Question 1513 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 1523 Marks
Evaluate the following integrals: $\int\text{e}^{2\text{x}}\sin\text{x }\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}\sin\text{x }\text{dx}$
Considering $\sin x$ as first function and $e^{2x}$ as second function
$\text{I}=\sin\text{x}\frac{\text{e}^{2\text{x}}}{2}-\int\cos\text{x}\frac{\text{e}^{2\text{x}}}{2}\text{dx}$
$\Rightarrow\text{I}=\sin\text{x}\frac{\text{e}^{2\text{x}}}{2}-\frac{1}{2}\int\cos\text{e}^{2\text{x}}\text{dx}$
$\Rightarrow\text{I}=\frac{\sin\text{x }\text{e}^{2\text{x}}}{2}-\frac{1}{2}\Big[\cos\text{x}\frac{\text{e}^{2\text{x}}}{2}-\int(-\sin\text{x})\frac{\text{e}^{2\text{x}}}{2}\text{dx}\Big]$
$\Rightarrow\text{I}=\frac{\sin\text{x }\text{e}^{2\text{x}}}{2}-\frac{\cos\text{x }\text{e}^{2\text{x}}}{4}-\frac{1}{2}\int\frac{\text{e}^{2\text{x}}\sin\text{x}}{2}\text{dx}$
$\text{I}=\frac{\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})}{4}-\frac{\text{I}}{4}$
$\Rightarrow5\text{I}=\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})$
$\Rightarrow\text{I}=\frac{\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})}{5}+\text{C}$
View full question & answer→Question 1533 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 1543 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 1553 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 1563 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 1573 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 1583 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 1593 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
Answer$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
Let $\text{x}\text{e}^\text{x}=\text{t}$
$\Rightarrow\big(1.\text{e}^\text{x}+\text{x}\text{e}^\text{x}\big)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(\text{x}+1)\text{e}^\text{x}\text{ dx}=\text{dt}$
Now, $\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
$=\int\frac{\text{dt}}{\cos^2\text{t}}$
$=\sec^2\text{t}\text{ dt}$
$=\tan(\text{t})+\text{C}$
$=\tan\big(\text{xe}^\text{x}\big)+\text{C}$
View full question & answer→Question 1603 Marks
Evaluate the following integrals:
$\int\text{x}^2\cos2\text{x dx}$
AnswerLet $\text{I}=\int\text{x}^2\cos2\text{x}\text{ dx}$
Using integration by parts,
$\text{I}=\text{x}^2\int\cos2\text{x dx}-\int(2\text{x}\int\cos2\text{x dx})\text{dx}$
$=\text{x}^2\frac{\sin2\text{x}}{2}-2\int\text{x}\Big(\frac{\sin2\text{x}}{2}\Big)\text{dx}$
$=\frac{1}{2}\text{x}^2\sin2\text{x}-\int\text{x}\sin2\text{x dx}$
$=\frac{1}{2}\text{x}^2\sin2\text{x}-[\text{x}\int\sin2\text{x dx}-\int(1\int\sin2\text{x dx})\text{dx}]$
$=\frac{1}{2}\text{x}^2\sin2\text{x}-\bigg[\text{x}\Big(\frac{-\cos2\text{x}}{2}\Big)-\int\Big(-\frac{\cos2\text{x}}{2}\Big)\text{dx}\bigg]$
$=\frac{1}{2}\text{x}^2\sin2\text{x}+\frac{\text{x}}{2}\cos2\text{x}-\frac{1}{2}\int(\cos2\text{x})\text{dx}$
$\text{I}=\frac{1}{2}\text{x}^2\sin2\text{x}+\frac{\text{x}}{2}\cos2\text{x}-\frac{1}{4}\sin2\text{x}+\text{C}$
View full question & answer→Question 1613 Marks
Evaluate the following integrals:
$\int\frac{\sin(\log\text{x})}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin(\log\text{x})}{\text{x}}\text{ dx}\ ....(1)$ 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,$\text{I}=\int\sin\text{t dt}$
$=-\cos\text{t}+\text{C}$
$=-\cos\big(\log\text{x}\big)+\text{C}$
View full question & answer→Question 1623 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 1633 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^3-3\text{x}^2+5\text{x}-7+\text{x}^2\text{a}^\text{x}}{2\text{x}^2}\text{dx}$
Answer$\int\frac{\text{x}^3-3\text{x}^2+5\text{x}-7+\text{x}^2\text{a}^\text{x}}{2\text{x}^2}\text{dx}$
$=\frac{1}{2}\int\frac{\text{x}^3}{\text{x}^2}\text{dx}-\frac{3}{2}\int\frac{\text{x}^2}{\text{x}^2}\text{dx}+\frac{5}{2}\int\text{x}\frac{\text{x}}{\text{x}^2}\text{dx}-\frac{7}{2}\int\text{x}^{-2}\text{dx}+\frac{1}{2}\int\frac{\text{x}^2\text{a}^\text{x}}{\text{x}^2}\text{dx}$
$=\frac{1}{2}\times\frac{\text{x}^2}{2}-\frac{3}{2}\text{x}+\frac{5}{2}\log\text{x}-\frac{7}{2}\text{x}^{-1}+\frac{1}{2}\frac{\text{a}^\text{x}}{\log\text{a}}+\text{C}$
$=\frac{1}{2}\Big[\frac{\text{x}^2}{2}-3\text{x}+5\log\text{x}-\frac{7}{\text{x}}+\frac{\text{a}^\text{x}}{\log\text{a}}\Big]+\text{C}$
$\therefore\ \int\frac{\text{x}^3-3\text{x}^2+5\text{x}-7+\text{x}^2\text{a}^\text{x}}{2\text{x}^2}\text{dx}$$=\frac{1}{2}\Big[\frac{\text{x}^2}{2}-3\text{x}+5\log\text{x}-\frac{7}{\text{x}}+\frac{\text{a}^\text{x}}{\log\text{a}}\Big]+\text{C}$
View full question & answer→Question 1643 Marks
Evalute the following integrals: $\int\frac{\text{e}^{2\text{x}}}{\text{e}^{2\text{x}}-2}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{2\text{x}}}{\text{e}^{2\text{x}}-2}\text{dx}$
Putting $e^{2x} = t$
$\Rightarrow2\text{e}^{2\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{2\text{x}}\text{dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{1}{\text{t}-2}\text{dt}$
$=\frac{1}{2}\text{ ln}|\text{t}-2|+\text{C}$
$=\frac{1}{2}\text{ ln}\big|\text{e}^{2\text{x}}-2\big|+\text{C }\big[\text{t}=\text{e}^{2\text{x}}\big]$
View full question & answer→Question 1653 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 1663 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 1673 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 1683 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 1693 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 1703 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 1713 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{t}$
$\frac{1}{\text{x}}\text{ dx}=\text{dt}$
$\text{dx}=\text{xdt}$
$\therefore\ \int\frac{(\log\text{x})^{\text{n}}}{\text{x}}\text{ dx}=\int(\text{t})^{\text{n}}\text{dt}$
$=\frac{\text{t}^{\text{n}+1}}{\text{n}+1}+\text{C}$
$=\frac{(\log\text{x})^{\text{n}+1}}{(\text{n}+1)}+\text{C}$
View full question & answer→Question 1723 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 1733 Marks
Evaluate the following integrals:
$\int \frac{1}{(\text{x}-1)\sqrt{2\text{x}+3}}\text{ dx}$
AnswerLet $\text{I}=\int \frac{1}{(\text{x}-1)\sqrt{2\text{x}+3}}\text{ dx}$
Let $2\text{x}+3=\text{t}^2$
$2\text{dx}=2\text{tdt}$
$\therefore\ \text{I}=\int\frac{\text{t dt}}{\big(\frac{\text{t}^2-3}{2}-1\big)\text{t}}$
$=2\int\frac{\text{dt}}{\text{t}^2-5}$
$=\frac{2}{2\sqrt{5}}\log\Big|\frac{\text{t}-\sqrt{5}}{\text{t}+\sqrt{5}}\Big|+\text{C}$
Thus, $\text{I}=\frac{1}{\sqrt{5}}\log\bigg|\frac{\sqrt{2\text{x}+3}-\sqrt{5}}{\sqrt{2\text{x}+3}+\sqrt{5}}\bigg|+\text{C}$
View full question & answer→Question 1743 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}+1)(\text{x}-2)}{\sqrt{\text{x}}}\text{dx}$
Answer$\int\frac{(\text{x}+1)(\text{x}-2)}{\sqrt{\text{x}}}\text{dx}$
$=\int\bigg(\frac{\text{x}^2-2\text{x}+\text{x}-2}{\sqrt{\text{x}}}\bigg)\text{dx}$
$=\bigg(\frac{\text{x}^2-\text{x}-2}{\sqrt{\text{x}}}\bigg)\text{dx}$
$=\int\Big(\text{x}^\frac{3}{2}-\text{x}^\frac{1}{2}-2\text{x}^\frac{-1}{2}\Big)\text{dx}$
$=\Bigg[\frac{\text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]-\Bigg[\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\Bigg]-2\Bigg[\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=\frac{2}{5}\text{x}^\frac{5}{2}-\frac{2}{3}\text{x}^\frac{3}{2}-4\text{x}^\frac{1}{2}+\text{C}$
View full question & answer→Question 1753 Marks
$\int\frac{\text{x}^3}{\text{x}-2}\text{dx}$
Answer$\text{Let I}=\int\frac{\text{x}^3}{\text{x}-2}\text{dx}$
Using long division method, we have
$\frac{\text{x}^3}{\text{x}-2}=\text{x}^2+2\text{x}+4+\frac{8}{\text{x}-2}$
$\text{I}=\int\Big(\text{x}^2+2\text{x}+4+\frac{8}{\text{x}-2}\Big)\text{dx}$
$=\int\text{x}^2\text{dx}+2\int\text{xdx}+4\int\text{dx}+8\int\frac{1}{\text{x}-2}\text{dx}$
$=\frac{\text{x}^3}{3}+\frac{2\text{x}^2}{2}+4\text{x}+8\log|\text{x}-2|+\text{c}$
$=\frac{\text{x}^3}{3}+\text{x}^2+4\text{x}+8\log|\text{x}-2|+\text{c}$
$\therefore\text{I}=\frac{\text{x}^3}{3}+\text{x}^2+4\text{x}+8\log|\text{x}-2|+\text{c}$
View full question & answer→Question 1763 Marks
$\int\sin\text{x}\sqrt{1+\cos2\text{x}}\text{ dx}$
Answer$\int\sin\text{x}\sqrt{1+\cos2\text{x}}\text{ dx}$
$=\int\sin\text{x}.\sqrt{2\cos^2\text{x}}\text{ dx}$ $[\therefore1+\cos2\text{x}=2\cos^2\text{x}]$
$=\sqrt{2}\int\sin\text{x}\cos\text{x dx}$
$=\frac{\sqrt{2}}{2}\int2\sin\text{x}\cos\text{x dx}$
$=\frac{1}{\sqrt{2}}\int\sin2\text{x dx}$
$=\frac{1}{\sqrt{2}}\Big[\frac{-\cos2\text{x}}{2}\Big]+\text{c}$
$=\frac{-1}{2\sqrt{2}}\cos2\text{x}+\text{c}$
View full question & answer→Question 1773 Marks
Evaluate the following integrals:
$\int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\text{dx}$
Answer$\int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\text{dx}$
$=\int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\times\frac{\text{cosec x}+\cot\text{x}}{\text{cosec x}+\cot\text{x}}\times\text{dx}$
$=\int\frac{\text{cosec x}(\text{cosec x}+\cot\text{x})}{\text{cosec}^2\text{x}-\cot^2\text{x}}\text{dx}$
$=\int(\text{cosec}^2\text{x}+\text{cosec x}\cot\text{x})\text{dx}$
$=\int\text{cosec}^2\text{x dx}+\int\text{cosec x dx}$
$=-\cot\text{x}-\text{cosec x}+\text{C}$
$\therefore\ \int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\text{dx}=-\cot\text{x}-\text{cosec x}+\text{C}$
View full question & answer→Question 1783 Marks
$\int(\text{e}^{\text{x}}+1)^2\text{e}^{\text{x}}\text{ dx}$
AnswerConsider $\text{I}=\int(\text{e}^{\text{x}}+1)^2\text{e}^{\text{x}}\text{dx}$
Let $(\text{e}^{\text{x}}+1)=\text{t}\rightarrow\text{e}^{\text{x}}\text{dx}=\text{dt}$
$\text{I}=\int(\text{e}^\text{x}+1)^2\text{e}^{\text{x}}\text{dx}$
$=\int(\text{t})^2\text{dt}$
$=\frac{\text{t}^3}{3}+\text{c}$
$=\frac{(\text{e}^{\text{x}}+1)^3}{3}+\text{c}$
View full question & answer→Question 1793 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}^2+4)\sqrt{\text{x}^2+1}}\text{ dx}$
Answer$\text{I}=\int\frac{1}{(\text{x}^2+4)\sqrt{\text{x}^2+1}}\text{ dx}$
Let $\text{x}^2+1=\text{u}^2$
$2\text{xdx}=2\text{udu}$
$\therefore\ \text{I}=\int\frac{\text{u}}{(\text{u}^2+3)\text{u}}\text{ du}$
$=\int\frac{1}{\text{u}^2+3}\text{ du}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{u}}{\sqrt{3}}\Big)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\bigg(\sqrt{\frac{\text{x}^2+1}{3}}\bigg)+\text{C}$
View full question & answer→Question 1803 Marks
Evaluate $\int\frac{1}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
AnswerConsider the integral
$\text{I}=\int\frac{1}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}$
$=\int\big(\sec^2\text{x}+\text{cosec}^2\text{x}\big)\text{dx}$
$=\int\sec^2\text{x dx}+\int\text{cosec}^2\text{x dx}$
$=\tan\text{x}-\cot\text{x}+\text{C}$
$=\frac{\sin\text{x}}{\cos\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}+\text{C}$
$=\frac{\sin^2\text{x}-\cos^2\text{x}}{\sin\text{x}\cos\text{x}}+\text{C}$
$=-\frac{2\cos2\text{x}}{\sin2\text{x}}+\text{C}$
$=-2\cot2\text{x}+\text{C}$
View full question & answer→Question 1813 Marks
$\int\cos^42\text{x dx}$
Answer$\int\cos^42\text{x dx}$
$=\int(\cos^22\text{x})^2\text{dx}$
$=\int\Big(\frac{1+\cos4\text{x}}{2}\Big)^2\text{dx}$ $\Big[\therefore\cos^2\text{x}=\frac{1+\cos2\text{x}}{2}\Big]$
$=\frac{1}{4}\int(1+\cos4\text{x})^2\text{dx}$
$=\frac{1}{4}\int(1+\cos^24\text{x}+2\cos4\text{x})\text{dx}$
$=\frac{1}{4}\Big[1+\Big(\frac{1+\cos8\text{x}}{2}\Big)+2\cos4\text{x}\Big]\text{dx}$
$=\frac{1}{4}\int\Big(\frac{3}{2}+\frac{\cos8\text{x}}{2}+2\cos4\text{x}\Big]\text{dx}$
$=\frac{1}{4}\Big[\frac{3\text{x}}{2}+\frac{\sin8\text{x}}{16}+\frac{2\sin4\text{x}}{4}\Big]+\text{C}$
$=\frac{3\text{x}}{8}+\frac{\sin8\text{x}}{64}+\frac{\sin4\text{x}}{8}+\text{C}$
View full question & answer→Question 1823 Marks
Evaluate:
$\int\frac{\text{e}^{\log\sqrt{\text{x}}}}{\text{x}}\text{dx}$
Answer$\int\frac{\text{e}^{\log\sqrt{\text{x}}}}{\text{x}}\text{dx}=\int\frac{\sqrt{\text{x}}}{\text{x}}\text{dx}$
$=\int\text{x}^\frac{1}{2}\times\text{x}^{-1}\text{dx}$
$=\int\text{x}^{\frac{1}{2}-1}\text{dx}$
$=\int\frac{\text{x}^{\frac{-1}{2}+1}+\text{c}}{\frac{-1}{2}+1}$
$=\frac{\text{x}^\frac{1}{2}}{\frac{1}{2}}$
$=\sqrt{\text{x}}+\text{c}$
View full question & answer→Question 1833 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{2\text{x}}}{1+\text{e}^\text{x}}\text{ dx}$
Answer$\int\frac{\text{e}^{2\text{x}}}{1+\text{e}^\text{x}}\text{ dx}$ $\Rightarrow\int\frac{\text{e}^\text{x}.\text{e}^\text{x}}{1+\text{e}^\text{x}}\text{ dx}$ then, Let $1+\text{e}^\text{x}=\text{t}$ $\Rightarrow\text{e}^\text{x}=\text{t}-1$ $\Rightarrow\text{e}^\text{x}\text{dx}=\text{dt}$ Now, $\int\frac{\text{e}^\text{x}.\text{e}^\text{x}}{1+\text{e}^\text{x}}\text{ dx}$$=\int\frac{(\text{t}-1)\text{dt}}{\text{t}}$
$=\Big(1-\frac{1}{\text{t}}\Big)\text{dt}$
$=\text{t}-\log|\text{t}|+\text{C}$
$=\big(1+\text{e}^\text{x}\big)-\log\big(1+\text{e}^\text{x}\big)+\text{C}$
Let $\text{C}+1=\text{C}^\text{n}$ $=\text{e}^\text{x}-\log\big(1+\text{e}^\text{x}\big)+\text{C}^\text{n}$
View full question & answer→Question 1843 Marks
Evaluate the following integrals:
$\int(\text{a}\tan\text{x}+\text{b}\cot \text{x})^2\text{dx}$
Answer$\int(\text{a}\tan\text{x}+\text{b}\cot \text{x})^2\text{dx}$
$=\int(\text{a}^2\tan^2\text{x}+\text{b}^2\cot^2\text{x}+2\text{ab}\tan\text{x}\cot\text{x})\text{dx}$
$=\text{a}^2\int\tan^2\text{x dx}+\text{b}^2\int\cot^2\text{x dx}+2\text{ab}\int\text{dx}$
$=\text{a}^2\int(\sec^2\text{x}-1)\text{dx}+\text{b}^2\int(\text{cosec}^2\text{x}-1)\text{dx}+2\text{ab}\int\text{dx}$
$=\text{a}^2[\tan\text{x}-\text{x}]+\text{b}^2[-\cot\text{x}-\text{x}]+2\text{ab}\text{x}+\text{C}$
$=\text{a}^2\tan\text{x}-\text{b}^2\cot\text{x}-(\text{a}^2+\text{b}^2-2\text{ab})\text{x}+\text{C}$
View full question & answer→Question 1853 Marks
Evaluate the following integrals:
$\int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}$
Answer$\int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}$
$=\int\sin^{-1}(\sin2\text{x} )\text{dx}$ $\Big[\because\ \sin2\text{x}=\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big]$
$=\int2\text{ x dx}$
$=2\int\text{x dx}$
$=\frac{2\text{x}^2}{2}+\text{C}$
$=\text{x}^2+\text{C}$
$\therefore\ \int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}=\text{x}^2+\text{C}$
View full question & answer→Question 1863 Marks
Evaluate the following integrals:
$\int\frac{3\text{x}^5}{1+\text{x}^{12}}\text{dx}$
AnswerLet $\text{I}=\int\frac{3\text{x}^5}{1+\text{x}^{12}}\text{dx}$
$=\int\frac{3\text{x}^5}{1+(\text{x}^6)^2}\text{dx}$
Let $\text{x}^6=\text{t}$
$\Rightarrow6\text{x}^5\text{dx = dt}$
$\Rightarrow\text{x}^5\text{dx}=\frac{\text{dt}}{6}$
$\text{I}=\frac{3}{6}\int\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{1}{2}\tan^{-1}(\text{t})+\text{C}$ $\Big[\text{Since}\int\frac{1}{\text{x}^2+1}\text{dx}=\tan^{-1}\text{x+C}\Big]$
$\text{I}=\frac{1}{2}\tan^{-1}(\text{x}^6)+\text{C}$
View full question & answer→Question 1873 Marks
$\int\frac{1}{\sqrt{2\text{x}+3}+\sqrt{2\text{x}-3}}\text{dx}$
Answer$\int\frac{\text{dx}}{(\sqrt{2\text{x}+3}+\sqrt{2\text{x}-3})}$
Rationalise the denominator
$=\int\frac{(\sqrt{2\text{x}+3}-\sqrt{2\text{x}}-3)}{(\sqrt{2\text{x}+3}+\sqrt{2\text{x}-3})(\sqrt{2\text{x}+3}-\sqrt{2\text{x}-3})}\text{dx}$
$=\int\frac{(\sqrt{2\text{x}+3}-\sqrt{2\text{x}-3})}{(2\text{x}+3)-(2\text{x}-3)}\text{dx}$
$=\frac{1}{6}\int(2\text{x}+3)^{\frac{1}{2}}\text{dx}-\frac{1}{6}\int(2\text{x}-3)^{\frac{1}{2}}\text{dx}$
$=\frac{1}{6}\Bigg[\frac{(2\text{x}+3)^{\frac{1}{2}+1}}{2\big(\frac{1}{2}+1\big)}\Bigg]-\frac{1}{6}\Bigg[\frac{(2\text{x}-3)^{\frac{1}{2}+1}}{2\big(\frac{1}{2}+1\big)}\Bigg]+\text{c}$
$=\frac{1}{18}\big\{(2\text{x}+3)^{\frac{3}{2}}-(2\text{x}-3)^{\frac{3}{2}}\big\}+\text{c}$
View full question & answer→Question 1883 Marks
Evaluate the following integrals:
$\int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{dx}$
Answer$\int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{dx}$
$=\int\frac{2\sin^2\text{x}}{2\cos^2\text{x}}\text{dx}$
$=\int\tan^2\text{x dx}$
$=\int(\sec^2\text{x}-1)\text{dx}$
$=\int\sec^2\text{x dx}-\int\text{dx}$
$=\tan\text{x}-\text{x}+\text{C}$
$\therefore\ \int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{dx}=\tan\text{x}-\text{x}+\text{C}$
View full question & answer→Question 1893 Marks
Evaluate the following integrals:
$\int\sin^5\text{x}\cos\text{x}\text{ dx}$
Answer$\int\sin^5\text{x}\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int\sin^5\text{x}\cos\text{x}\text{ dx}$
$=\int\text{t}^5\text{dt}$
$=\frac{\sin^6\text{x}}{6}+\text{C}$
$=\sin\text{x}+\frac{\sin^5\text{x}}{5}-\frac{2}{3}\sin^3\text{x}$
View full question & answer→Question 1903 Marks
Evalute the following integrals:
$\int\frac{\cos\text{x}}{\cos(\text{x}-\text{a})}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{\cos(\text{x}-\text{a})}\text{dx}$
Putting x - a = t
⇒ x = a + t
⇒ dx = dt
$\therefore\text{I}=\int\frac{\cos(\text{a}+\text{t})\text{dt}}{\cos\text{t}}$
$=\frac{\cos\text{a}\cos\text{t}}{\cos\text{t}}-\frac{\sin\text{a}\sin\text{t}}{\cos\text{t}}\text{dt}$
$=\int\big(\cos\text{a}-\sin\text{a}\tan\text{t}\big)\text{dt}$
$=\text{t}\cos\text{a}-\sin\text{a ln}|\sec\text{t}|+\text{C}$
$=(\text{x}-\text{a})\cos\text{a}-\sin\text{a ln}|\sec(\text{a}-\text{a})|+\text{C}\ \big[\text{t}=\text{x}-\text{a}\big]$
View full question & answer→Question 1913 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 1923 Marks
Write a value of $\int\text{x}^2\sin\text{x}^3\text{ dx}$
AnswerLet $\text{I}=\int\text{x}^2\sin\text{x}^3\text{ dx}$
Let $\text{x}^3=\text{t}$
$=3\text{x}^2\text{dx}=\text{dt}$
$=\text{x}^2\text{dx}=\frac{1}{3}\text{dt}$
$\therefore\ \text{I}=\frac{1}{3}\int\sin\text{t dt}$
$=\frac{1}{3}(-\cos\text{t})+\text{C}$
Hence, $\text{I}=\frac{-1}{3}\cos\text{x}^3+\text{C}$
View full question & answer→Question 1933 Marks
$\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{dx}$
$=\int\frac{\sec^2\text{x}}{(1-\tan\text{x})^2}\text{dx}$
$=\int\frac{\sec^2\text{x dx}}{(1-\tan\text{x})^2}$
$-\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\sec^2\text{x dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{dt}}{\text{t}^2}$
$=-\int\text{t}^{-2}\text{dt}$
$=-\bigg[\frac{\text{t}^{-2+1}}{-2+1}\bigg]+\text{c}$
$=\frac{1}{\text{t}}+\text{c}$
$=\frac{1}{1-\tan\text{x}}+\text{c}$
View full question & answer→Question 1943 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2+6\text{x}+13}\text{dx}$
AnswerWe have $\text{x}^2+6\text{x}+13=\text{x}^2+6\text{x}+3^2-3^2+13=(\text{x}+3)^2+4$
Sol, $\int\frac{\text{dx}}{\text{x}^2+6\text{x}+13}=\int\frac{1}{(\text{x}+3)^2+2^2}\text{dx}$
Let $\text{x}+3=\text{t}$ Then $\text{dx = dt}$
Therefore, $\int\frac{\text{dx}}{\text{x}^2+6\text{x}+13}=\int\frac{\text{dt}}{\text{t}^2+2^2}=\frac{1}{2}\tan^{-1}\frac{\text{t}}{2}+\text{c}$ [by 7.4 (3)]
$=\frac{1}{2}\tan^{-1}\frac{\text{x}+3}{2}+\text{C}$
View full question & answer→Question 1953 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x}}{\sec2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sec\text{x}}{\sec2\text{x}}\text{dx},$ then
$\text{I}=\int\frac{\frac{1}{\cos\text{x}}}{\frac{1}{\cos2\text{x}}}\text{dx}$
$=\int\frac{\cos2\text{x}}{\cos\text{x}}\text{dx}$
$=\int\frac{2\cos^2\text{x}-1}{\cos\text{x}}\text{dx}$
$=\int2\cos\text{ x dx}-\int\frac{1}{\cos\text{x}}\text{dx}$
$=2\int\cos\text{dx}-\int\sec\text{x dx}$
$=2\sin\text{x}-\log|\sec\text{x}+\tan\text{x}|+\text{C}$
$\because\text{I}=2\sin\text{x}-\log|\sec\text{x}+\tan\text{x}|+\text{C}$
View full question & answer→Question 1963 Marks
$\int\frac{1+\cos\text{x}}{1-\cos\text{x}}\text{dx}$
Answer$\int\Big(\frac{1+\cos\text{x}}{1-\cos\text{x}}\Big)\text{dx}$
$=\int\bigg(\frac{2\cos^2\frac{\text{x}}{2}}{2\sin^2\frac{\text{x}}{2}}\bigg)\text{dx}$ $\big[\therefore1+\cos\text{x}=2\cos^2\frac{\text{x}}{2} \&1-\cos\text{x}=2\sin^2\frac{\text{x}}{2}\big]$
$=\int\cot^2\frac{\text{x}}{2}\text{dx}$
$=\int\Big(\text{cosec}^2\frac{\text{x}}{2}-1\Big)\text{dx}$
$=\frac{-\cot\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}-\text{x+c}$
$=-2\cot\big(\frac{\text{x}}{2}\big)-\text{x+c}$
View full question & answer→Question 1973 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 1983 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 1993 Marks
Evaluate the following integrals:
$\int\text{x}^2\text{e}^{\text{x}^3}\cos\big(\text{e}^{\text{x}^3}\big)\text{dx}$
AnswerLet $\text{I}=\int\text{x}^2\text{e}^{\text{x}^3}\cos\big(\text{e}^{\text{x}^3}\big)\text{dx}\ ....(1)$ Let $\text{e}^{\text{x}^3}=\text{t}$ then, $\text{d}\big(\text{e}^{\text{x}^3}\big)=\text{dt}$ $\Rightarrow3\text{x}^2\text{e}^{\text{x}^3}\text{dx}=\text{dt}$ $\Rightarrow\text{x}^2\text{e}^{\text{x}^3}\text{dx}=\frac{\text{dt}}{3}$Putting $\text{e}^{\text{x}^3}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{3}$ in equation (1), we get
$\text{I}=\int\cos\text{t}\frac{\text{dt}}{3}$ $=\frac{\sin\text{t}}{3}+\text{C}$ $=\frac{\sin\big(\text{e}^{\text{x}^3}\big)}{3}+\text{C}$ $\text{I}=\frac{1}{3}\sin\big(\text{e}^{\text{x}^3}\big)+\text{C}$
View full question & answer→Question 2003 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→