Question 513 Marks
Evaluate the following integrals:$\int\text{x}^2\text{e}^{-\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\text{x}^2\text{e}^{-\text{x}}\text{dx}$
Using integration by parts,
$\text{I}=\text{x}^2\int\text{e}^{-\text{x}}\text{dx}-\int(2\text{x}\int\text{e}^{-\text{x}}\text{dx})$
$=-\text{x}^2\text{e}^{-\text{x}}-\int(2\text{x})(-\text{e}^{-\text{x}})$
$=-\text{x}^2\text{e}^{-\text{x}}+2\int\text{xe}^{-\text{x}}\text{dx}$
$=-\text{x}^2\text{e}^{-\text{x}}+2[\text{x}\int\text{e}^{-\text{x}}\text{dx}-\int(1\times\int\text{e}^{-\text{x}}\text{dx})\text{dx}]$
$=-\text{x}^2\text{e}^{-\text{x}}+2[\text{x}(-\text{e}^{-\text{x}})-\int(-\text{e}^{-\text{x}})\text{dx}]$
$=-\text{x}^2\text{e}^{-\text{x}}-2\text{xe}^{-\text{x}}+2\int\text{e}^{-\text{x}}\text{dx}$
$\text{I}=-\text{x}^2\text{e}^{-\text{x}}-2\text{xe}^{-\text{x}}-2\text{e}^{-\text{x}}+\text{C}$
$\text{I}=-\text{e}^{-\text{x}}(\text{x}^2+2\text{x}+2)+\text{C}$
View full question & answer→Question 523 Marks
Evaluate the following integrals:$\int\log_{10}\text{x dx}$
AnswerLet $\text{I}=\int\log_{10}\text{x dx}$
$=\int\frac{\log\text{x}}{\log10}\text{dx}$
$=\frac{1}{\log10}\int1\times\log\text{x dx}$
Using integration by parts,
$=\frac{1}{\log10}\Big[\log\text{x}\int\text{dx}-\int\Big(\frac{1}{\text{x}}\int\text{dx}\Big)\text{dx}\Big]$
$=\frac{1}{\log10}\Big[\text{x}\log\text{x}-\int\big(\frac{\text{x}}{\text{x}}\big)\text{dx}\Big]$
$=\frac{1}{\log10}[\text{x}\log\text{x}-\text{x}]$
$\text{I}=\frac{\text{x}}{\log10}(\log\text{x}-1)$
View full question & answer→Question 533 Marks
Evaluate the following integrals:$\int\text{x}^2\sin^2\text{x dx}$
AnswerLet $\text{I}=\int\text{x}^2\sin^2\text{x dx}$
$=\int\text{x}^2\Big(\frac{1-\cos2\text{x}}{2}\Big)\text{dx}$
$=\int\frac{\text{x}^2}{2}\text{dx}-\int\Big(\frac{\text{x}^2\cos2\text{x}}{2}\Big)\text{dx}$
$=\frac{\text{x}^3}{6}-\frac{1}{2}[\int\text{x}^2\cos2\text{x dx}]$
$=\frac{\text{x}^3}{6}-\frac{1}{2}[\text{x}^2\int\cos2\text{x dx}-\int(2\text{x}\int\cos2\text{x dx})\text{dx]}$
$=\frac{\text{x}^3}{6}-\frac{1}{2}\Big(\text{x}^2\frac{\sin2\text{x}}{2}\Big)+\frac{1}{2}\times2\int\Big(\text{x}\frac{\sin2\text{x}}{2}\Big)\text{dx}$
$=\frac{\text{x}^3}{6}-\frac{1}{4}\text{x}^2\sin2\text{x}+\frac{1}{2}[\text{x}\int\sin2\text{x dx}-\int(1\int\sin2\text{x dx})\text{dx}]$
$=\frac{\text{x}^3}{6}-\frac{1}{4}\text{x}^2\sin2\text{x}+\frac{1}{2}\Big[\text{x}\Big(-\frac{\cos2\text{x}}{2}\Big)-\int\Big(-\frac{\cos2\text{x}}{2}\Big)\text{dx}\Big]$
$=\frac{\text{x}^3}{6}-\frac{1}{4}\text{x}^2\sin2\text{x}-\frac{1}{4}\text{x}\cos2\text{x}+\frac{1}{4}\frac{\sin2\text{x}}{2}+\text{C}$
$\text{I}=\frac{\text{x}^3}{6}-\frac{1}{4}\text{x}^2\sin2\text{x}-\frac{1}{4}\text{x}\cos2\text{x}+\frac{1}{8}\sin2\text{x}+\text{C}$
View full question & answer→Question 543 Marks
$\int\frac{\text{x}+3}{(\text{x}+1)^4}\text{dx}$
Answer$\int\bigg[\frac{\text{x}+3}{(\text{x}+1)^4}\bigg]\text{dx}$
$=\int\bigg[\frac{\text{x}+1+2}{(\text{x}+1)^4}\bigg]\text{dx}$
$=\int\bigg[\frac{(\text{x}+1)}{(\text{x}+1)^4}+\frac{2}{(\text{x}+1)^4}\bigg]\text{dx}$
$=\int\frac{\text{dx}}{(\text{x}+1)^3}+2\int\frac{\text{dx}}{(\text{x}+1)^4}$
$=\int(\text{x}+1)^{-3}\text{dx}+2\int(\text{x}+1)^{-4}\text{dx}$
$=\bigg[\frac{(\text{x}+1)^{-3+1}}{-3+1}\bigg]+2\bigg[\frac{(\text{x}+1)^{-4+1}}{-4+1}\bigg]+\text{c}$
$=-\frac{1}{2}(\text{x}+1)^{-2}-\frac{2}{3}(\text{x}+1)^{-3}+\text{c}$
$=-\frac{1}{2(\text{x}+1)^2}-\frac{2}{3(\text{x}+1)^3}+\text{c}$
View full question & answer→Question 553 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{16-6\text{x}-\text{x}^2}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\sqrt{16-6\text{x}-\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{16-(\text{x}^2+6\text{x})}}$
$=\int\frac{\text{dx}}{\sqrt{16-(\text{x}^2+6\text{x}+3^2-3^2)}}$
$=\int\frac{\text{dx}}{\sqrt{16+9-(\text{x}+3)^2}}$
$=\int\frac{\text{dx}}{\sqrt{5^2-(\text{x}+3)^2}}$
$=\sin^{-1}\Big(\frac{\text{x}+3}{5}\Big)+\text{C}$
View full question & answer→Question 563 Marks
Evaluate the following integrals:$\int\frac{\text{x}+7}{3\text{x}^2+25\text{x}+28}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}+7}{3\text{x}^2+25\text{x}+28}\text{ dx}$ $=\int\frac{\text{x}+7}{3\text{x}^2+21\text{x}+4\text{x}+28}\text{ dx}$ $=\int\frac{\text{x}+7}{3\text{x}(\text{x}+7)+4(\text{x}+7)}\text{ dx}$ $=\int\frac{\text{x}+7}{(3\text{x}+4)(\text{x}+7)}\text{ dx}$ $=\int\frac{1}{(3\text{x}+4)}\text{ dx}$$=\frac{1}{3}\ln|3\text{x}+4|+\text{C}$
View full question & answer→Question 573 Marks
Evaluate the following integrals:
$\int\frac{1}{1-\cos\text{x}}\text{dx}$
Answer$\int\frac{1}{1-\cos\text{x}}\text{dx}$
$=\int\frac{1}{1-\cos\text{x}}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}\times\text{dx}$
$=\int\frac{1+\cos\text{x}}{1-\cos^2\text{x}}\times\text{dx}$
$=\int\frac{1+\cos\text{x}}{\sin^2\text{x}}\times\text{dx}$
$=\int\frac{1}{\sin^2\text{x}}\text{dx}+\int\frac{\cos\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\text{cosec}^2\text{x dx}+\int\cot\text{x}\times\text{cosec x dx}$
$=-\cot\text{x}-\text{cosec x}+\text{C}$
$\therefore\ \int\frac{1}{1-\cos\text{x}}\text{dx}=-\cot\text{x}-\text{cosec x}+\text{C}$
View full question & answer→Question 583 Marks
Evaluate the following integrals:
$\int\text{x}^2\sqrt{\text{a}^6-\text{x}^6}\text{dx}$
AnswerLet $\text{I}=\int\text{x}^2\sqrt{\text{a}^6-\text{x}^6}\text{dx}$
Let $\text{x}^3=\text{t}$
$\Rightarrow3\text{x}^2\text{dx}=\text{dt}$
$\therefore\ \text{I}=\frac{1}{3}\int\sqrt{\text{a}^6-\text{t}^2}\text{dt}$
$=\frac{1}{3}\begin{Bmatrix}\frac{\text{t}}{2}\sqrt{\text{a}^6-\text{t}^2}+\frac{\text{a}^6}{2}\sin^{-1}\Big(\frac{\text{t}}{\text{a}^3}\Big)\end{Bmatrix}+\text{C}$
$\therefore\ \text{I}=\frac{\text{x}^3}{6}\sqrt{\text{a}^6-\text{x}^6}+\frac{\text{a}^6}{6}\sin^{-1}\Big(\frac{\text{x}^3}{\text{a}^3}\Big)+\text{C}$
View full question & answer→Question 593 Marks
$\int\frac{1}{(7\text{x}-5)^2}+\frac{1}{\sqrt{5\text{x}-4}}\text{dx}$
AnswerLet $\text{I}=\int\bigg[\frac{1}{(7\text{x}-5)^3}+\frac{1}{\sqrt{5\text{x}-4}}\bigg]\text{dx}.$ Then,
$\text{I}=\int(7\text{x}-5)^{-3}\text{dx}+\int(5\text{x}-4)^{\frac{-1}{2}}\text{dx}$
$=\frac{(7\text{x}-5)^{-2}}{7\times(-2)}+\frac{(5\text{x}-4)^{\frac{1}{2}}}{5\times\frac{1}{2}}+\text{c}$
$=-\frac{(7\text{x}-5)^{6-2}}{14}+\frac{2}{5}\sqrt{(5\text{x}-4)}+\text{c}$
$\text{I}=\frac{-1}{14}(7\text{x}-5)^{-2}+\frac{2}{5}\times\sqrt{5\text{x}-4}+\text{c}.$
View full question & answer→Question 603 Marks
Evaluate the following integrals:
$\int\Big(\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\tan^{-1}\text{x}+\frac{1}{1+\text{x}^2}\Big)\text{dx}$
Here, $\text{f(x)}=\tan^{-1}\text{x}$ and $\text{f}'\text{(x)}=\frac{1}{1+\text{x}^2}$
and we know thet,
$\int\text{e}^{\text{ax}}(\text{af}(\text{x})+\text{f}'(\text{x}))\text{dx}=\text{e}^{\text{ax}}\text{f(x)+C}$
$\therefore\int\text{e}^{\text{x}}\Big(\tan^{-1}\text{x}+\frac{1}{1+\text{x}^2}\Big)\text{dx}=\text{e}^{\text{x}}\tan^{-1}\text{x + C}$
Thus,
$\text{I}=\text{e}^{\text{x}}\tan^{-1}\text{x + C}$
View full question & answer→Question 613 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{\tan^{-1}\text{x}}.(1+\text{x}^2)}\text{dx}$
Answer$\int\frac{\text{dx}}{\sqrt{\tan^{-1}\text{x}}(1+\text{x}^2)}$
$\text{Let }\tan^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{1+\text{x}^2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
$\text{Now,}\int\frac{\text{dx}}{\sqrt{\tan^{-1}\text{x}}(1+\text{x}^2)}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\tan^{-1}\text{x}}+\text{C}$
View full question & answer→Question 623 Marks
Evaluate the following integrals:
$\int\cot^3\text{x }\text{cosec}^2\text{x}\text{ dx}$
Answer$\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}$
$\text{Let},\cot\text{x}=\text{t}$
$\Rightarrow-\text{cosec}^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{cosec}^2\text{x}\text{ dx}=-\text{dt}$
$\text{Now},\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}$
$=\int\text{t}^3(-\text{dt})$
$=\frac{-\text{t}^4}{4}+\text{C}$
$=\frac{-\cot^4\text{x}}{4}+\text{C}$
View full question & answer→Question 633 Marks
Evaluate the following integrals:
$\int\sqrt{3+2\text{x}-\text{x}^2}\text{dx}$
Answer$\int\sqrt{3+2\text{x}-\text{x}^2}\text{dx}=\int\sqrt{4-(\text{x}-1)^2}\text{dx}$ Let X - 1 = t, so that dx = dt Thus,$\int\sqrt{3+2\text{x}-\text{x}^2}\text{dx}=\int\sqrt{4-\text{t}^2}\text{dt}$ $=\frac{1}{2}\text{t}\sqrt{4-\text{t}^2}+\frac{4}{2}\sin^{-1}\Big(\frac{\text{t}}{2}\Big)+\text{C}$$=\frac{1}{2}(\text{x}-1)\sqrt{3+2\text{x}-\text{x}^2}+2\sin^{-1}\Big(\frac{\text{x}-1}{2}\Big)+\text{C}$
View full question & answer→Question 643 Marks
Evaluate the following integrals:
$\int2\text{x}\sec^3\big(\text{x}^2+3\big)\tan\big(\text{x}^2+3\big)\text{dx}$
Answer $\int2\text{x}\sec^3\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)\text{dx}$ $=\int\sec^2\big(\text{x}^2+3\big).\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)2\text{x}\text{ dx}$ Let $\sec\big(\text{x}^2+3\big)=\text{t}$ $\Rightarrow\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big).2\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big).2\text{x}\text{ dx}=\text{dt}$Now, $\int\sec^2\big(\text{x}^2+3\big).\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)2\text{x}\text{ dx}$
$=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^2}{3}+\text{C}$
$=\frac{\sec^2(\text{x}^2+3)}{3}+\text{C}$
View full question & answer→Question 653 Marks
Evaluate the following integrals:
$\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
Answer$\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
$\text{Let},\text{x}-\cos\text{x}=\text{t}$
$\Rightarrow(1+\sin\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\sin\text{x})\text{dx}=\text{dt}$
$\text{Now,}\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\text{x}-\cos\text{x}}+\text{C}$
View full question & answer→Question 663 Marks
Evaluate the following integrals:
$\int\frac{\log\text{x}^2}{\text{x}}\text{dx}$
Answer$\int\frac{\log\text{x}^2\text{dx}}{\text{x}}$
$=\int\frac{2\log\text{x}}{\text{x}}\text{dx}$
$=2\int\frac{\log\text{x}}{\text{x}}\text{dx}$
$\text{Let }\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{Now, }2\int\frac{\log\text{x}}{\text{x}}\text{dx}$
$=2\int\text{t dt}$
$2\Big[\frac{\text{t}^2}{2}\Big]+\text{C}$
$=\text{t}^2+\text{C}$
$=(\log\text{x})^2+\text{C}$
View full question & answer→Question 673 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 683 Marks
Write a value of $\int\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})\text{dx}$
$\because\ \int\text{e}^{\text{x}}\big(\text{f(x})+\text{f}'(\text{x})\big)\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}$
Here, $\text{f(x)}=\sin\text{x}$ and $\text{f}'(\text{x})=\cos\text{x}$
$\therefore\ \text{I}=\int\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})\text{dx}=\text{e}^{\text{x}}\sin\text{x}+\text{C}$
View full question & answer→Question 693 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\Big(\frac{\sin\text{x}\cos\text{x}-1}{\sin^2\text{x}}\Big)\text{dx}$
Answer Let $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{\sin\text{x}\cos\text{x}-1}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\big[\cot\text{x}-\text{cosec}^2\text{x}\big]\text{dx}$
Here, $\text{f(x)}=\cot\text{x}$
$\Rightarrow\text{f}'\text{(x)}=-\text{cosec}^2\text{x}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\cot\text{x}=\text{t}$
Diff both sides w.r.t x
$\text{e}^{\text{x}}(\cot\text{x}-\text{cosec}^2\text{x})\text{dx = dt}$
$\therefore\text{I}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\cot\text{x + C}$
View full question & answer→Question 703 Marks
$\int(\text{e}^{\text{x}}+\frac{1}{\text{e}^{\text{x}}})^2\text{dx}$
Answer$\int(\text{e}^{\text{x}}+\frac{1}{\text{e}^{\text{x}}})^2\text{dx}$
$=\int(\text{e}^{2\text{x}}+\frac{1}{\text{e}^{2\text{x}}}2\text{e}^\text{x}\times\frac{1}{\text{e}^{\text{x}}})\text{dx}$
$=\int(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}+2)\text{dx}$
$=\frac{\text{e}^{2\text{x}}}{2}+\frac{\text{e}^{-2\text{x}}}{-2}+2\text{x}+\text{c}$
$=\frac{\text{e}^{2\text{x}}}{2}-\frac{\text{e}^{-2\text{x}}}{2}+2\text{x}+\text{c}$
View full question & answer→Question 713 Marks
Evalute the following integrals:
$\int\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}\text{dx}$
Answer $\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}=\frac{2\cos\text{x}-3\sin\text{x}}{2(3\cos\text{x}+2\sin\text{x})}$
Let $3\cos\text{x}+2\sin\text{x}=\text{t}$
$(-3\sin\text{x}+2\cos\text{x})\text{dx}=\text{dt}$
$\int\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}\text{dx}=\int\frac{\text{dt}}{2\text{t}}$
$=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{2}\log|\text{t}|+\text{C}$
$=\frac{1}{2}\log|2\sin\text{x}+3\cos\text{x}|=\text{C}$
View full question & answer→Question 723 Marks
Evaluate the following integrals:$\int\frac{\text{x}}{\sqrt{4-\text{x}^4}}\text{ dx}$
Answer$\int\frac{\text{x}\text{ dx}}{\sqrt{4-\text{x}^4}}$
$\Rightarrow\int\frac{\text{x}\text{ dx}}{\sqrt{2^2-(\text{x}^2)^2}}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x}\text{ dx}=\text{dt}$
$\text{x}\text{ dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\text{x}\text{ dx}}{\sqrt{2^2-(\text{x}^2)^2}}$
$\frac{1}{2}\int\frac{\text{dt}}{\sqrt{2^2-\text{t}^2}}$
$=\frac{1}{2}\times\sin^{-1}\Big(\frac{1}{2}\Big)+\text{C}$
$=\frac{1}{2}\times\sin^{-1}\Big(\frac{\text{x}^2}{2}\Big)+\text{C}$
View full question & answer→Question 733 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+\text{x}+1}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+\text{x}+1}}\text{ dx}$
Let $\text{x}+1=\frac{1}{\text{t}}$
$\text{dt}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=-\int\frac{\frac{1}{\text{t}^2}\text{ dt}}{\frac{1}{\text{t}}\sqrt{\Big(\frac{1}{\text{t}^2}+\frac{1}{\text{t}}-1\Big)}}$
$=-\int\frac{\text{dt}}{\sqrt{1+\text{t}-\text{t}^2}}$
$=-\int\frac{\text{dt}}{\sqrt{\frac{5}{4}-\big(\frac{1}{4}-\text{t}+\text{t}^2\big)}}$
$=-\int\frac{\text{dt}}{\sqrt{\frac{5}{4}-\big(\text{t}-\frac{1}{2}\big)^2}}$
$=-\sin^{-1}\Bigg(\frac{\text{t}-\frac{1}{2}}{\frac{\sqrt{5}}{2}}\Bigg)+\text{C}$
$\therefore\ \text{I}=-\sin^{-1}\Big(\frac{2\text{t}-1}{\sqrt{5}}\Big)+\text{C}$ $\Big[\text{When}\text{t}=\frac{1}{\text{x}+1}\Big]$
View full question & answer→Question 743 Marks
Evaluate the following integrals:
$\int\frac{5\cos^3\text{x}+6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\text{dx}$
Answer $\int\bigg(\frac{5\cos^3\text{x}+6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int\bigg(\frac{5\cos^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}+\frac{6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int\Big(\frac{5}{2}\frac{\cos\text{x}}{\sin^2\text{x}}+3\frac{\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\frac{5}{2}\int\Big(\frac{\cos\text{x}}{\sin\text{x}}\times\frac{1}{\sin\text{x}}\Big)\text{dx}+3\int\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\text{dx}$
$=\frac{5}{2}\int(\text{cosec x}\cot\text{x})\text{dx}+3\int\sec\text{x}\tan\text{x dx}$
$=\frac{5}{2}(-\text{cosec x})+3\sec\text{x}+\text{C}$
$=-\frac{5}{2}\text{cosec x}+3\sec\text{x}+\text{C}$
View full question & answer→Question 753 Marks
Evaluate the following integrals:
$\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\sqrt{16+\text{t}^2}\text{dt}$
$=\int\sqrt{4^2+\text{t}^2}\text{dt}$
$=\frac{\text{t}}{2}\sqrt{16+\text{t}^2}+\frac{16}{2}\log\big|\text{t}+\sqrt{16+\text{t}^2}\big|+\text{C}$
$\therefore\ \text{I}=\frac{\log\text{x}}{2}\sqrt{16+(\log\text{x})^2}\\+8\log\Big|\log\text{x}+\sqrt{16+(\log\text{x})^2}\Big|+\text{C}$
View full question & answer→Question 763 Marks
Evaluate the following integrals:
$\int\frac{5\text{x}^4+12\text{x}^3+7\text{x}^2}{\text{x}^2+\text{x}}\text{dx}$
Answer$\int\frac{5\text{x}^4+12\text{x}^3+7\text{x}^2}{\text{x}^2+\text{x}}\text{dx}$
$=\int\frac{5\text{x}^4+7\text{x}^3+5\text{x}^3+7\text{x}^2}{\text{x}^2+\text{x}}\text{dx}$
$=\int\frac{5\text{x}^3+7\text{x}^2+5\text{x}^2+7\text{x}}{\text{x}+1}\text{dx}$
$=\int\frac{5\text{x}^2(\text{x}+1)+7\text{x}(\text{x}+1)}{\text{x}+1}\text{dx}$
$=\int(5\text{x}^2+7\text{x})\text{dx}$
$=\frac{5\text{x}^3}{3}+\frac{7\text{x}^2}{2}+\text{C}$
View full question & answer→Question 773 Marks
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\sin\Big(\text{x}-\frac{\pi}{6}\Big)\sin\Big(\text{x}+\frac{\pi}{6}\Big)}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\sin\Big(\text{x}-\frac{\pi}{6}\Big)\sin\Big(\text{x}+\frac{\pi}{6}\Big)}\text{dx}$
$=\int\frac{\sin2\text{x}}{\sin^2\text{x}-\sin^2\frac{\pi}{6}}\text{dx}$
$\big[\because \sin(\text{A}+\text{B})\sin(\text{A}-\text{B})=\sin^2\text{A}-\sin^2\text{B}\big]$
$=\int\frac{\sin2\text{x}}{\sin^2\text{x}-\frac{1}{4}}\text{dx}$
Putting $\sin^2\text{x}-\frac{1}{4}=\text{t}$
$\Rightarrow2\sin\text{x}\cos\text{ x dx}=\text{dt}$
$\Rightarrow\sin2\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}\big|\sin^2\text{x}-\frac{1}{4}\big|+\text{C}\ \Big[\because\text{t}=\sin^2\text{x}-\frac{1}{4}\Big]$
View full question & answer→Question 783 Marks
Write a value of $\int\frac{1}{1+2\text{e}^{\text{x}}}\text{dx}$
AnswerLet $\int\frac{1}{1+2\text{e}^{\text{x}}}\text{dx}$
Dividing and multiplying by $e^x$
$\text{I}=\int\frac{\frac{1}{\text{e}^{\text{x}}}\text{dx}}{\frac{1}{\text{e}^{\text{x}}}+2}$
$=\int\frac{\text{e}^{-\text{x}}\text{dx}}{\text{e}^{-\text{x}}+2}$
Let $\text{e}^{-\text{x}}+2=\text{t}$
$-\text{e}^{-\text{x}}\text{dx}=\text{dt}$
$\text{e}^{-\text{x}}\text{dx}=-\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$=-\log|\text{e}^{-\text{x}}+2|+\text{C}$ $(\because\text{t}=\text{e}^{-\text{x}}+2)$
View full question & answer→Question 793 Marks
Evaluate the following integrals:
$\int\text{x}^3\cos\text{x}^4\text{dx}$
Answer$\text{I}=\int\text{x}^3.\cos\big(\text{x}^4\big)\text{dx}$
Let $\text{x}^4=\text{t}$ then,
$\Rightarrow4\text{x}^3\text{ dx}=\text{dt}$
$\Rightarrow\text{x}^3\text{dx}=\frac{\text{dt}}{4}$
$\Rightarrow\text{x}^3\text{dx}= \frac{\text{dt}}{\text{4} }$
Now, $\int\text{x}^3.\cos\big(\text{x}^4\big)\text{dx}$
$=\frac{1}{4}\int\cos(\text{t})\text{dt}$
$=\frac{1}{4}[\sin(\text{t})]+\text{C}$
$=\frac{1}{4}\big[\sin\text{x}^4\big]+\text{C}$
View full question & answer→Question 803 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2\text{dx}}{\text{x}^6-\text{a}^6}\text{dx}$
Answer$\int\frac{\text{x}^2\text{dx}}{\text{x}^6-\text{a}^6}$
Let $\text{x}^3=\text{t}$
$\Rightarrow3\text{x}^2\text{dx = dt}$
$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$
Now, $\int\frac{\text{x}^2\text{dx}}{\text{x}^6-\text{a}^6}$
$=\frac{1}{3}\int\frac{\text{dt}}{\text{t}^2-(\text{a}^3)^2}$
$=\frac{1}{3}\times\frac{1}{2\text{a}^3}\log\bigg|\frac{\text{t}-\text{a}^3}{\text{t}+\text{a}^3}\Big|+\text{C}$
$=\frac{1}{6\text{a}^3}\log\Big|\frac{\text{x}^3-\text{a}^3}{\text{x}^3+\text{a}^3}\Big|+\text{C}$
View full question & answer→Question 813 Marks
Evaluate the following intregals:
$\int\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}\ \text{dx}$
We express
$\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{C}}{\text{x}+3}$
$\Rightarrow3\text{x}-2=\text{A}(\text{x}+1)(\text{x}+3)+\text{B}(\text{x}+3)+\text{C}(\text{x}+1)^2$
Equating the coefficient of $x^2, x$ and constants, we get
0 = A + C and 3 = 4A + B + 2C and -2 = 3A + 3B + C
$\text{or }\text{A}=\frac{11}{4}\text{ and }\text{B}=-\frac{5}{2}\text{ and }\text{C}=-\frac{11}{4}$
$\therefore\text{I}=\int\bigg(\frac{\frac{11}{4}}{\text{x}+1}+\frac{-\frac{5}{2}}{(\text{x}+1)^2}+\frac{-\frac{11}{4}}{\text{x}+3}\bigg)\ \text{dx}$
$=\frac{11}{4}\int\frac{1}{\text{x}+1}\text{ dx }-\frac{5}{2}\int\frac{1}{(\text{x}+1)^2}\text{ dx }-\frac{11}{4}\int\frac{1}{\text{x}+3}\text{ dx}$
$=\frac{11}{4}\log|\text{x}+1|+\frac{5}{2(\text{x}+1)}-\frac{11}{4}\log|\text{x}+3|+\text{C}$
Hence,
$\int\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}\ \text{dx}=\frac{11}{4}\log|\text{x}+1|+\frac{5}{2(\text{x}+1)}-\frac{11}{4}\log|\text{x}+3|+\text{C}$
View full question & answer→Question 823 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+1}}\text{ dx}$
AnswerWe have
$\text{I}=\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+1}}\text{ dx}$
Putting $\text{x}-1=\frac{1}{\text{t}}$
$\text{dx}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=\int\frac{-\frac{1}{\text{t}^2}\text{ dt}}{\big(\frac{1}{\text{t}}\big)\sqrt{\big(1+\frac{1}{\text{t}}}\big)^2+1}$
$=\int\frac{-\frac{1}{\text{t}}\text{ dt}}{\sqrt{1+\frac{1}{\text{t}^2}+\frac{2}{\text{t}}+1}}$
$=\int\frac{-\frac{1}{\text{t}}\text{ dt}}{\sqrt{\frac{\text{t}^2+1+2\text{t}+\text{t}^2}{\text{t}}}}$
$=\int\frac{-\text{dt}}{\sqrt{2\text{t}^2+2\text{t}+1}}$
$=-\frac{1}{\sqrt{2}}\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\frac{1}{2}}}$
$=-\frac{1}{\sqrt{2}}\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\frac{1}{4}-\frac{1}{4}+\frac{1}{2}}}$
$=-\frac{1}{\sqrt{2}}\int\frac{\text{dt}}{\big(\text{t}+\frac{1}{2}\big)^2+\big(\frac{1}{2}\big)^2}$
$=-\frac{1}{\sqrt{2}}\log\begin{vmatrix}\text{t}+\frac{1}{2}+\sqrt{\Big(\text{t}+\frac{1}{2}\Big)^2+\frac{1}{4}}\end{vmatrix}+\text{C}$ where $\text{t}=\frac{1}{\text{x}-1}$
View full question & answer→Question 833 Marks
Evaluate the following integrals:
$\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$
Answer $\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$
Let $\text{e}^\text{x}-1=\text{t}^2$
$\Rightarrow\text{e}^\text{x}=\text{t}^2+1$
$\text{e}^\text{x}=2\text{t}\frac{\text{dt}}{\text{dx}}$
$\text{dx}=\frac{2\text{t dt}}{\text{e}^\text{x}}$
$\text{dx}=\frac{2\text{t dt}}{\text{t}^2+1}$
Now, $\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$
$=\int\frac{\text{t. 2t dt}}{\text{t}^2+1}$
$=2\int\frac{\text{t}^2\text{ dt}}{\text{t}^2+1}$
$=2\int\Big(\frac{\text{t}^2+1-1}{\text{t}^2+1}\Big)\text{ dt}$
$=2\int\text{dt}-2\int\frac{\text{dt}}{\text{t}^2+1}$
$=2\text{t}-2\tan^{-1}(\text{t})+\text{C}$
$=2\sqrt{\text{e}^\text{x}-1}-2\tan^{-1}\big(\sqrt{\text{e}^\text{x}-1}\big)+\text{C}$
View full question & answer→Question 843 Marks
Write a value of $\int\frac{\log\text{x}^{\text{n}}}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\log\text{x}^{\text{n}}}{\text{x}}\text{ dx}$
Let $\log\text{x}^{\text{n}}=\text{t}$
$\frac{\text{nx}^{\text{n}-1}}{\text{x}^{\text{n}}}\text{ dx}=\text{dt}$
$\frac{\text{n}}{\text{x}}=\text{ dt}$
$\therefore\ \text{I}=\text{n}\int\text{t}\text{ dt}$
$=\text{n}\Big(\frac{\text{t}^2}{2}\Big)+\text{C}$
Putting the value of t
$\text{I}=\frac{\text{n}(\log\text{x}^{\text{n}})^2}{2}+\text{C}$
View full question & answer→Question 853 Marks
Evaluate the following integrals:
$\int\frac{2\text{x}^4+7\text{x}^3+6\text{x}^2}{\text{x}^2+2\text{x}}\text{dx}$
Answer$\int\bigg(\frac{2\text{x}^4+7\text{x}^3+6\text{x}^2}{\text{x}^2+2\text{x}}\bigg)\text{dx}$
$=\int\frac{\text{x}^2(2\text{x}^2+7\text{x}+6}{\text{x}(\text{x}+2)}$
$=\int\frac{\text{x}\big[2\text{x}^2+4\text{x}+3\text{x}+6\big]}{(\text{x}+2)}\text{dx}$
$=\int\frac{\text{x}(2\text{x}(\text{x}+2)+3(\text{x}+2))}{(\text{x}+2)}\text{dx}$
$=\int\frac{\text{x}(2\text{x}+3)(\text{x}+2)}{(\text{x}+2)}\text{dx}$
$=\int(2\text{x}^2+3\text{x})\text{dx}$
$=2\int\text{x}^2\text{dx}+3\int\text{x }\text{dx}$
$=2\Big[\frac{\text{x}^3}{3}\Big]+3\Big[\frac{\text{x}^2}{2}\Big]+\text{C}$
$=\frac{2}{3}\text{x}^3+\frac{3}{2}\text{x}^2+\text{C}$
View full question & answer→Question 863 Marks
$\int\frac{1}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\text{dx}.$ Then,
$\text{I}=\int\frac{1}{\sqrt{\text{x}+1}+\sqrt{\text{x}}}\times\frac{\sqrt{\text{x}+1}-\sqrt{\text{x}}}{\sqrt{\text{x}+1}-\sqrt{\text{x}}}\times\text{dx}$
$=\int\frac{\sqrt{\text{x+1}}-\sqrt{\text{x}}}{(\sqrt{\text{x}+1})^2-(\sqrt{\text{x}})^2}\times\text{dx}$
$=\int\frac{\sqrt{\text{x}+1}-\sqrt{\text{x}}}{\text{x}+1-\text{x}}\times\text{dx}$
$=\int(\sqrt{\text{x}+1}-\sqrt{\text{x}})\times\text{dx}$
$=\int(\text{x}+1)^{\frac{1}{2}}\text{dx}-\int\text{x}^{\frac{1}{2}}\text{dx}$
$=\frac{2}{3}(\text{x}+1)^{\frac{3}{2}}-\frac{2}{3}\text{x}^{\frac{3}{2}}+\text{c}$
$\therefore\text{I}=\frac{2}{3}(\text{x}+1)^{\frac{3}{2}}-\frac{2}{3}\text{x}^{\frac{3}{2}}+\text{c}$.
View full question & answer→Question 873 Marks
Evalute the following integrals:
$\int\frac{\text{a}}{\text{b}+\text{ce}^\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{a}}{\text{b}+\text{ce}^\text{x}}\text{dx}$
Dividing numerator and denomimator by $e^x$
$\Rightarrow\text{I}=\int\frac{\text{ae}^{-\text{x}}}{\text{be}^{-\text{x}}+\text{c}}\text{dx}$
Putting $e^{-x} = t$
$\Rightarrow-\text{e}^{-\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{-\text{x}}\text{dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{a}}{\text{bt}+\text{c}}\text{dt}$
$=\frac{-\text{a}}{\text{b}}\text{ ln}|\text{bt}+\text{c}|+\text{C}$
$\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\text{ ln}|\text{ax}+\text{b}|+\text{C}\Big]$
$=\frac{-\text{a}}{\text{b}}\text{ ln}|\text{be}^{-\text{x}}+\text{c}|+\text{C}\ \big[\because\text{t}=\text{e}^{-\text{x}}+\text{C}\big]$
View full question & answer→Question 883 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}}(\log\text{x})^2\text{dx}$
AnswerLet I $=\int\frac{1}{\text{x}}(\log\text{x})^2\text{dx}\ .....(1)$
Let $\log\text{x}=\text{t}$ then,
$\text{d}(\log\text{x})=\text{dt}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
Putting $\log\text{x}=\text{t}$ and $\frac{1}{\text{x}}\text{dx}=\text{dt}$ in equation (1), we get
$\text{I}=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^3}{3}+\text{C}$
$=\frac{(\log\text{x})^3}{3}+\text{C}$
$\therefore\text{I}=\frac{1}{3}(\log\text{x})^3+\text{C}$
View full question & answer→Question 893 Marks
Evaluate the following integrals:
$\int\sec^6\text{x }\tan\text{x}\text{ dx}$
Answer$\int\sec^6\text{x }\tan\text{x}\text{ dx}$
$\int\sec^6\text{x}.\sec\text{x}\tan\text{x}\text{ dx}$
Let $\sec\text{x}=\text{t}$
$\sec\text{x}\tan\text{x}\text{ dx}=\text{dt}$
Now, $\int\sec^6\text{x}.\sec\text{x}\tan\text{x}\text{ dx}$
$=\int\text{t}^6\text{dt}$
$=\frac{\text{t}^6}{6}+\text{C}$
$=\frac{\sec^6\text{x}}{6}+\text{C}$
View full question & answer→Question 903 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
Answer$\int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1}{\sqrt{\text{x}}}+\frac{1}{\sqrt{\text{x}}\text{x}}\Big)\text{dx}$
$=\int\text{x}^{\frac{-1}{2}}+\int\text{x}^{\frac{-3}{2}}\text{dx}$
$=2\text{x}^{\frac{1}{2}}-2\text{x}^{\frac{-1}{2}}+\text{C}$
$=2\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}+\text{C}$
$\therefore\ \int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}=2\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}+\text{C}$
View full question & answer→Question 913 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2(\text{xe}^\text{x})}\text{ dx}$
Answer$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2(\text{xe}^\text{x})}\text{ dx}$ Let $\text{xe}^\text{x}=\text{t}$ $\Rightarrow(1.\text{e}^\text{x}+\text{xe}^\text{x})=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow(\text{x}+1)\text{e}^\text{x}\text{dx}=\text{dt}$ Now, $\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2({\text{xe}^\text{x}})}=\text{dx}$$=\int\frac{\text{dt}}{\sin^2\text{t}}$
$=\int\text{cosec}^2\text{t}\text{ dt}$
$=-\cot(\text{t})+\text{C}$
$=-\cot(\text{xe}^\text{x})+\text{C}$
View full question & answer→Question 923 Marks
Evaluate the following integrals:
$\int\text{x e}^{\text{x}^2}=\text{dx}$
AnswerLet $\text{I}=\int\text{x e}^{\text{x}^2}=\text{dx}\ ....(1)$ Let $\text{x}^2=\text{t}$ then, $\text{d}(\text{x}^2)=\text{dt}$ $\Rightarrow2\text{x dx}=\text{dt}$ $\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ Putting $\text{x}^2=\text{t}$ and $\text{x dx}=\frac{\text{dt}}{2}$ in equation (1),we get,
$\text{I}=\int\text{e}^\text{t}\frac{\text{dt}}{2}$
$=\frac{1}{2}\text{e}^\text{t}+\text{C}$
$=\frac{1}{2}\text{e}^{\text{x}^2}+\text{C}$
$\text{I}=\frac{1}{2}\text{e}^{\text{x}^2}+\text{C}$
View full question & answer→Question 933 Marks
Evaluate the following integrals:
$\int\frac{(1+\text{x})^3}{\sqrt{\text{x}}}\text{dx}$
Answer$\int\frac{(1+\text{x})^3}{\sqrt{\text{x}}}\text{dx}$
$=\int\frac{1}{\sqrt{\text{x}}}\text{dx}+\int\frac{\text{x}^3}{\sqrt{\text{x}}}\text{dx}+\int\frac{3\text{x}^2}{\sqrt{\text{x}}}\text{dx}+\int\frac{3\text{x}}{\sqrt{\text{x}}}\text{dx}$
$=\int\text{x}^{\frac{-1}{2}}\text{dx}+\int\text{x}^{\frac{5}{2}}\text{dx}+3\int\text{x}^{\frac{3}{2}}\text{dx}+3\int\text{x}^{\frac{1}{2}}\text{dx}$
$=\frac{\text{x}^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}+\frac{3\text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{3\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{C}$
$=\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\frac{\text{x}^{\frac{7}{2}}}{\frac{7}{2}}+\frac{\text{3x}^{\frac{5}{2}}}{\frac{5}{2}}+\frac{\text{3x}^{\frac{3}{2}}}{\frac{3}{2}}+\text{C}$
$=2\text{x}^{\frac{1}{2}}+\frac{2}{7}\text{x}^{\frac{7}{2}}+\frac{6}{5}\text{x}^{\frac{5}{2}}+\frac{6}{3}\text{x}^{\frac{3}{2}}+\text{C}$
$=2\text{x}^{\frac{1}{2}}+\frac{2}{7}\text{x}^{\frac{7}{2}}+\frac{6}{5}\text{x}^{\frac{5}{2}}+2\text{x}^{\frac{3}{2}}+\text{C}$
View full question & answer→Question 943 Marks
Write a value of $\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{dx}$
We know that,
$\int\text{e}^{\text{x}}\int\text{f}(\text{x})+\text{f}'(\text{x})=\text{e}^{\text{x}}\text{f}(\text{x})+\text{C}$
Hence, $\text{f}'(\text{x})=-\frac{1}{\text{x}^2}$
Then, $\int\text{e}^{\text{ax}}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{e}^{\text{x}}\cdot\frac{1}{\text{x}}+\text{C}$
$\therefore\ \text{I}=\text{e}^{\text{x}}\cdot\frac{1}{\text{x}}+\text{C}$
View full question & answer→Question 953 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^6+1}{\text{x}^2+1}\text{dx}$
Answer$\int\Big(\frac{\text{x}^6+1}{\text{x}^2+1}\Big)\text{dx}$
$=\int\bigg[\frac{(\text{x}^2)^3+1^3}{\text{x}^2+1}\bigg]\text{dx}$ $[\text{A}^3+\text{B}^3=(\text{A+B})(\text{A}^2-\text{AB}+\text{B}^2)]$
$=\int\frac{(\text{x}^2+1)(\text{x}^4-\text{x}^2+1)}{(\text{x}^2+1)}\text{dx}$
$=\int(\text{x}^4-\text{x}^2+1)\text{dx}$
$=\int\text{x}^4\text{dx}+\int\text{x}^2\text{dx}+\int1\text{dx}$
$=\frac{\text{x}^{4+1}}{4+1}-\frac{\text{x}^{2+1}}{2+1}+\text{x}+\text{C}$
$=\frac{\text{x}^5}{5}-\frac{\text{x}^3}{3}+\text{x}+\text{C}$
View full question & answer→Question 963 Marks
Integrate the following integrals:
$\int\sin4\text{x}\cos7\text{x dx}$
Answer $\int\sin4\text{x}\cos7\text{x dx}$
$=\frac{1}{2}\int2\cos7\text{x}\sin4\text{x dx}$
$=\frac{1}{2}\int\big[\sin(7\text{x}+4\text{x})-\sin(7\text{x}-4\text{x})\big]\text{dx}$ $[\therefore2\cos\text{A}\sin\text{B}=\sin(\text{A}+\text{B})-\sin(\text{A}-\text{B})\big]$
$=\frac{1}{2}\int\big(\sin(11\text{x})-\sin(3\text{x})\big)\text{dx}$
$=\frac{1}{2}\Big[-\frac{\cos(11\text{x})}{11}+\frac{\cos(3\text{x})}{3}\Big]+\text{c}$
$=-\frac{\cos(11\text{x})}{22}+\frac{\cos(3\text{x})}{6}$
View full question & answer→Question 973 Marks
Write a value of $\int\frac{1}{\text{x}(\log\text{x})^{\text{n}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{x}(\log\text{x})^{\text{n}}}\text{ dx}$
Let $\log\text{x}=\text{t}$
$\frac{1}{\text{x}}\text{ dx}=\text{dt}$
$\text{dx}=\text{xdt}$
$\therefore\ \text{I}=\int\frac{1}{\text{t}^{\text{n}}}\text{ dt}$
$=\frac{\text{t}^{-\text{n}+1}}{-\text{n}+1}+\text{C}$
$\text{I}=\frac{(\log\text{x})^{1-\text{n}}}{1-\text{n}}+\text{C}$
View full question & answer→Question 983 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{2\text{x}-\text{x}^2}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\sqrt{2\text{x}-\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{2\text{x}-\text{x}^2-1+1}}$
$=\int\frac{\text{dx}}{\sqrt{1-(\text{x}^2-2\text{x}+1)}}$
$=\int\frac{\text{dx}}{1-(\text{x}-1)^2}$
$=\sin(\text{x}-1)+\text{C}$ $\Big[\because\ \int\frac{\text{dx}}{\sqrt{\text{a}^2-\text{x}^2}}=\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
View full question & answer→Question 993 Marks
Evaluate the following integrals:
$\int\Big(\frac{\text{x}+1}{\text{x}}\Big)(\text{x}+\log\text{x})^2\text{dx}$
Answer$\int\Big(\frac{\text{x}+1}{\text{x}}\Big)(\text{x}+\log\text{x})^2\text{dx}$ Let $\text{x}+\log\text{x}=\text{t}$ $\Rightarrow\Big(1+\frac{1}{\text{x}}\Big)=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\Big(\frac{\text{x}+1}{\text{x}}\Big)\text{dx}=\text{dt}$Now, $\int\Big(\frac{\text{x}+1}{\text{x}}\Big)(\text{x}+\log\text{x})^2\text{dx}$
$=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^2}{3}+\text{C}$
$=\frac{(\text{x}+\log\text{x})^3}{3}+\text{C}$
View full question & answer→Question 1003 Marks
Evaluate the following integrals:$\int\text{x}^2\cos\text{x dx}$
AnswerLet $\text{I}=\int\text{x}^2\cos\text{x dx}$
Using integration by parts,
$\text{I}=\text{x}^2\int\cos\text{x dx}-\int(2\text{x}\int\cos\text{x dx})\text{dx}$
$=\text{x}^2\sin\text{x}-2\int\text{x}\sin\text{x dx}$
$=\text{x}^2\sin\text{x}-2[\text{x}\int\sin\text{x dx}-\int(1\int\sin\text{x dx})\text{dx}]$
$=\text{x}^2\sin\text{x}-2[\text{x}(-\cos\text{x})-\int(-\cos\text{x})\text{dx}]$
$=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\int(\cos\text{x})\text{dx}$
$\text{I}=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x}+\text{C}$
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