Question 1013 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{1-\cos2\text{x}}}\text{dx}$
Answer$\int\frac{1}{\sqrt{1-\cos2\text{x}}}\text{dx}$$\int\frac{1}{\sqrt{2\sin^2\text{x}}}\text{dx}\ \big[\because 1-\cos 2\text{x}=2\sin^2\text{x}\big]$
$=\frac{1}{\sqrt{2}}\int\text{cosec x dx}$
$=\frac{1}{\sqrt{2}}\text{ln}|\text{cosec x}-\cot\text{x}|=\text{C}$
$=\frac{1}{\sqrt{2}}\text{ln}\Big|\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}\Big|+\text{C}$
$=\frac{1}{\sqrt{2}}\text{ln}\bigg|\frac{2\sin^2\frac{\text{x}}{2}}{\sin\text{x}}\bigg|+\text{C} \Big[\because 1-\cos\text{x}=2\sin^2\frac{\text{x}}{2}\Big]$
$=\frac{1}{\sqrt{2}}\text{ln}\Bigg|\frac{2\sin^2\frac{\text{x}}{2}}{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}\Bigg|+\text{C}\ \Big[\because\sin\text{x}=2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\Big]$
$=\frac{1}{\sqrt{2}}\text{ln}\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 1023 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{5\text{x}^2-2\text{x}}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\sqrt{5\text{x}^2-2\text{x}}}$
$=\int\frac{\text{dx}}{\sqrt{5\big(\text{x}^2-\frac{2}{5}\text{x}\big)}}$
$=\frac{1}{\sqrt5}\int\frac{\text{dx}}{\sqrt{\text{x}^2-\frac{2}{5}\text{x}+\big(\frac{1}{5}\big)^2-\big(\frac{1}{5}\big)^2}}$
$=\frac{1}{\sqrt5}\int\frac{\text{dx}}{\big(\text{x}-\frac{1}{5}\big)^2-\big(\frac{1}{5}\big)^2}$
$=\frac{1}{\sqrt5}\log\bigg|\text{x}-\frac{1}{5}+\sqrt{\big(\text{x}-\frac{1}{5}\big)^2+\big(\frac{1}{5}\big)^2}\bigg|+\text{C}$
$=\frac{1}{\sqrt5}\log\Big|\frac{5\text{x}-1}{5}+\frac{\sqrt{5\text{x}^2-2\text{x}}}{\sqrt5}\Big|+\text{C}$
View full question & answer→Question 1033 Marks
Evaluate the following integrals:
$\int\frac{1}{1-\sin\text{x}}\text{dx}$
Answer$\int\frac{1}{1-\sin\text{x}}\text{dx}$$=\int\frac{(1+\sin\text{x})}{(1-\sin\text{x})\times(1+\sin\text{x})}\text{dx}$
$=\Big(\frac{1+\sin\text{x}}{1-\sin^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1+\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1}{\cos^2\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\Big)\text{dx}$
$=\int(\sec^2\text{x}+\sec\text{x}\tan\text{x})\text{dx}$
$=\tan\text{x}+\sec\text{x}+\text{C}$
View full question & answer→Question 1043 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 1053 Marks
Evaluate the following integrals:
$\int\sin^4\text{x}\cos^3\text{x}\text{ dx}$
Answer$\int\sin^4\text{x}\cos^3\text{x}\text{ dx}$
$=\int\sin^4\text{x}\cdot\cos^2\text{x }\cos\text{x}\text{ dx}$
$=\int\sin^4\text{x}\big(1-\sin^2\text{x}\big)\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int\sin^4\text{x}\big(1-\sin^2\text{x}\big)\cos\text{x}\text{ dx}$
$=\int\text{t}^4(1-\text{t}^2)\text{dt}$
$=\int(\text{t}^4+\text{t}^6)=\text{dt}$
$=\frac{\text{t}^5}{5}-\frac{\text{t}^7}{7}+\text{C}$
$=\frac{\sin^5\text{x}}{5}-\frac{\sin^7\text{x}}{7}+\text{C}$
View full question & answer→Question 1063 Marks
Evaluate the following integrals:$\int\frac{1}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-4}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-2^2}}$
$=\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\Big(\text{x}^{\frac{1}{3}}\Big)^2-2^2}}$
Let $\text{x}^{\frac{1}{3}}=\text{t}$
$\Rightarrow\frac{1}{3}\text{x}^{\frac{-2}{3}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{1}{3\text{x}^{\frac{2}{3}}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{\text{dx}}{\text{x}^{\frac{2}{3}}}=3\text{dt}$
Now, $\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-2^2}}$
$=3\int\frac{\text{dt}}{\sqrt{\text{t}^2-2^2}}$
$=3\log\Big|\text{t}+\sqrt{\text{t}^2-2^2}\Big|+\text{C}$
$=3\log\Bigg|\text{x}^{\frac{1}{3}}+\sqrt{\text{x}^{\frac{2}{3}}-4}\Bigg|+\text{C}$
View full question & answer→Question 1073 Marks
Evalute the following integrals:
$\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\text{dx}$
AnswerLet $\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\text{dx}\ .....\text{(i)}$
Let $\tan\text{x}+2=\text{t}$ then,
$\text{d}(\tan\text{x}+2)=\text{dt}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{1}{\sec^2\text{x}}\text{dt}$
Putting $\tan\text{x}+2=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$ In equation (i), we get,
$\text{I}=\int\frac{\sec^2\text{x}}{\text{t}}\times\frac{1}{\sec^2\text{x}}\text{dt}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\tan\text{x}+2|+\text{C}$
$\Rightarrow\text{I}=\log|\tan\text{x}+2|+\text{C}$
View full question & answer→Question 1083 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2}{\text{x}^6+\text{a}^6}\text{dx}$
Answer$\int\frac{\text{x}^2}{\text{x}^6+\text{a}^6}\text{dx}$
$\Rightarrow\int\frac{\text{x}^2\text{dx}}{(\text{x}^3)^2+(\text{a}^3)^2}$
Let $\text{x}^3=\text{t}$
$\Rightarrow3\text{x}^3\text{dx = dt}$
$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$
Now $\int\frac{\text{x}^2}{\text{x}^6+\text{a}^6}\text{dx}$
$=\frac{1}{3}\int\frac{\text{dt}}{\text{t}^2+(a^3)^2}$
$=\frac{1}{3\text{a}^3}\tan^{-1}\Big(\frac{\text{t}}{\text{a}^3}\Big)+\text{C}$
$=\frac{1}{3\text{a}^3}\tan^{-1}\Big(\frac{\text{x}^3}{\text{a}^3}\Big)+\text{C}$
View full question & answer→Question 1093 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}(\tan\text{x}-\log\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\tan\text{x}-\log\cos\text{x})\text{dx}$$=\int\text{e}^{\text{x}}\tan\text{x dx}-\int\text{e}^{\text{x}}\log\cos\text{x dx}$
Integrating by parts
$=\int\text{e}^{\text{x}}\tan\text{x dx}-\Big\{\text{e}^{\text{x}}\log\cos\text{x}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\log\cos\text{x}\Big)\text{dx}\Big\}$
$=\int\text{e}^{\text{x}}\tan\text{x dx}-\big\{\text{e}^{\text{x}}\log\cos\text{x}+\int\text{e}^{\text{x}}\tan\text{x dx}\big\}$
$=\int\text{e}^{\text{x}}\tan\text{x dx}-\text{e}^{\text{x}}\log\cos\text{x}-\int\text{e}^{\text{x}}\tan\text{x dx}+\text{C}$
$=-\text{e}^{\text{x}}\log\cos\text{x}+\text{C}$
$=\text{e}^{\text{x}}\log\sec\text{x}+\text{C}$ $\big[\because\log\sec\text{x}=-\log\cos\text{x}\big]$
View full question & answer→Question 1103 Marks
Evaluate the following integrals:$\int\frac{\log\text{x}}{\text{x}^{\text{n}}}\text{dx}$
Answer$\int\frac{\log\text{x}}{\text{x}^{\text{n}}}\text{dx}=\int(\log\text{x})\Big(\frac{1}{\text{x}^{n}}\Big)\text{dx}$
by integration by parts
$\int(\log\text{x})\Big(\frac{1}{\text{x}^{\text{n}}}\Big)\text{dx}=\log\text{x}\int\Big(\frac{1}{\text{x}^{\text{n}}}\Big)\text{dx}-\int\Big(\frac{\text{d}(\log\text{x})}{\text{dx}}\Big)\Big(\int\Big(\frac{1}{\text{x}^{n}}\Big)\text{dx}\Big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)-\int\frac{1}{\text{x}}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)\text{dx}=\log\text{x}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)-\int\Big(\frac{\text{x}^{-\text{n}}}{1-\text{n}}\Big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)-\Big(\frac{1}{1-\text{n}}\Big)\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)=\log\text{x}\Big(\frac{\text{x}^{1-\text{n}}}{1-\text{n}}\Big)-\bigg(\frac{\text{x}^{1-\text{n}}}{[1-\text{n}]^2}\bigg)+\text{C}$
View full question & answer→Question 1113 Marks
Evaluate the following integrals:
$\int\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{1+\cos4\text{x}}}\text{dx}$
Answer$\int\Big(\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{1+\cos4\text{x}}}\Big)\text{x}$
$=\int\frac{\cos(2\text{x})}{\sqrt{2\cos^2(2\text{x})}}\text{dx}$ $\Big[\therefore1+\cos\text{A}=2\cos^2\Big(\frac{\text{A}}{2}\Big)\text{ and }\cos^2\text{A}-\sin^2\text{A}=\cos\text{2A}\Big]$
$=\frac{1}{\sqrt2}\int\Big(\frac{\cos2\text{x}}{\cos\text{2x}}\Big)\text{dx}$
$=\frac{1}{\sqrt{2}}[\text{x}]+\text{C}$
$=\frac{\text{x}}{\sqrt{2}}+\text{C}$
View full question & answer→Question 1123 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
Here, $\text{f(x)}=\log\sin\text{x}$ Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{f}'\text{(x)}=\cot\text{x}$
let $\text{e}^{\text{x}}\log\sin\text{x = t}$
Diff. both sides w.r.t x
$\text{e}^{\text{x}}\log(\sin\text{x})+\text{e}^{\text{x}}\times\frac{1}{\sin\text{x}}\times\cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\big[\text{e}^{\text{x}}\log(\sin\text{x})+\text{e}^{\text{x}}\cot\text{x}\big]\text{dx = dt}$
$\Rightarrow\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx = dt}$
$\therefore \int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx} =\int\text{dt}$
$=\text{t + C}$
$=\text{e}^{\text{x}}\log\sin\text{x}+\text{C}$
View full question & answer→Question 1133 Marks
Evaluate the following integrals:
$\int\tan^32\text{x}\sec2\text{x dx}$
Answer$\int\tan^32\text{x}\sec2\text{x}=\tan^22\text{x}\tan2\text{x}\sec2\text{x}$
$=\big(\sec^22\text{x}-1\big)\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 dx}-\int\tan2\text{x}\sec2\text{x dx}$
$=\int\sec^22\text{x}\tan2\text{x}\sec2\text{x dx}-\frac{\sec2\text{x}}{2}+\text{C}$
Let $2\text{x}=\text{t}$
$\therefore\ 2\sec2\text{x}\tan2\text{x dx}=\text{dt}$
$\therefore\ \int\tan^32\text{x}\sec2\text{x dx}=\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 1143 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}}\Big(\text{x}^3-\frac{2}{\text{x}}\Big)\text{dx}$
Answer $\int\sqrt{\text{x}}\Big(\text{x}^3-\frac{2}{\text{x}}\Big)\text{dx}$
$=\int\Big(\text{x}^{\frac{7}{2}}-\frac{2}{\sqrt{\text{x}}}\Big)\text{dx}$
$=\int\Big(\text{x}^{\frac{7}{2}}-{2}{\text{x}^{-\frac{1}{2}}}\Big)\text{dx}$
$=\frac{\text{x}^{\frac{7}{2}+1}}{\frac{7}{2}+1}-2\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=\frac{2}{9}\text{x}^{\frac{9}{2}}-4\text{x}^{\frac{1}{2}}+\text{C}$
$=\frac{2}{9}\text{x}^{\frac{9}{2}}-4\sqrt{\text{x}}+\text{C}$
View full question & answer→Question 1153 Marks
Evaluate the following integrals:
$\int \frac{1}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
Answer Let $\text{I}=\int \frac{1}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
Let $\text{x}+2=\text{t}^2$
$\therefore\ \text{I}=\int\frac{2\text{tdt}}{(\text{t}^2-3)\text{t}}$
$=2\int\frac{\text{dt}}{\text{t}^2-3}$
$=\frac{2}{\sqrt{3}}\log\Big|\frac{\text{t}-\sqrt{3}}{\text{t}+\sqrt{3}}\Big|+\text{C}$
Thus, $\text{I}=\frac{1}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}-2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}$
View full question & answer→Question 1163 Marks
Evaluate the following integrals:
$\int\text{x}^2\tan^{-1}\text{x dx} $
AnswerLet $\text{I}=\int\text{x}^2\tan^{-1}\text{x dx}$
$=\tan^{-1}\text{x}\int\text{x}^2\text{dx}-\int\Big(\frac{1}{1+\text{x}^2}\int\text{x}^2\text{dx}\Big)$
$=\tan^{-1}\text{x}\Big(\frac{\text{x}^3}{3}\Big)-\frac{1}{3}\frac{\text{x}^3}{1+\text{x}^2}\text{dx}$
$=\frac{1}3{\text{x}^3}\tan^{-1}\text{x}-\frac{1}{3}\int\Big(\text{x}-\frac{\text{x}}{1+\text{x}^2}\Big)\text{dx}$
$=\frac{1}{3}\text{x}^3\tan^{-1}\text{x}-\frac{1}{3}\times\frac{\text{x}^2}{2}+\frac{1}3{}\int\frac{\text{x}}{1+\text{x}^2}\text{dx}$
$\text{I}=\frac{1}{3}\text{x}^3\tan^{-1}\text{x}-\frac{1}{6}\text{x}^2+\frac{1}{6}\log\big|1+\text{x}^2\big|+\text{C}$
View full question & answer→Question 1173 Marks
Evaluate the following integrals:$\int\frac{\sin8\text{x}}{\sqrt{9+\sin^44\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin8\text{x}}{\sqrt{9+(\sin4\text{x})^4}}\text{ dx}$
Let $\sin^24\text{x}=\text{t}$
$\Rightarrow2\sin4\text{x}.\cos4\text{x}(4)\text{dx}=\text{dt}$
$\Rightarrow4\sin8\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\sin8\text{x}\text{ dx}=\frac{\text{dt}}{4}$
$\text{I}=\frac{1}{4}\int\frac{\text{dt}}{\sqrt{(3)^2+\text{t}^2}}$
$\text{I}=\frac{1}{4}\log\Big|\text{t}+\sqrt{(3)^2+\text{t}^2}\Big|+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\log\Big|\text{x}+\sqrt{\text{a}^2+\text{x}^2}\Big|+\text{C}\Big]$
$\text{I}=\frac{1}{4}\log\Big|\sin^24\text{x}+\sqrt{9+\sin^44\text{x}}\Big|+\text{C}$
View full question & answer→Question 1183 Marks
Evalute the following integrals:
$\int\frac{1}{\text{x}\log\text{x}\log(\log\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.
Let $\text{I}=\int\frac{1}{\text{x}\log\text{x}\log(\log\text{x})}\text{dx}$
Putting $\log(\log\text{x})=\text{t}$
$\Rightarrow\frac{1}{\text{x}\log\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}\log\text{x}}\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\log\text{x})|+\text{C}\ \big[\because\text{t}=\log(\log\text{x})\big]$
View full question & answer→Question 1193 Marks
Evaluate the following integrals:
$\int\frac{\sin2\text{x}}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
Answer$\int\frac{\sin2\text{x}}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
$\text{Let a}+\text{b}\cos2\text{x}=\text{t}$
$\Rightarrow-\text{b}\sin(2\text{x})\text{dx}\times2=\text{dt}$
$\Rightarrow\sin(2\text{x})\text{dx}=\frac{-\text{dt}}{2\text{b}}$
$\text{Now,}\int\frac{\sin(2\text{x})}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
$=-\frac{1}{2\text{b}}\int\frac{\text{dt}}{\text{t}^2}$
$=\frac{-1}{2\text{b}}\int\text{t}^{-2}\text{dt}$
$=\frac{-1}{2\text{b}}\Big[\frac{\text{t}^{-2+1}}{-2+1}\Big]+\text{C}$
$=\frac{1}{2\text{b}}\times\frac{1}{\text{t}}+\text{C}$
$=\frac{1}{2\text{b}(\text{a}+\text{b}\cos2\text{x})}+\text{C}$
View full question & answer→Question 1203 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\big[\sec\text{x}+\log(\sec\text{x}+\tan\text{x})\big]\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\big[\sec\text{x}+\log(\sec\text{x}+\tan\text{x})\big]\text{dx}$
$=\int\text{e}^{\text{x}}\sec\text{x dx}+\int\text{e}^{\text{x}}\log(\sec\text{x}+\tan\text{x})\text{dx}$
Integrating by parts
$=\int\text{e}^{\text{x}}\sec\text{x dx}+\text{e}^{\text{x}}\log(\sec\text{x}\tan\text{x})-\int\text{e}^{\text{dx}}\Big\{\frac{\text{d}}{\text{dx}}\log(\sec\text{x}+\tan\text{x})\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}\sec\text{x dx}+\text{e}^{\text{x}}\log(\sec\text{x}+\tan\text{x})-\int\text{e}^{\text{x}}\sec\text{x dx}$
$=\text{e}^{\text{x}}\log(\sec\text{x}+\tan\text{x})+\text{C}$
View full question & answer→Question 1213 Marks
Evaluate the following integrals:
$\int\text{x}\frac{\tan^{-1}\text{x}^2}{1+\text{x}^4}\text{ dx}$
Answer Let $\text{I}=\int\text{x}\frac{\tan^{-1}\text{x}^2}{1+\text{x}^4}\text{ dx}\ ....(1)$ Let $\tan^{-1}\text{x}^2=\text{t}$ then, $\text{d}\big(\tan^{-1}\text{x}^2\big)=\text{dt}$ $\Rightarrow\frac{1\times2\text{x}}{1+(\text{x}^2)^2}\text{ dx}=\text{dt}$ $\Rightarrow\frac{1\times\text{x}}{1+\text{x}^4}\text{ dx}=\frac{\text{dt}}{2}$ Putting, $\tan^{-1}\text{x}^2=\text{t}$ and $\frac{\text{x}}{1+\text{x}^4}\text{ dx}=\frac{\text{dt}}{2}$ in equation (1),we get,
$\text{I}=\int\text{t}\frac{\text{dx}}{2}$
$=\frac{1}{2}\int\text{t dt}$
$=\frac{1}{2}\times\frac{\text{t}^2}{2}+\text{C}$
$\text{I}=\frac{\text{t}^2}{4}+\text{C}$
$=\frac{(\tan^{-1}\text{x}^2)^2}{4}+\text{C}$
$\text{I}=\frac{1}{4}\big(\tan^{-1}\text{x}^2\big)^2+\text{C}$
View full question & answer→Question 1223 Marks
$\int\frac{2\text{x}+3}{(\text{x}-1)^2}\text{dx}$
Answer$\int\bigg(\frac{2\text{x}+3}{(\text{x}-1)^2}\bigg)\text{dx}$
$=\int\bigg[\frac{2\text{x}-2+2+3}{(\text{x}-1)^2}\bigg]\text{dx}$
$=\int\bigg[\frac{2(\text{x}-1)+5}{(\text{x}-1)^2}\bigg]\text{dx}$
$=2\int\frac{\text{dx}}{(\text{x}-1)}+5\int(\text{x}-1)^{-2}\text{dx}$
$=2\text{ln}|\text{x}-1|+5\bigg[\frac{(\text{x}-1)^{-2+1}}{-2+1}\bigg]+\text{C}$
$=2\text{ln}|\text{x}-1|-\frac{5}{\text{x}-1}+\text{c}$
View full question & answer→Question 1233 Marks
Evaluate the following integrals:
$\int\text{x}\cos^2\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\cos^2\text{x dx}$
Using integration by parts,
$\text{I}=\text{x}\int\cos^2\text{x dx}-\int(1\int\cos^2\text{x dx})\text{dx}$
$=\text{x}\int\Big(\frac{\cos2\text{x}+1}{2}\Big)\text{dx}-\int\bigg(\int\Big(\frac{1+\cos2\text{x}}{2}\Big)\text{dx}\bigg)\text{dx}$
$=\frac{\text{x}}{2}\Big[\frac{\sin2\text{x}}{2}+\text{x}\Big]-\frac{1}{2}\int\Big(\text{x}+\frac{\sin2\text{x}}{2}\Big)\text{dx}$
$=\frac{\text{x}}{4}\sin2\text{x}+\frac{\text{x}^2}{2}-\frac{1}{2}\times\frac{\text{x}^2}{2}-\frac{1}{4}\Big(-\frac{\cos2\text{x}}{2}\Big)+\text{C}$
$\text{I}=\frac{\text{x}}{4}\sin2\text{x}+\frac{\text{x}^2}{4}+\frac{1}{8}\cos2\text{x}+\text{C}$
View full question & answer→Question 1243 Marks
Evaluate the following integrals:
$\int\frac{(1+\sqrt{\text{x}})^2}{\sqrt{\text{x}}}\text{dx}$
Answer$\int\frac{(1+\sqrt{\text{x}})^2}{\sqrt{\text{x}}}\text{dx}$
$=\int\frac{1+\text{x}+2\sqrt{\text{x}}}{\text{x}^{\frac{1}{2}}}\text{dx}$
$\int\text{x}^{\frac{-1}{2}}+\int\text{x}^{\frac{1}{2}}\text{dx}+2\int\text{dx}$
$=2\sqrt{\text{x}}+\frac{2}{3}\text{x}^{\frac{3}{2}}+2\text{x}+\text{C}$
$\therefore\ \int\frac{(1+\sqrt{\text{x}})^2}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\frac{2}{3}\text{x}^{\frac{3}{2}}+2\text{x}+\text{C}$
View full question & answer→Question 1253 Marks
Evaluate the following integrals:$\int(\text{x}+1)\log\text{x dx}$
Answer $\int(\text{x}+1).\log\text{x dx}$
$=\log \text{x}\int(\text{x}+1)\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\log\text{x})\int(\text{x}+1)\text{dx}\Big\}\text{dx}$
$=\log\text{x}\Big[\frac{\text{x}^2}{2}+\text{x}\Big]-\int\frac{1}{\text{x}}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)-\int\big(\frac{\text{x}}{2}+1\big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)-\Big(\frac{\text{x}^2}{4}+\text{x}\Big)+\text{C}$
View full question & answer→Question 1263 Marks
Evaluate the following integrals:
$\int\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
AnswerLet first function be $\sin^{-1}\text{x}$ and second dunction be $\frac{\text{x}}{\sqrt{1-\text{x}^2}}.$
First we find the intergral of the second function, i.e, $\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}.$
Put $\text{t}=1-\text{x}^{2}.$ Then $\text{dt}=-2\text{x dx}$
Therefore, $\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}=-\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}=-\sqrt{\text{t}}=-\sqrt{1-\text{x}^2}$
Hence, $\int\frac{\text{x}\sin^{-1}}{\sqrt{1-\text{x}^2}}\text{dx}=(\sin^{-1}\text{x})\Big(-\sqrt{1-\text{x}^2}\Big)-\int\frac{1}{\sqrt{1-\text{x}^2}}\Big(-\sqrt{1-\text{x}^2}\Big)\text{dx}$
$=-\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+\text{x + C}=\text{x}-\sqrt{1-\text{x}^2}\sin^{-1}\text{x+C}$
View full question & answer→Question 1273 Marks
Evaluate the following integrals:
$\int \frac{\text{1}}{\sqrt{\text{x}} + \text{x}} \text{ dx}$
Answer$\int \frac{\text{dx}}{\sqrt{\text{x}} + \text{x}}\text{dx}$
$=\int \frac{\text{dx}}{\sqrt{\text{x}}\big(1 + \sqrt{\text{x}}\big)}$
Let $1 + \sqrt{\text{x}} = \text{t}$
$\Rightarrow \frac{1}{2\sqrt{\text{x}}} = \frac{\text{dt}}{\text{dx}}$
$\Rightarrow \frac{\text{dx}}{\sqrt{\text{x}}} = 2\text{dt}$
Now, $\int \frac{\text{dx}}{\sqrt{\text{x}}\big(1 + \sqrt{\text{x}}\big)}$
$= \int \frac{2\text{dt}}{\text{t}}$
$= 2\int\frac{\text{dt}}{\text{t}}$
$= 2\log|\text{t}| + \text{C}$
$= 2\log \big(1 + \sqrt{\text{x}}\big) + \text{C}$
View full question & answer→Question 1283 Marks
Write a value of $\int\tan^3\text{x}\sec^2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\tan^3\text{x}\sec^2\text{x}\text{ dx}$ Let $\tan\text{x}=\text{t}$ $\sec^2\text{x dx}=\text{dt}$ $\therefore\ \text{I}=\int\text{t}^3\text{ dt}$ $=\frac{\text{t}^4}{4}+\text{C}$$=\frac{\tan^{4}\text{x}}{4}+\text{C}$ $(\because\text{t}=\tan\text{x})$
View full question & answer→Question 1293 Marks
Evaluate the following integrals:
$\frac{\text{x}+1}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\frac{\text{x}+1}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{(\text{x}+1)(1+\text{xe}^\text{x}-\text{xe}^\text{x})}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{\int(\text{x}+1)(1+\text{xe}^\text{x})}{\text(1+\text{xe}^\text{x})}\ \text{dx}-\int\frac{(\text{x}+1)(\text{xe}^\text{x})}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{(\text{x}+1)}{\text{x}}\ \text{dx}-\int\frac{\text{e}^\text{x}(\text{x}+1)}{1+\text{xe}^\text{x}}\ \text{dx}$
$=\int\frac{(\text{x}+1)\text{e}^\text{x}}{\text{xe}^\text{x}}\ \text{dx}-\int\frac{\text{e}^\text{x}(\text{x}+1)}{1+\text{xe}^\text{x}}\ \text{dx}$
$=\log|\text{xe}^\text{x}|-\log|1+\text{xe}^\text{x}|+\text{C}$
$\therefore\text{I}=\log\Big|\frac{\text{xe}^\text{x}}{1+\text{xe}^\text{x}}\Big|+\text{C}$
View full question & answer→Question 1303 Marks
Evaluate the following integrals:
$\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$
Answer$\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$ Let $\text{x}+\tan^{-1}\text{x}=\text{t}$ $\Big(1+\frac{1}{1+\text{x}^2}\Big)=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\Big(\frac{\text{x}^2-1+1}{\text{x}^2+1}\Big)\text{dx}=\text{dt}$ $\Rightarrow\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}=\text{dt}$Now, $\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$
$=\int5^\text{t}\text{dt}$
$=\frac{5^\text{t}}{\log5}+\text{C}$
$=\frac{5^{\text{x}+\tan^{-1}\text{x}}}{\log5}+\text{C}$
View full question & answer→Question 1313 Marks
Write a value of $\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{\sin^2\text{x}+\cos^2\text{x}+\sin2\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$=\int\text{dx}$
$\therefore\ \text{I}=\text{x}+\text{C}$
View full question & answer→Question 1323 Marks
Write a value of $\int\log_\text{e}\text{x}\text{ dx}$
Answer$\int\log_\text{e}\text{x}\text{ dx}$
$=\int1\cdot\log_\text{e}\text{x dx}$
$=\log_\text{e}\text{x}\int1\text{ dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\log_\text{e}\text{x})\int1\text{dx}\Big\}\text{dx}$
$=\log_\text{e}\text{x}\int1\cdot\text{dx}-\int\frac{1}{\text{x}}\cdot\text{x dx}$
$=\log_\text{e}\text{x}\cdot\text{x}-\int\text{dx}$
$=\text{x}\log_\text{e}\text{x}-\text{x}+\text{C}$
$=\text{x}(\log_\text{e}\text{x}-1)+\text{C}$
View full question & answer→Question 1333 Marks
Evaluate the following intregals:
$\int\frac{1}{\cos\text{x}(\sin\text{x}+2\cos\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos\text{x}(\sin\text{x}+2\cos\text{x})}\ \text{dx}$
$=\int\frac{1}{\sin\text{x}\cos\text{x}+2\cos^2\text{x}}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\ \text{dx}$
Let $2+\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$\text{I}=\log|2+\tan\text{x}|+\text{C}$
View full question & answer→Question 1343 Marks
Evaluate the following integrals:
$\int\cos^7\text{x}\text{ dx}$
Answer$\int\cos^7\text{x}\text{ dx}$
$=\int\cos^6\text{x}\cdot\cos\text{x}\text{ dx}$
$=\int(\cos^2\text{x})^3\cos\text{x}\text{ dx}$
$=\int(1-\sin^2\text{x})^3\cdot\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int(1-\sin^2\text{x})^3\cdot\cos\text{x}\text{ dx}$
$=\int(1-\text{t}^2)^3\text{dt}$
$=\int\big(1-\text{t}^6-3\text{t}^2+3\text{t}^4)\text{dt}$
$=\Big[\text{t}-\frac{\text{t}^7}{\text{7}}-\frac{3\text{t}^3}{3}+\frac{3\text{t}^5}{5}\Big]+\text{C}$
$=\sin\text{x}-\frac{1}{7}\sin^7\text{x}-\sin^3\text{x}+\frac{3}{5}\sin^5\text{x}+\text{C}$
View full question & answer→Question 1353 Marks
Evaluate the following integrals:
$\int (3\text{x}\sqrt{\text{x}}+4\sqrt{\text{x}}+5)\text{dx}$
Answer$\int(3\text{x}\sqrt{5}+4\sqrt{\text{x}}+5)\text{dx}$
$=\int3\text{x}\sqrt{5}\text{dx}+\int4\sqrt{\text{x}}\text{dx}+\int5\text{dx}$
$=\int3\text{x}^{\frac{3}{2}}\text{dx}+4\int\text{x}^{\frac{1}{2}}\text{dx}+5\int\text{dx}$
$=\frac{\text{x}\frac{3}{2}+1}{\frac{3}{2}+1}+\frac{4\text{x}^{\frac{1}{2}}}{\frac{1}{2}+1}+5\text{x}+\text{C}$
$=\frac{6}{5}\text{x}^{\frac{5}{2}}+\frac{8}{3}\text{x}^{\frac{3}{2}}+5\text{x}+\text{C}$
View full question & answer→Question 1363 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{(\text{x}^2+4)\sqrt{\text{x}^2+9}}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}}{(\text{x}^2+4)\sqrt{\text{x}^2+9}}\text{ dx}$
Let $\text{x}^2+9=\text{u}^2$
$2\text{xdx}=2\text{udu}$
$\therefore\ \text{I}=\int\frac{\text{u}}{(\text{u}^2-5)\text{u}}\text{ du}$
$=\int\frac{\text{du}}{\text{u}^2-5}$
$=\frac{1}{2\sqrt{5}}\log\Big(\frac{\text{u}-\sqrt{5}}{\text{u}+\sqrt{5}}\Big)+\text{C}$
$=\frac{1}{2\sqrt{5}}\log\bigg(\frac{\sqrt{\text{x}^2+9}-\sqrt{5}}{\sqrt{\text{x}^2+9}+\sqrt{5}}\bigg)+\text{C}$
View full question & answer→Question 1373 Marks
Evaluate the following integrals:$\int\text{e}^{\text{}x}\Big(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{}x}\Big(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big(\frac{1}{1+\cos\text{x}}+\frac{\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\bigg(\frac{1}{2\cos\frac{\text{x}}{2}}+\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\text{e}^{\text{}x}\big(\frac{1}{2}\sec^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}\big)\text{dx}$
Putting $\text{e}^{\text{x}}\tan\frac{\text{x}}{2}=\text{t}$
Diff. both sides w.r.t.x
$\text{e}^{\text{x}}.\tan\big(\frac{\text{x}}{2}\big)+\text{e}^{\text{x}}\times\frac{1}{2}\sec^{2}\frac{\text{x}}{2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}\big[\tan\frac{\text{x}}{2}+\frac{1}{2}\sec^2\big(\frac{\text{x}}{2}\big)\big]\text{dx}=\text{dt}$
$\therefore\int\text{e}^{\text{x}}\big(\frac{1}{2}\sec^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}\big)\text{dx}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\tan\big(\frac{\text{x}}{2}\big)+\text{C}$
View full question & answer→Question 1383 Marks
Integrate the following integrals:
$\int\cos3\text{x}\cos4\text{x dx}$
AnswerLet I $=\int\cos3\text{x}\cos4\text{x dx}.$ Then,
$\text{I}=\frac{1}{2}\int(2\cos3\text{x}\cos4\text{x})\text{dx}$
$=\frac{1}{2}\int(\cos7\text{x}+\cos(-\text{x}))\text{dx}$
$=\frac{1}{2}\int\cos7\text{x}+\frac{1}{2}\int\cos\text{dx}$ $[\because\cos(-0)=\cos0]$
$=\frac{\sin7\text{x}}{2\times7}+\frac{\sin\text{x}}{2}+\text{C}$
$=\frac{1}{14}\times\sin7\text{x}+\frac{1}{2}\sin\text{x}+\text{C}$
$\therefore\text{I}=\frac{1}{14}\times\sin7\text{x}+\frac{1}{2}\times\sin\text{x}+\text{C}$
View full question & answer→Question 1393 Marks
Evaluate the following integrals:
$\int5^{5^{5^{\text{x}}}}5^{5^{\text{x}}}5^\text{x}\text{ dx}$
Answer$\int5^{5^{5^{\text{x}}}}5^{5^{\text{x}}}5^\text{x}\text{ dx}$
Let $5^\text{x}=\text{t}$
$\Rightarrow5^\text{x}\log5=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow5^\text{x}\text{dx}=\frac{\text{dt}}{\log5}$
Now, $\int5^{5^{5^{\text{x}}}}5^{5^{\text{x}}}5^\text{x}\text{ dx}$
$=5\int5^{5^{\text{t}}}.5^\text{t}.\frac{\text{dt}}{\log5}$
Again let $5^\text{t}=\text{p}$
$\Rightarrow5^\text{t}\log5=\frac{\text{dp}}{\text{dt}}$
$\Rightarrow5^\text{t}\text{dt}=\frac{\text{dp}}{\log5}$
Again $\int5^{5^\text{t}}.5^\text{x}.\frac{\text{dt}}{\log5}$
$=\int5^\text{p}.\frac{\text{dp}}{(\log5)^2}$
$=\frac{5^\text{p}}{(\log5)^3}+\text{C}$
$=\frac{5^{5^{5^{\text{x}}}}}{(\log5)^3}+\text{C}$
View full question & answer→Question 1403 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}^3+8)(\text{x}-1)}{\text{x}^2-2\text{x}+4}\text{dx}$
Answer$\int\frac{(\text{x}^3+8)(\text{x}-1)}{(\text{x}^2-2\text{x}+4)}\text{dx}$
$=\int\frac{(\text{x}^3+2^3)(\text{x}-1)}{(\text{x}^2-2\text{x}+4)}\text{dx}$
$=\int\frac{(\text{x}+2)(\text{x}^2-2\text{x}+4)(\text{x}-1)}{(\text{x}^2-2\text{x}+4)}\text{dx}$ $\big[\therefore\ \text{a}^3+\text{b}^3=(\text{a + b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=\int(\text{x}+2)(\text{x}-1)\text{dx}$
$=\int(\text{x}^2-\text{x}+2\text{x}-2)\text{dx}$
$=(\text{x}^2+\text{x}-2)\text{dx}$
$=\int\text{x}^2\text{dx}+\int\text{x dx}-2\int1\text{dx}$
$=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{2}-2\text{x}+\text{C}$
View full question & answer→Question 1413 Marks
Evaluate the following integrals:
$\int\sqrt{16\text{x}^2+25}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{16\text{x}^2+25}\text{dx}$
$=\int\sqrt{(4\text{x})^2+5^2}\text{dx}$
$=4\int\sqrt{\text{x}^2+\Big(\frac{5}{4}\Big)^2}\text{dx}$
$=4\begin{Bmatrix}\frac{\text{x}}{2}\sqrt{\text{x}^2+\Big(\frac{5}{4}\Big)^2}+\frac{\big(\frac{5}{4}\big)^2}{2}\log\Bigg|\text{x}+\sqrt{\text{x}^2+\Big(\frac{5}{4}\Big)^2}\Bigg|+\text{C}\end{Bmatrix}$
$\therefore\ \text{I}=2\text{x}\sqrt{\text{x}^2+\frac{25}{16}}+\frac{25}{8}\log\bigg|\text{x}+\sqrt{\text{x}^2+\frac{25}{16}}\bigg|+\text{C}$
View full question & answer→Question 1423 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{1+\cos\text{x}}}\text{dx}$
AnswerWe have,
$\int\frac{1}{\sqrt{1+\cos\text{x}}}\text{dx}$
$\int\frac{1}{\sqrt{2\cos^2\frac{\text{x}}{2}}}\text{dx}$
$=\int\frac{1}{\sqrt{2}\cos\frac{\text{x}}{2}}\text{dx}$
$=\frac{1}{\sqrt{2}}\int\sec\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{\sqrt{2}}\int\text{cosec}\Big(\frac{\pi}{2}+\frac{\text{x}}{2}\Big)\text{dx}$
$=\frac{2}{\sqrt{2}}\log\Big|\tan\Big(\frac{\pi}{4}+\frac{\text{x}}{4}\Big)\Big|+\text{C}$
$\because\int\frac{1}{\sqrt{1+\cos\text{x}}}\text{dx}=\sqrt{2}\log\Big|\tan\Big(\frac{\pi}{4}+\frac{\text{x}}{4}\Big)\Big|+\text{C}$
View full question & answer→Question 1433 Marks
Evaluate the following integrals:
$\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Answer$\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$ Let $\sqrt{\text{x}}=\text{t}$ $\Rightarrow\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\frac{\text{dx}}{\sqrt{\text{x}}}=2\text{dt}$ Now, $\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$$=2\int\sec^2\text{t dt}$
$=2\tan(\text{t})+\text{C}$
$=2\tan\big(\sqrt{\text{x}}\big)+\text{C}$
View full question & answer→Question 1443 Marks
Write a value of $\int\frac{\text{a}^{\text{x}}}{3+\text{a}^{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{a}^{\text{x}}}{3+\text{a}^{\text{x}}}\text{ dx}$
Let $3+\text{a}^{\text{x}}=\text{t}$
$\text{a}^{\text{x}}\log\text{a dx}=\text{dt}$
$\text{a}^{\text{x}}\text{dx}=\frac{\text{dt}}{\log\text{a}}$
$\text{I}=\int\frac{\text{dt}}{\log\text{a}\cdot\text{t}}=\frac{1}{\log\text{a}}\log\text{t}+\text{C}$
$\text{I}=\frac{1}{\log\text{a}}\log(3+\text{a}^{\text{x}})+\text{C}$
View full question & answer→Question 1453 Marks
Evalute the following integrals:
$\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$$=\int\frac{\cos^2\text{x}-\sin^2\text{x}}{(\cos\text{x}+\sin\text{x})^2}\text{dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}+\sin\text{x}}\text{dx}$
Putting $\cos\text{x}+\sin\text{x}=\text{t}$
$\Rightarrow-\sin\text{x}+\cos\text{x}=\frac{\text{dt}}{\text{dt}}$
$\Rightarrow(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}|\cos\text{x}+\sin\text{x}|+\text{C}\ \big[\because\text{t}=\cos\text{x}+\sin\text{x}\big]$
View full question & answer→Question 1463 Marks
Evaluate the following integrals:
$\int\big\{\sqrt{\text{x}}\big(\text{ax}^2+\text{bx}+\text{c}\big)\big\}\text{dx}$
Answer$\int\sqrt{\text{x}}\Big(\text{ax}^2+\text{bx}+\text{c}\Big)\text{dx}$
$=\int\text{x}^{\frac{1}{2}}\Big(\text{ax}^2+\text{bx}+\text{c}\Big)\text{dx}$
$=\int\Big(\text{ax}^{2+\frac{1}{2}}+\text{bx}^{\frac{1}{2}+1}+\text{cx}^{\frac{1}{2}}\Big)\text{dx}$
$=\text{a}\int\text{x}^{\frac{5}{2}}\text{dx}+\text{b}\int\text{x}^{\frac{3}{2}}\text{dx}+\text{c}\int\text{x}^{\frac{1}{2}}\text{dx}$
$=\text{a}\begin{bmatrix}\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}\end{bmatrix}+\text{b}\begin{bmatrix}\frac{\text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\end{bmatrix}+\text{c}\begin{bmatrix}\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\end{bmatrix}+\text{C}$
$=\frac{2\text{a}}{7}\text{x}^{\frac{7}{2}}+\frac{2\text{b}}{5}\text{x}^{\frac{3}{2}}+\frac{2\text{c}}{3}\text{x}^{\frac{3}{2}}+\text{C}$
View full question & answer→Question 1473 Marks
Evaluate the following integrals:
$\int\tan^5\text{x }\sec^4\text{x}\text{dx}$
Answer$\int\tan^5\text{x }\sec^4\text{x}\text{dx}$
$=\int\tan^5\text{x }\sec^2\text{x}.\sec^2\text{x}\text{dx}$
$=\int\tan^5\text{x}(1+\tan^2\text{x})\sec^2\text{xdx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{xdx}=\text{dt}$
Now, $\int\tan^5\text{x}(1+\tan^2\text{x})\sec^2\text{xdx}$
$=\int\text{t}^5(1+\text{t}^2)\text{dt}$
$=\int(\text{t}^5+\text{t}^7)\text{dt}$
$=\frac{\text{t}^6}{6}+\frac{\text{t}^8}{8}+\text{C}$
$=\frac{\tan^6\text{x}}{6}+\frac{\tan^8\text{x}}{8}+\text{C}$
View full question & answer→Question 1483 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}^2-2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\text{x}^2-2\text{x}}\text{dx}$
$\Rightarrow\text{I}=\int\sqrt{\text{x}^2-2\text{x}+1-1}\text{dx}$
$\Rightarrow\text{I}=\int\sqrt{(\text{x}-1)^2-1^2}\text{dx}$
$\because\ \int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2-\text{a}^2}-\frac{\text{a}^2}{2}\ln\big(|\text{x}+\sqrt{\text{x}^2-\text{a}^2}|\big)+\text{C}$
$\therefore\ \text{I}=\frac{(\text{x}-1)}{2}\sqrt{(\text{x}-1)^2-1}-\frac{1}{2}\ln\big|(\text{x}-1)+\sqrt{\text{x}^2-2\text{x}}\big|+\text{C}$
View full question & answer→Question 1493 Marks
Evaluate the following integrals:
$\int\text{e}^\text{x}\sqrt{\text{e}^{2\text{x}}+1}\text{dx}$
AnswerLet $\text{I}=\int\text{e}^\text{x}\sqrt{\text{e}^{2\text{x}}+1}\text{dx}$
Putting $\text{e}^\text{x}=\text{t}$
$\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{\text{t}}{2}\sqrt{\text{t}^2+1}+\frac{1^2}{2}\ln\Big|\text{t}+\sqrt{\text{t}^2+1}\Big|+\text{C}$
$\Big[\because\ \int\sqrt{\text{x}^2+\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2+\text{a}^2}+\frac{1}{2}\ln\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|+\text{C}\Big]$
$=\frac{\text{e}^\text{x}}{2}\sqrt{\text{e}^{2\text{x}}+1}+\frac{1}{2}\ln\Big|\text{e}^\text{x}+\sqrt{\text{e}^{2\text{x}}+1}\Big|+\text{C}\ \big(\because\ \text{t}=\text{e}^\text{x}\big)$
View full question & answer→Question 1503 Marks
Evaluate the following integrals:
$\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
Answer$\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
$\text{Let, }\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\text{Now,}\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
$=\int\text{t}^\frac{3}{2}\text{dt}$
$=\Bigg[\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]+\text{C}$
$=\frac{2}{5}\text{t}^\frac{5}{2}+\text{C}$
$=\frac{2}{5}\tan^\frac{5}{2}\text{x}+\text{C}$
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