x}\cos^{2}\text{x}}$
$=\frac{\Big(\sin^{4}\text{x}+\cos^{4}\text{x}\Big)\Big(\sin^{2}\text{x}+\cos^{2}\text{x}\Big)\Big(\sin^{2}\text{x}-\cos^{2}\text{x}\Big)}{\Big(\sin^{2}\text{x}-\sin^{2}\text{x}\cos^{2}\text{x}\Big)+\Big(\cos^{2}\text{x}-\sin^{2}\text{x}\cos^{2}\text{x}\Big)}$
$=\frac{\Big(\sin^{4}\text{x}+\cos^{4}\text{x}\Big)\Big(\sin^{2}\text{x}-\cos^{2}\text{x}\Big)}{\Big(\sin^{2}\text{x}\Big(1-\cos^{2}\text{x}\Big)+\cos^{2}\text{x}\Big(1-\sin^{2}\text{x}\Big)}$
$=\frac{-\Big(\sin^{4}\text{x}+\cos^{4}\text{x}\Big)\Big(\cos^{2}\text{x}-\sin^{2}\text{x}\Big)}{\Big(\sin^{4}\text{x}+\cos^{4}\text{x}\Big)}$
$=-\cos2\text{x}$
$\therefore\int\frac{\sin^{8}\text{x}-\cos^{8}\text{x}}{1-2\sin^{2}\text{x}\cos^{2}\text{x}}\text{dx}=\int-\cos2\text{x}\ \text{dx}=-\frac{\sin2\text{x}}{2}+\text{C}$
View full question & answer→Question 2023 Marks
$\int\frac{1}{2-3\text{x}}+\frac{1}{\sqrt{3\text{x}-2}}\text{dx}$
Answer$\int\Big(\frac{1}{2-3\text{x}}+\frac{1}{\sqrt{3\text{x}-2}}\Big)\text{dx}$
$=\int\frac{\text{dx}}{2-3\text{x}}+\int(3\text{x}-2)^{-\frac{1}{2}}\text{dx}$
$=\frac{\ln(2-3\text{x})}{-3}+\Bigg[\frac{(3\text{x}-2)^{-\frac1{2}+1}}{3\big(-\frac{1}{2}+1\big)}\Bigg]+\text{c}$
$=\frac{\ln(2-3\text{x})}{-3}+\frac{2}{3}(3\text{x}-2)^{\frac{1}{2}}+\text{c}$
$=-\frac{1}{3}\ln(2-3\text{x})+\frac{2}{3}\sqrt{3\text{x}-2}+\text{c}$
View full question & answer→Question 2033 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{1-\text{x}^2}(2+3\sin^{-1}\text{x})}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{1-\text{x}^2}(2+3\sin^{-1}\text{x})}\text{dx}$
Putting $\sin^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}=\text{dt}$
$\therefore\text{I}=\frac{1}{2+3\text{t}}\text{dt}$
$=\frac{1}{3}\text{ln}|2+3\text{t}|+\text{C}$
$=\frac{1}{3}\text{ln}|2+3\sin^{-1}\text{tx}|+\text{C }\big[\because\text{t}=\sin^{-1}\text{x}\big]$
View full question & answer→Question 2043 Marks
Find the integrals of the functions in Exercises:
$\tan^32\text{x}\sec2\text{x}$
Answer$\tan^32\text{x }\sec2\text{x}=\tan^22\text{x }\tan2\text{x }\sec2\text{x}$
$=(\sec^22\text{x}-1)\tan2\text{x }\sec2\text{x}$
$=\sec^22\text{x }\tan2\text{x }\sec2\text{x}-\tan2\text{x }\sec2\text{x}$
$\therefore\int\tan^32\text{x }\sec2\text{x}\text{ dx}$
$=\int\sec^22\text{x }\tan2\text{x }\sec2\text{x}\text{ dx}-\int\tan2\text{x }\sec2\text{x}\text{ dx}$
$=\int\sec^22\text{x }\tan2\text{x }\sec2\text{x}\text{ dx}-\frac{\sec2\text{x}}{2}+\text{C}$
$\text{Let }\sec2\text{x}=\text{t}$
$\therefore2\sec2\text{x }\tan2\text{x}\text{ dx}=\text{dt}$
$\therefore\int\tan^32\text{x }\sec2\text{x}\text{ dx}\text=\frac{1}{2}\int\text{t}^2\text{dt}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{\text{t}^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{(\sec2\text{x})^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
View full question & answer→Question 2053 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\cos\text{x}-\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\cos\text{x}-\sin\text{x})\text{dx}$
$=\int\text{e}^{\text{x}}\cos\text{x dx}-\int\text{e}^{\text{x}}\sin\text{x dx}$
Integrating by parts
$=\text{e}^{\text{x}}\cos\text{x}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\cos\text{x}\Big)\text{dx}-\int\text{e}^{\text{x}}\sin\text{x dx}$
$=\text{e}^{\text{x}}\cos\text{x}+\int\text{e}^{\text{x}}\sin\text{x dx}-\int\text{e}^{\text{x}}\sin\text{x dx}$
$=\text{e}^{\text{x}}\cos\text{x}+\text{C}$
$\therefore\int\text{e}^{\text{x}}(\cos\text{x}-\sin\text{x})\text{dx}=\text{e}^{\text{x}}\cos\text{x}+\text{C}$
View full question & answer→Question 2063 Marks
Evaluate the following integrals:
$\int\frac{\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
Answer$\int\frac{\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
$=\int\frac{\tan\text{x}}{(\sec\text{x}+\tan\text{x})}\times\Big(\frac{\sec\text{x}-\tan\text{x}}{\sec\text{x}-\tan\text{x}}\Big)\text{dx}$
$=\int\frac{\tan\text{x}(\sec\text{x}-\tan\text{x})}{(\sec^2\text{x}-\tan^2\text{x})}\text{dx}$
$=\int\Big(\frac{\sec\text{x}\tan\text{x}-\tan^2\text{x}}{1}\Big)\text{dx}$
$=\int\sec\text{x}\tan\text{x dx}-\int(\sec^2\text{x}-1)\text{dx}$
$=\sec\text{x}-\tan\text{x}+\text{x}+\text{C}$
View full question & answer→Question 2073 Marks
Evaluate $\int\frac{1}{\text{x}(1+\log\text{x})}\text{ dx}$
Answer$\text{I}=\int\frac{1}{\text{x}(1+\log\text{x})}\text{ dx}$
Let $(1+\log\text{x})=\text{t}$
Or, $\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$\text{I}=\log|\text{t}|+\text{C}$
$\therefore\ \text{I}=\log|1+\log\text{x}|+\text{C}$
View full question & answer→Question 2083 Marks
$\int\frac{1+\cos4\text{x}}{\cot\text{x}-\tan\text{x}}\text{dx}$
Answer$\int\Big(\frac{1+\cos4\text{x}}{\cot\text{x}-\tan\text{x}}\Big)\text{dx}$
$=\int\frac{(1+\cos4\text{x})}{\big(\frac{\cos\text{x}}{\sin\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}\big)}\text{dx}$
$=\int\frac{2\cos^22\text{x}\times\sin\text{x}\cos\text{x}}{(\cos^2\text{x}-\sin^2\text{x})}\text{dx}$
$=\int\frac{\cos^22\text{x}\times2\sin\text{x}\cos\text{x}}{\cos2\text{x}}\text{dx}$
$=\int\cos2\text{x}\sin2\text{x}\text{ dx}$
$=\frac{1}{2}\int2\sin2\text{x}\cos2\text{x dx}$
$=\frac{1}{2}\int\sin4\text{x dx}$
$=\frac{1}{2}\Big[-\frac{\cos4\text{x}}{4}\Big]+\text{c}$
$=-\frac{1}{8}\cos4\text{x}+\text{c}$
View full question & answer→Question 2093 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int\limits_{2}^{8}|\text{x}-5|\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{2}^{8}|\text{x}-5|\ \text{dx}$ $\text{putting}\ \text{x}-5=0\ \ \Rightarrow\ \text{x}=5\in(2,8)$ $\therefore\ \ \text{from eq. (i)},\ \text{I}=\int\limits_{2}^{5}|\text{x}-5|\ \text{dx}+\int\limits_{5}^{8}|\text{x}-5| \text{dx}$ $=\int\limits_{2}^{5}-(\text{x}-5)\ \text{dx}+\int\limits_{5}^{8}(\text{x}-5)\text{dx}$ $=-\bigg(\frac{\text{x}^{2}}{2}-5\text{x}\bigg)^{-2}_{-5}+\bigg(\frac{\text{x}^{2}}{2}-5\text{x}\bigg)^{5}_{-2}$$=-\bigg[\bigg(\frac{25}{2}-25\bigg)-(10-2)\bigg]+\bigg[(32-40)-\bigg(\frac{25}{2}-25\bigg)\bigg]$
$=25-\frac{25}{2}-8-8-\frac{25}{2}+25=34-\frac{50}{2}=34-25=9$
View full question & answer→Question 2103 Marks
Evaluate the following intregals:
$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$
Consider,
$\text{X}=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)+\text{B}$
$\text{x}=\text{A}(2\text{x}+1)+\text{B}$
$\Rightarrow\text{x}=(2\text{A})\text{x}+\text{A}+\text{B}$
Equating coefficient of like terms
$2\text{A}=1$
$\Rightarrow\text{A}=\frac{1}{2}$
And
$\text{A}+\text{B}=0$
$\Rightarrow\frac{1}{2}+\text{B}=0$
$\Rightarrow\text{B}=-\frac{1}{2}$
$\therefore\text{I}=\int\frac{\big(\frac{1}{2}(2\text{x}+1)-\frac{1}{2}\big)}{\sqrt{\text{x}^2+\text{x}+1}}\text{dx}$
View full question & answer→Question 2113 Marks
Evalute the following integrals:
$\int\frac{\cos2\text{x}+\text{x}+1}{\text{x}^2+\sin2\text{x}+2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos2\text{x}+\text{x}+1}{\text{x}^2+\sin2\text{x}+2\text{x}}\text{dx}$
Putting $\text{x}^2+\sin2\text{x}+2\text{x}=\text{t}$
$\Rightarrow2\text{x}+2\cos2\text{x}+2=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(\text{x}+\cos2\text{x}+1)\text{dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{2}\text{ln}|\text{t}|+\text{C}$
$=\frac{1}{2}\text{ln}|\text{x}^2+\sin2\text{x}+2\text{x}|+\text{C}$
$\big[\because\text{t}=\text{x}^2+\sin2\text{x}+2\text{x}\big]$
View full question & answer→Question 2123 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos2\text{x}\text{ dx}$ Then,
Integrating by parts
$\text{I}=\Big[\text{x}^2\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}2\text{x}\frac{\sin2\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\text{x}^2\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\Big[-\text{x}\frac{\cos2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0+\int_{0}^\limits{\frac{\pi}{2}}-1\frac{\cos2\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\text{x}^2\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\Big[-\text{x}\frac{\cos2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\Big[\frac{\sin2\text{x}}{4}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0-\frac{\pi}{4}-0$
$\Rightarrow\text{I}=-\frac{\pi}{4}$
View full question & answer→Question 2133 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x cosec x}}{\log(\tan\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.
Let $\text{I}=\int\frac{\sec\text{x cosec x}}{\log(\tan\text{x})}\text{dx}$
Putting $\log\tan\text{x}=\text{t}$
$\Rightarrow\frac{\sec^2\text{x}}{\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x cosec x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\tan\text{x})|+\text{C}$
View full question & answer→Question 2143 Marks
Evaluate the following integrals:
$\int(\tan\text{x}+\cot\text{x})^2\text{dx}$
Answer$\int(\tan\text{x}+\cot\text{x})^2\text{dx}$
$=\int(\tan^2\text{x}+\cot^2\text{x}+2\tan\text{x}\cot\text{x})\text{dx}$
$=\int(\tan^2\text{x}+\cot^2\text{x}+2)\text{dx}$
$=\int\big[(\sec^2\text{x}-1)+(\text{cosec}^2\text{x}-1)+2\big]\text{dx}$
$=\int(\sec^2\text{x}+\text{cosec}^2\text{x})\text{dx}$
$=\tan\text{x}-\cot\text{x}+\text{C}$
View full question & answer→Question 2153 Marks
Evaluate the following:
$\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{dx}$ $=\int\frac{(\sin\text{x}+\cos\text{x})}{\sqrt{\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}}}\text{dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2}}\text{dx}$ $=\int1\text{dx}=\text{x}+\text{C}$
View full question & answer→Question 2163 Marks
Evaluate the definite integral in Exercise:
$\int\limits_{0}^{\frac{\pi}{2}}\cos^{2}\text{x}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\cos^{2}\text{x}\ \text{dx}$$\int\cos^{2}\text{x}\ \text{dx}=\int\bigg(\frac{1+\cos2\text{x}}{2}\bigg)\text{dx}=\frac{\text{x}}{2}+\frac{\sin2\text{x}}{4}=\frac{1}{2}\bigg(\text{x}+\frac{\sin2\text{x}}{2}\bigg)=\text{F}\text{(x)}$ By second fundamental theorem of calculus, we obtin $\text{I}=\bigg[\text{F}\bigg(\frac{\pi}{2}\bigg)-\text{F}(0)\bigg]$$=\frac{1}{2}\Bigg[\bigg(\frac{\pi}{2}-\frac{\sin\pi}{2}\bigg)-\bigg(0+\frac{\sin0}{2}\bigg)\Bigg]$
$=\frac{1}{2}\bigg[\frac{\pi}{2}+0-0-0\bigg]$
$=\frac{\pi}{4}$
View full question & answer→Question 2173 Marks
Evaluate the following integrals:
$\int\sqrt{\tan\text{x}}\sec^4\text{x}\text{ dx}$
Answer$\int\sqrt{\tan\text{x}}\sec^4\text{x}\text{ dx}$
$=\int\sqrt{\tan\text{x}}\cdot\sec^2\text{x}\cdot\sec^2\text{x}\text{ dx}$
$=\int\sqrt{\tan\text{x}}\cdot(1+\tan^2\text{x})\sec^2\text{x}\text{ dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x}\text{ dx}=\text{ dt}$
Now, $\int\sqrt{\tan\text{x}}\cdot(1+\tan^2\text{x})\sec^2\text{x}\text{ dx}$
$=\int\sqrt{\text{t}}(1+\text{t}^2)\text{dt}$
$=\int\Big(\sqrt{\text{t}}+\text{t}^{\frac{5}{2}}\Big)\text{dt}$
$=\int\Big(\text{t}^{\frac{1}{2}}+\text{t}^{\frac{5}{2}}\Big)\text{dt}$
$=\frac{2}{3}\text{t}^{\frac{3}{2}}+\frac{2}{7}\text{t}^{\frac{7}{2}}+\text{C}$
$=\frac{2}{3}\tan^{\frac{3}{2}}\text{x}+\frac{2}{7}\tan^{\frac{7}{2}}\text{x}+\text{C}$
View full question & answer→Question 2183 Marks
Integrate the function in Exercise:
$\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
AnswerLet $\text{I}=\frac{\text{x}\cos^{-1}\text{x}}{\sqrt{1-\text{x}^2}}$
$\text{I}=\frac{-1}{2}\int\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}.\cos^{-1}\text{x}\text{dx}$
Taking $\cos^{-1} x$ as first function and $\Bigg(\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}\Bigg)$ as second function and integrating by parts, we obtain.
$\text{I}=\frac{-1}{2}\Bigg[\cos^{-1}\text{x}\int\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\cos^{-1}\text{x}\Bigg)\int\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}\Bigg\}\text{dx}\Bigg]$
$=\frac{-1}{2}\Bigg[\cos^{-1}\text{x}.2\sqrt{1-\text{x}^2}-\int\frac{-1}{\sqrt{1-\text{x}^2}}.2\sqrt{1-\text{x}^2}\text{dx}\Bigg]\frac{-1}{2}\Bigg[\cos^{-1}\text{x}.2\sqrt{1-\text{x}^2}-\int\frac{-1}{\sqrt{1-\text{x}^2}}.2\sqrt{1-\text{x}^2}\text{dx}\Bigg]$
$=\frac{-1}{2}\Big[2\sqrt{1-\text{x}^2}\cos^{-1}\text{x}-\int2\text{dx}\Big]$
$=\frac{-1}{2}\Big[2\sqrt{1-\text{x}^2}\cos^{-1}\text{x}+2\text{x}\Big]+\text{C}$
$=-\Big[\sqrt{1-\text{x}^2}\cos^{-1}\text{x}+\text{x}\Big]+\text{C}$
View full question & answer→Question 2193 Marks
Write a value of $\int\frac{\sin\text{x}}{\cos^3\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}}{\cos^3\text{x}}\text{ dx}$
Let $\cos\text{x}=\text{t}$
$-\sin\text{x dx}=\text{dt}$
$\sin\text{x dx}=-\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}^3}$
$=-\int\text{t}^{-3}\text{dt}$
$=-\Big[\frac{\text{t}^{-3+1}}{-3+1}\Big]+\text{C}$
$=\frac{1}{2\text{t}^2}+\text{C}$
$=\frac{1}{2\cos^2\text{x}}+\text{C}$ $(\because\text{t}=\cos\text{x})$
$=\frac{1}{2}\sec^2\text{x}+\text{C}$
View full question & answer→Question 2203 Marks
By using the properties of definite integrals, evaluate the integral in Exercise:
$\int^{\frac{\pi}{2}}_{0}\big(2\log\sin\text{x}-\log\sin2\text{x}\big)\text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\big(2\log\sin\text{x}-\log\sin2\text{x}\big)\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\big(\log\sin^{2}\text{x}-\log\sin2\text{x}\big)\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{\sin^{2}\text{x}}{\sin2\text{x}}\bigg)\text{dx}$ $=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{\sin^{2}\text{x}}{2\sin\text{x}\cos\text{x}}\bigg)\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{1}{2}\tan\text{x}\bigg)\text{dx}$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{1}{2}\tan\bigg(\frac{\pi}{2}-\text{x}\bigg)\bigg)\text{dx}\ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\text{dx}=\int\limits_{0}^{\text{a}}\text{f}\text{(a}-\text{x})\text{dx}=\bigg]$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{1}{2}\cot\text{x}\bigg)\text{dx}$ Adding eq.(i) and (ii) $21=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\log\bigg(\frac{1}{2}\tan\text{x}\bigg)+\log\bigg(\frac{1}{2}\cot\text{x}\bigg)\bigg]\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\log\bigg(\frac{1}{2}\tan\text{x}\bigg)\bigg(\frac{1}{2}\cot\text{x}\bigg)\bigg]\text{dx}$ $\Rightarrow\ \ 21=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\log\frac{1}{4}\bigg]\text{dx}=\log\frac{1}{4}\text{(x)}^{\frac{\pi}{2}}_{0}=\big(\log1-\log4\big)\frac{\pi}{2}=-\frac{\pi}{2}\log4$$\Rightarrow\ \ \text{I}=-\frac{\pi}{4}\log2^{2}=-\frac{2\pi}{4}\log2=-\frac{\pi}{2}\log2$
View full question & answer→Question 2213 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{2\pi}\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{2\pi}\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\text{dx}$ Then,
Integrating by parts,
$\text{I}=\Big[2\text{e}^{\text{x}}\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0-\int_{0}^\limits{2\pi}2\text{e}^{\text{x}}\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\text{dx}$
Integrating second term by parts,
$\text{I}=\Big[2\text{e}^{\text{x}}\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0+\bigg\{\Big[4\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0\\+\int_{0}^\limits{2\pi}-4\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\text{ dx}\bigg\}$
$\Rightarrow\text{I}=\Big[2\text{e}^{\text{x}}\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0+\Big[4\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0-4\text{I}$
$\Rightarrow5\text{I}=-2\text{e}^{2\pi}\frac{1}{\sqrt{2}}-2\frac{1}{\sqrt{2}}-4\text{e}^{2\pi}\frac{1}{\sqrt{2}}-4\frac{1}{\sqrt{2}}$
$\Rightarrow5\text{I}=-3\sqrt{2}\text{e}^{2\pi}-3\sqrt{2}$
$\Rightarrow\text{I}=-\frac{3\sqrt{2}}{5}\big(\text{e}^{2\pi}+1\big)$
View full question & answer→Question 2223 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 2233 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2}\cos^2\Big(\frac{1}{\text{x}}\Big)\text{dx}$
Answer$\int\frac{1}{\text{x}^2}\cos^2\Big(\frac{1}{\text{x}}\Big)\text{dx}$
Let $\frac{1}{\text{x}}=\text{t}$
$\Rightarrow-\frac{1}{\text{x}^2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}^2}\text{ dx}=-\text{dt}$
Now, $\int\frac{1}{\text{x}^2}\cos^2\Big(\frac{1}{\text{x}}\Big)\text{dx}$
$=-\int\cos^2\text{t}\ \text{dt}$
$=-\int\Big(\frac{1+\cos2\text{t}}{2}\Big)\text{dt}$
$=-\frac{1}{2}\int(1+\cos2\text{t})\text{dt}$
$=-\frac{1}{2}\Big[\text{t}+\frac{\sin2\text{t}}{2}\Big]+\text{C}$
$=-\frac{1}{2}\Bigg[\frac{1}{\text{x}}+\frac{\sin\big(\frac{2}{\text{x}}\big)}{2}\Bigg]+\text{C}$
$=-\frac{1}{2}\Big(\frac{1}{\text{x}}\Big)-\frac{1}{4}\sin\Big(\frac{2}{\text{x}}\Big)+\text{C}$
View full question & answer→Question 2243 Marks
Find the integrals of the functions in Exercises:
$\cos2\text{x}\cos4\text{x}\cos6\text{x}$
AnswerIt is known that, $\cos\text{A}\cos\text{B}=\frac{1}{2}\big\{\cos(\text{A}+\text{B})+\cos(\text{A}-\text{B}) \big\}$
$\therefore\int \cos2\text{x}(\cos4\text{x}\cos6\text{x) dx}$
$=\int\cos2\text{x}\bigg[\frac{1}{2}\{\cos(\text{4x+6x})+\cos(\text{4x}-\text{6x}) \big\}\bigg]\text{ dx}$
$=\frac{1}{2}\int\big\{\cos2\text{x}\cos10\text{x}+\cos2\text{x}\cos(-2\text{x})\big\}\text{ dx}$
$=\frac{1}{2}\int\big\{\cos2\text{x}\cos10\text{x}+\cos^22\text{x}\big\}\text{ dx}$
$=\frac{1}{2}\int\Bigg[\bigg\{\frac{1}{2}\cos(2\text{x}+10\text{x})+\cos(2\text{x}-10\text{x})\bigg\}+\bigg(\frac{1+\cos4\text{x}}{2}\bigg)\Bigg]\text{dx}$
$=\frac{1}{4}\int(\cos12\text{x}+\cos8\text{x}+1+\cos4\text{x})\text{dx}$
$=\frac{1}{4}\bigg[\frac{\sin12\text{x}}{12}+\frac{\sin8\text{x}}{8}+\text{x}+\frac{\sin4\text{x}}{4}\bigg]+\text{C}$
View full question & answer→Question 2253 Marks
Evaluate the following integrals:
$\int\sec^42\text{x}\text{ dx}$
Answer$\int\sec^42\text{x}\text{ dx}$
$=\int\sec^22\text{x}.\sec^22\text{x}\text{ dx}$
$=\int(1+\tan^22\text{x}).\sec^22\text{x}\text{ dx}$
Let $\tan2\text{x}=\text{t}$
$\sec^22\text{x}.2\text{dx}=\text{dt}$
$\sec^22\text{x}.\text{dx}=\frac{\text{dt}}{2}$
Now, $\int(1+\tan^22\text{x}).\sec^22\text{x}\text{ dx}$
$=\frac{1}{2}\int(1+\text{t}^2)\text{dt}$
$=\frac{1}{2}\Big[\text{t}+\frac{\text{t}^3}{3}\Big]+\text{C}$
$=\frac{\text{t}}{2}+\frac{\text{t}^3}{6}+\text{C}$
$=\frac{\tan(2\text{x})}{2}+\frac{\tan^3(2\text{x})}{6}+\text{C}$
View full question & answer→Question 2263 Marks
Evalute the following integrals:
$\int\big\{1+\tan\text{x}\tan(\text{x}+\theta)\big\}\text{dx}$
AnswerSince,
$\tan(\text{A}-\text{B})=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}$
$\therefore\tan(\text{x}+\theta-\text{x})=\frac{\tan(\text{x}+\theta)-\tan\text{x}}{1+\tan(\text{x}+\theta)\tan\text{x}}$
$\Rightarrow 1+\tan(\text{x}+\theta)\tan\text{x}=\frac{\tan(\text{x}+\theta)-\tan\text{x}}{\tan\theta}$
$\Rightarrow\int1+\tan(\text{x}+\theta)\tan\text{x dx}$
$=\frac{1}{\tan\theta}\big[\int\tan(\text{x}+\theta)\text{dx}-\int\tan\text{x dx}\big]$
$=\frac{1}{\tan\theta}\big[-\log|\cos(\text{x}+\theta)++\log|\cos\text{x}|\big]+\text{C}$
$=\frac{1}{\tan\theta}\big[\log|\cos\text{x}|-\log|\cos(\text{x}+\theta)|\big]+\text{C}$
$=\frac{1}{\tan\theta}\log\Big|\frac{\cos\text{x}}{\cos(\text{x}+\theta)}\Big|+\text{C}$
View full question & answer→Question 2273 Marks
Prove the following Exercise:
$\int^{1}_{0}\sin^{-1}\text{x dx}=\frac{\pi}{2}-1$
Answer$\text{Let I}=\int^{1}\limits_{0}\sin^{-1}\text{x dx}$
$\text{I}=\int^{1}\limits_{0}\sin^{-1}\text{x.}\ 1.\text{ dx}$
Integrating by Parts, we obtain
$\text{I}=\Big[\sin^{-1}\text{x}.\text{x}\Big]^{1}_{0}-\int^{1}\limits_{0}\frac{1}{\sqrt{1-\text{x}^{2}}}.\text{x dx}$
$=\Big[\text{x}\sin^{-1}\text{x}\Big]^{1}_{0}+\frac{1}{2}\int^{1}\limits_{0}\frac{(-2\text{x)}}{\sqrt{1-\text{x}^{2}}}\text{dx}$
$\text{Let} 1-\text{x}^{2}=\text{t}\Rightarrow-2\text{x dx}=\text{dt}$
when $\text{x}=0,\text{t}=1$ and when $\text{x}=1,\text{t}=0$
$\text{I}=\Big[\text{x}\sin^{-1}\text{x}\Big]^{1}_{0}+\frac{1}{2}\int^{0}\limits_{1}\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\Big[\text{x}\sin^{-1}\text{x}\Big]^{1}_{0}+\frac{1}{2}\Big[2\sqrt{\text{t}}\Big]^{0}_{1}$
$=\sin^{-1}(1)+\Big[-\sqrt{1}\Big]$
$=\frac{\pi}{2}-1$
Hence, the given result is Proved.
View full question & answer→Question 2283 Marks
Write a value of $\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{ dx}$
Let $\text{x}+\log\sin\text{x}=\text{t}$
$(1+\cot\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
Hence,
$\text{I}=\log|\text{x}+\log\sin\text{x}|+\text{C}$
View full question & answer→Question 2293 Marks
Evaluate the following:
$\int\frac{\cos\text{x}-\cos2\text{x}}{1-\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}-\cos2\text{x}}{1-\cos\text{x}}\text{dx}$
$=\int\frac{2\sin\frac{3\text{x}}{2}\cdot\sin\frac{\text{x}}{2}}{2\sin^2\frac{\text{x}}{2}}\text{dx}$
$=\int\frac{\sin\frac{3\text{x}}{2}}{\sin\frac{\text{x}}{2}}\text{dx}$
$=\int\frac{3\sin\frac{\text{x}}{2}-4\sin^3\frac{\text{x}}{2}}{\sin\frac{\text{x}}{2}}\text{dx}$ $\big[\because\sin3\text{x}=3\sin\text{x}-4\sin^3\text{x}\big]$
$=3\int\text{dx}-4\int\sin^2\frac{\text{x}}{2}\text{dx}$
$=3\int\text{dx}-4\int\frac{1-\cos\text{x}}{2}\text{dx}$ $=\int\text{dx}+2\int\cos\text{xdx}=\text{x}+2\sin\text{x}+\text{C}$
View full question & answer→Question 2303 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 2313 Marks
Evaluate the following integrals:
$\int\text{x}^3\sin\big(\text{x}^4+1\big)\text{dx}$
Answer$\int\text{x}^3.\sin\big(\text{x}^4+1\big)\text{dx}$
Let $\text{x}^4+1=\text{t}$
$\Rightarrow4\text{x}^3=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{x}^3\text{ dx}=\frac{\text{dt}}{4}$
Now, $\int\text{x}^3.\sin\big(\text{x}^4+1\big)\text{dx}$
$=\frac{1}{4}\int\sin(\text{t})\text{dt}$
$=\frac{1}{4}[-\cos\text{t}]+\text{C}$
$=\frac{1}{4}[-\cos(\text{x}^4+1)]+\text{C}$
View full question & answer→Question 2323 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 2333 Marks
Evaluate the following integrals:
$\int\text{x}\sin2\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\sin2\text{x dx}$
Using integration by parts,
$\text{I}=\text{x}\int\sin2\text{x dx}-\int(1\int\sin2\text{x dx})\text{dx}$
$=\text{x}\Big(-\frac{\cos2\text{x}}{2}\Big)-\int\Big(-\frac{\cos2\text{x}}{2}\Big)\text{dx}$
$=-\frac{\text{x}}{2}\cos2\text{x}+\frac{1}{2}\int\cos2\text{x dx}$
$=-\frac{\text{x}}{2}\cos2\text{x}+\frac{1}{2}\frac{\sin2\text{x}}{2}+\text{C}$
$\text{I}=-\frac{\text{x}}{2}\cos2\text{x}+\frac{1}{4}\sin2\text{x}+\text{C}$
View full question & answer→Question 2343 Marks
Evaluate the following integrals:
$\int\sec\text{x}.\text{log} (\sec\text{x}+\tan\text{x})\text{dx}$
Answer$\int\sec\text{x}.\text{log} (\sec\text{x}+\tan\text{x})\text{dx}$
$\text{Let }\text{ log}(\sec\text{x}+ \tan\text{x})=\text{t}$
$\Rightarrow \frac{(\sec\text{x} \tan\text{x}+\sec^{2}\text{x})} {(\sec\text{x} +\tan\text{x})}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow \frac{\sec\text{x} (\sec\text{x}+\tan\text{x})} {(\sec\text{x} +\tan\text{x})}\text{dx}=\text{dt}$
$\text{Now},\int\sec \text{x}.\text{log}(\sec\text{x}+\tan\text{x})\text{dx}$
$=\int \text{t}.\text{dt}$
$=\frac{\text{t}^{2}}{2}+\text{C}$
$=\frac{[\text{log}(\sec\text{x}+\tan\text{x})]^2}{2}+\text{C}$
View full question & answer→Question 2353 Marks
By Using properties of definite integral, evaluate the following integral in Exercise:
$\int^{\text{a}}_{0}\frac{\sqrt{\text{x}}}{\sqrt{\text{x}+\sqrt{\text{a}-\text{x}}}}\text{dx}$
Answer$\text{Let}\ \text{I}= \int^{\text{a}}\limits_{0}\frac{\sqrt{\text{x}}}{\sqrt{\text{x}+\sqrt{\text{a}-\text{x}}}}\text{dx}\ \ ...(\text{i})$ $\Rightarrow\ \ \ \text{I}=\int^{\text{a}}\limits_{0}\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{a}-\text{x}}+\sqrt{\text{a}-\text{(a}-\text{x)}}}\text{dx}=\int^{\text{a}}\limits_{0}\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{a}-\text{x}}+\sqrt{\text{x}}}\text{dx}\ \ ...(\text{ii})$ Adding eq.(i) and (ii),$ 21=\int^{\text{a}}\limits_{0}\bigg(\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}+\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{a}-\text{x}}+\sqrt{\text{x}}}\bigg)\text{dx}=\int^{\text{a}}\limits_{0}\bigg(\frac{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\bigg)\text{dx}=\int^{\text{a}}\limits_{0}1\text{dx}=\text{(x)}^{\text{a}}_{0}=\text{a}$
$\Rightarrow\ \ \text{I}=\frac{\text{a}}{2}$
View full question & answer→Question 2363 Marks
Evaluate the following integrals:$\int\frac{\sec^2\text{x}}{1-\tan^2\text{x}}\text{dx}$
Answer$\int\frac{\sec^2\text{x dx}}{1-\tan^2\text{x}}$
Let $\tan\text{x = t}$
$\Rightarrow\sec^2\text{x dx = dt}$
Now, $\int\frac{\sec^2\text{x dx}}{1-\tan^2\text{x}}$
$=\int\frac{\text{dt}}{1-\text{t}^2}$
$=\frac{1}{2}\log\bigg|\frac{1+\text{t}}{1-\text{t}}\bigg|+\text{C}$
$=\frac{1}{2}\log\bigg|\frac{1+\tan\text{x}}{1-\tan\text{x}}\bigg|+\text{C}$
View full question & answer→Question 2373 Marks
Find the integrals of the functions in Exercises:
$\cos^42\text{x}$
Answer$\cos^42\text{x}=(\cos^22\text{x})^2$
$=\bigg(\frac{1+\cos4\text{x}}{2}\bigg)^2$
$=\frac{1}{4}\Big[1+\cos^24\text{x}+2\cos4\text{x}\Big]$
$=\frac{1}{4}\Bigg[1+\bigg(\frac{1+\cos8\text{x}}{2}\bigg)+2\cos4\text{x}\Bigg]$
$=\frac{1}{4}\Bigg[1+\frac{1}{2}+\frac{\cos8\text{x}}{2}+2\cos4\text{x}\Bigg]$
$=\frac{1}{4}\Bigg[\frac{3}{2}+\frac{\cos8\text{x}}{2}+2\cos4\text{x}\Bigg]$
$\therefore\int\cos^42\text{x}\text{ dx}=\int\bigg(\frac{3}{8}+\frac{\cos8\text{x}}{8}+\frac{\cos4\text{x}}{2}\bigg)\text{ dx}$
$=\frac{3}{8}\text{x}+\frac{\sin8\text{x}}{64}+\frac{\sin4\text{x}}{8}+\text{C}$
View full question & answer→Question 2383 Marks
Evaluate the following integrals:
$\int\frac{\big\{\text{e}^{\sin^{-1}\text{x}}\big\}^2}{\sqrt{1-\text{x}^2}}\text{dx}$
AnswerLet $\text{e}^{\sin^{-1}\text{z}}=\text{t}$
Differentiating both sides w.r.t. x,
$\text{e}^{\sin^{-1}\text{x}}\times \frac{1}{\sqrt{1-\text{x}^2}}\text{dx}=\text{dt}$
Now, $\int \frac{(\text{e}^{\sin-1_\text{x}})^2}{\sqrt{1-\text{x}^2}}\text{dx}$
$\int \text{e}^{\sin^{-1}\text{z}}\times \frac{\text{e}^{\sin^{-1}\text{x}}}{\sqrt{1-\text{x}^2}}\text{dx}$
$\int \text{t}.\text{dt}$
$=\frac{\text{t}^2}{2}+\text{C}$
$=\frac{(\text{e}^{\sin^{-1}\text{x}})^2}{2}+\text{C}$
View full question & answer→Question 2393 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 2403 Marks
Evaluate the following integrals:
$\int3\sqrt{\cos^2\text{x}}\sin\text{x }\text{dx}$
AnswerLet I $=\int3\sqrt{\cos^2\text{x}}\sin\text{x }\text{dx}\ ....(1)$
Let $\cos\text{x}=\text{t}$ then,
$\text{d}(\cos\text{x})=\text{dt}$
$\Rightarrow-\sin\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\text{dx}=-\frac{\text{dt}}{\sin\text{x}}$
Putting $\cos\text{x}=\text{t}$ and $\text{dx}=-\frac{\text{dt}}{\sin\text{x}}$ in equation (1), we get
$\text{I}=\int3\sqrt{\text{t}^2}\sin\text{x}\times\frac{-\text{dt}}{\sin\text{x}}$
$=-\int\text{t}^\frac{2}{3}\sin\text{x}\frac{\text{dt}}{\sin\text{x}}$
$=-\int\text{t}^\frac{2}{3}\text{dt}$
$=-\frac{3}{5}\times\text{t}^\frac{5}{3}+\text{C}$
$=-\frac{3}{5}(\cos\text{x})^\frac{5}{3}+\text{C}$
$\therefore\text{I}=-\frac{3}{5}(\cos\text{x})^\frac{5}{3}+\text{C}$
View full question & answer→Question 2413 Marks
Find the integrals of the functions in Exercises:
$\frac{\sin^2\text{x}}{1+\cos\text{x}}$
Answer$\frac{\sin^2\text{x}}{1+\cos\text{x}}=\frac{\bigg(2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\bigg)^2}{2\cos^2\frac{\text{x}}{2}}$ $\ \ \ \ \ \bigg[\sin\text{x}=2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2};\cos\text{x}=2\cos^2\frac{\text{x}}{2}-1\bigg]$ $=\frac{4\sin^2\frac{\text{x}}{2}\cos^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}$ $=2\sin^2\frac{\text{x}}{2}$ $=1-\cos\text{x}$ $\therefore\int\frac{\sin^2\text{x}}{1+\cos\text{x}}\text{ dx}=\int(1-\cos\text{x})\text{ dx}$ $=\text{x}-\sin\text{x}+\text{C}$
View full question & answer→Question 2423 Marks
Evaluate the following:
$\int\frac{\text{e}^{6\log\text{x}}-\text{e}^{5\log\text{x}}}{\text{e}^{4\log\text{x}}-\text{e}^{3\log\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{6\log\text{x}}-\text{e}^{5\log\text{x}}}{\text{e}^{4\log\text{x}}-\text{e}^{3\log\text{x}}}\text{dx}$
$=\int\frac{\text{e}^{\log\text{x}^6}-\text{e}^{\log\text{x}^5}}{\text{e}^{\log\text{x}^4}-\text{e}^{3\log\text{x}^3}}\text{dx}$ $\big[\because\ \text{a}\log\text{b}-\text{b}-\log\text{b}^{\text{a}}\big]$
$=\int\frac{\text{x}^6-\text{x}^5}{\text{x}^4-\text{x}^3}\text{dx}$ $\big[\because\ \text{e}^{\log\text{x}}=\text{x}\big]$
$=\int\frac{\text{x}^3-\text{x}^2}{\text{x}-1}\text{dx}=\int\frac{\text{x}^2(\text{x}-1)}{\text{x}-1}\text{dx}$ $=\int\text{x}^2\text{dx}=\frac{\text{x}^3}{3}+\text{C}$
View full question & answer→Question 2433 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 2443 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 2453 Marks
Evaluate the following integrals:
$\int\frac{\cos\text{x}}{1-\cos\text{x}}\text{dx}\text{ or }\int\frac{\cot\text{x}}{\text{cosec x}-\cot\text{x}}\text{dx}$
Answer$\int\frac{\cot\text{x}}{\text{cosec x}-\cot\text{x}}\text{dx}$
$=\int\frac{\frac{\cos\text{x}}{\sin\text{x}}}{\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}}\text{dx}$
$=\int\Big(\frac{\cos\text{x}}{1-\cos\text{x}}\Big)\times\frac{(1+\cos\text{x})}{(1+\cos\text{x})}\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\cos^2\text{x}}{1-\cos^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\cos^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{\cos\text{x}}{\sin\text{x}}\times\frac{1}{\sin\text{x}}+\frac{\cos^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\big[(\cot\text{x}\text{ cosec x})+\cot^2\text{x}\big]\text{dx}$
$=\int\big[\cot\text{x}\text{ cosec x}+\text{cosec}^2\text{x}-1\big]\text{dx}$
$=-\text{cosec x}-\cot\text{x}-\text{x}+\text{C}$
View full question & answer→Question 2463 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 2473 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 2483 Marks
Integrate the rational function in exercise:
$\frac{\cos\text{x}}{(1-\sin\text{x})(2-\sin\text{x})}$
[Hint: Put sin x = t]
Answer$\text{I}=\int\frac{\cos\text{x}}{(1-\sin\text{x})(2-\sin\text{x})}\text{dx}\dots(\text{i})$
Putting $\sin \text{x = t}$
$\Rightarrow \ \cos\text{x}=\frac{\text{dx}}{\text{dt}}$
$\Rightarrow \cos \text{x dx = dt}$
$\therefore$ From eq. (i),
$\text{I}=\int\frac{1}{(1-\text{t})(2-\text{t})}\text{dt}=\int\frac{(2-\text{t})-(1-\text{t})}{(1-\text{t})(2-\text{t})}\text{dt}$
$=\int\Bigg(\frac{(2-\text{t})}{(1-\text{t})(2-\text{t})}-\frac{(1-\text{t})}{(1-\text{t})(2-\text{t})}\Bigg)\text{dt}=\int\Bigg(\frac{1}{(1-\text{t})}-\frac{1}{(2-\text{t})}\Bigg)\text{dt}$
$=\int\frac{1}{1-\text{t}}\text{dt}-\int\frac{1}{2-\text{t}}\text{dt}=\frac{\text{log}|1-\text{t}|}{-1\rightarrow\text{Coeff. of t}}-\frac{\text{log}|2-\text{t}}{-1}+\text{c}$
$=-\text{log}|1-\text{t}|+\text{log}|2-\text{t}|+\text{c}=\text{log}\Bigg|\frac{2-\text{t}}{1-\text{t}}\Bigg|+\text{c}=\text{log}\Bigg|\frac{2-\sin\text{x}}{1-\sin\text{x}}\Bigg|+\text{c}$
View full question & answer→Question 2493 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 2503 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, $\int_{\frac{\pi}{4}}^\limits{\frac{\pi}{2}}\cot\text{x}\text{ dx}$
$=\big[\log(\sin\text{x})\big]^{\frac{\pi}{2}}_\frac{\pi}{4}$
$=\Big[\log\Big(\sin\frac{\pi}{2}\Big)-\log\Big(\sin\frac{\pi}{4}\Big)\Big]$
$=\Big[\log1-\log\frac{1}{\sqrt{2}}\Big]$
$=\big[0-\log\text{2}\big]$
$=\log\sqrt{2}$ $[\because\log\text{a}^{\text{n}}-\text{n}\log\text{a}\big]$
$=\frac{1}{2}\log2$
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