Question 2515 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-1}\text{ dx}\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-1}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^3+\text{x}+1}{\text{x}^2-1}\ \text{dx}$
$=\int\Big(\text{x}+\frac{2\text{x}+1}{\text{x}^2-1}\Big)\text{dx}$
Now,
$\frac{2\text{x}+1}{\text{x}^2-1}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{\text{x}-1}$
$\Rightarrow2\text{x}+1=\text{A}(\text{x}-1)+\text{B}(\text{x}+1)$
put x = 1
$\Rightarrow3=2\text{B}\Rightarrow\text{B}=\frac{3}{2}$
put x = -1
$\Rightarrow-1=-2\text{A}\Rightarrow\text{A}=\frac{1}{2}$
$\therefore\text{I}=\int\text{xdx}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}+\frac{3}{2}\int\frac{\text{dx}}{\text{x}-1}$
$\text{I}=\frac{\text{x}^2}{2}+\frac{1}{2}\log|\text{x}+1|+\frac{3}{2}\log|\text{x}-1|+\text{C}$
View full question & answer→Question 2525 Marks
$\int\frac{1}{\text{x}^{\frac{1}{3}}\big(\text{x}^{\frac{1}{3}}-1\big)}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{x}^{\frac{1}{3}}\big(\text{x}^{\frac{1}{3}}-1\big)}\text{dx}$
$=\int\frac{1}{\text{x}^{\frac{2}{3}}-\text{x}^{\frac{1}{3}}}\text{dx}$
Let $\text{x}=\text{t}^{3}$
On differentiating both sides, we get
$\text{dx}=3\text{t}^{2}\text{dt}$
$\therefore\ \text{I}\int\frac{3\text{t}^{2}}{(\text{t})^{\frac{2}{3}}-(\text{t}^{3})^{\frac{1}{3}}}\text{dt}$
$=\int\frac{3\text{t}^{2}}{\text{t}^2-\text{t}}\text{dt}$
$=3\int\frac{\text{t}}{\text{t}-1}\text{dt}$
$=3\int\frac{(\text{t}-1)+1}{\text{t}-1}\text{dt}$
$=3\int\Big[(1)+\frac{1}{\text{t}-1}\Big]\text{dt}$
$=\big[1+\log(\text{t}-1)\big]+\text{C}$
$=3\text{x}^\frac{1}{3}+3\log\big({\text{x}^\frac{1}{3}-1\big)}+\text{C}$
Hence, $\int\frac{1}{\text{x}^{\frac{1}{3}}\big(\text{x}^{\frac{1}{3}}-1\big)}\text{dx}=3\text{x}^\frac{1}{3}+3\log\big({\text{x}^\frac{1}{3}-1\big)}+\text{C}$
View full question & answer→Question 2535 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{7-6\text{x}-\text{x}^2}}\text{ dx}$
Answer$7 - 6x - x^2$ can be written as $7 - (x^2 + 6x + 9 - 9)$.
Therefore, $7-(\text{x}^2+6\text{x}+9-9)$
$=16-(\text{x}^2+6\text{x}+9)$ $=16-(\text{x}+3)^2$
$=(4)^2-(\text{x}+3)^2$
$\therefore\ \int\frac{1}{\sqrt{7-6\text{x}-\text{x}^2}}\text{ dx}$
$=\int\frac{1}{\sqrt{(4)^2-(\text{x}+3)^2}}\text{ dx}$ Let x + 3 = t $\Rightarrow\text{dx}=\text{dt}$
$\Rightarrow\int\frac{1}{\sqrt{(4)^2-(\text{x}+3)^2}}\text{ dx}=\int\frac{1}{\sqrt{(4)^2-(\text{t})^2}}\text{ dt}$
$=\sin^{-1}\Big(\frac{\text{t}}{4}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\text{x}+3}{4}\Big)+\text{C}$
View full question & answer→Question 2545 Marks
Evaluate the follwing intregals:
$\int\frac{2\text{x}}{(\text{x}^2+1)(\text{x}^2+2)^2}\ \text{dx}$
Answer$\int\frac{2\text{x}}{(\text{x}^2+1)(\text{x}^2+2)^2}\ \text{dx}$
Let $\text{x} ^2=\text{y}$
$\Rightarrow2\text{x dx}=\text{dy}$
$\Rightarrow\text{dx}=\frac{\text{dy}}{2\text{x}}$
$\int\frac{2\text{x}}{(\text{x}^2+1)(\text{x}^2+2)^2}\ \text{dx}$
$=\int\frac{\text{dy}}{(\text{y}+1)(\text{y}+2)^2}$
Let $\frac{1}{(\text{y}+1)(\text{y}+2)^2}=\frac{\text{A}}{\text{y}+1}+\frac{\text{B}}{\text{y}+2}+\frac{\text{C}}{(\text{y}+2)^2}\ ...(1)$
$\Rightarrow1=\text{A}(\text{y}+2)^2+\text{B}(\text{y}+1)(\text{y}+2)+\text{C}(\text{y}+1)\ ...(2)$
Putting y = -2 in (2)
$1 = C (-2 + 1)$
$⇒ C = -1$
Putting y = -1 in (2)
$1 = A (-1 + 2)^2$
$\Rightarrow 1 = A (1)$
$\Rightarrow A = 1$
Putting y = 0 in (2)
$1 = 4A + B(2) + C$
$⇒ 1 = 4 + 2B - 1$
$⇒ -2 = 2B$
$⇒ B = -1$
Substituting the values of A, B and C in (1)
$\frac{1}{(\text{y}+1)(\text{y}+2)^2}=\frac{1}{\text{y}+1}-\frac{1}{\text{y}+2}-\frac{1}{(\text{y}+2)^2}$
$\Rightarrow\int\frac{\text{dy}}{(\text{y}+1)(\text{y}+2)^2}=\int\frac{\text{dy}}{\text{y}+1}-\int\frac{\text{dy}}{\text{y}+2}-\int\frac{\text{dy}}{(\text{y}+2)^2}$
$=\log|\text{y}+1|-\log|\text{y}+2|+\frac{1}{\text{y}+2}+\text{C}$
Hence, $\int\frac{2\text{x}}{(\text{x}^2+1)(\text{x}^2+2)^2}\ \text{dx}=\log|\text{x}^2+1|-\log|\text{x}^2+2|+\frac{1}{\text{x}^2+2}+\text{C}$
View full question & answer→Question 2555 Marks
Evaluate the following integrals:
$\int\cos^3\sqrt{\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\cos^3\sqrt{\text{x}}\text{dx}$
Let $\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt }$
$=2\int\text{t}\cos^3\text{t dt}$
$=2\int\text{t}\Big(\frac{3\cos\text{t}+\cos3\text{t}}{4}\Big)\text{dt}$
$=\frac{1}{2}\int\text{t}(3\cos\text{t}+\cos3\text{t})\text{dt}$
Using integral\tion by parts,
$\text{I}=\frac{1}{2}\Big[\text{t}\Big(3\sin\text{t}+\frac{1}{3}\sin3\text{t}\Big)+\int\Big(1\times3\sin\text{t}+\frac{\sin3\text{t}}{3}\Big)\text{dt}\Big]$
$=\frac{1}{2}\Big[\text{t}\Big(\frac{9\sin\text{t}+\sin3\text{t}}{3}\Big)+3\cos\text{t}+\frac{\cos3\text{t}}{9}\Big]+\text{C}$
$=\frac{1}{18}\big[27\text{t}\sin\text{t}+3\text{t}\sin3\text{t}+9\cos\text{t}+\cos3\text{t}\big]+\text{C}$
$\text{I}=\frac{1}{18}\big[27\sqrt{\text{x}}\sin\sqrt{\text{x}}+3\sqrt{\text{x}}\sin3\sqrt{\text{x}}+9\cos\sqrt{\text{x}}+\cos3\sqrt{\text{x}}\big]+\text{C}$
View full question & answer→Question 2565 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}\ \text{dx}$
AnswerConsider the integrals
$\text{I}=\int\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}\ \text{dx}$
Let $y = x^2$
Thus,
$\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}=\frac{\text{y}+1}{(\text{y}+4)(\text{y}+25)}$
$\Rightarrow\frac{\text{y}+1}{(\text{y}+4)(\text{y}+25)}=\frac{\text{A}}{\text{y}+4}+\frac{\text{B}}{\text{y}+25}$
$\Rightarrow\frac{\text{y}+1}{(\text{y}+4)(\text{y}+25)}=\frac{\text{A}(\text{y}+25)+\text{B}(\text{y}+4)}{(\text{y}+4)(\text{y}+25)}$
$\Rightarrow\text{y}+1=\text{Ay}+25\text{A}+\text{By}+4\text{B}$
Compairing the coefficient, we have,
$A + B = 1$ and $25A + 4B = 1$
Solving the above equations, we have,
$\text{A}=\frac{-1}{7}\text{ and }\text{B}=\frac{8}{7}$
Thus, $\int\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}\ \text{dx}$
$=\int\frac{\frac{1}{7}}{\text{x}^2+4}\ \text{dx}+\int\frac{\frac{8}{7}}{\text{x}^2+25}\ \text{dx}$
$=\frac{-1}{7}\int\frac{1}{\text{x}^2+4}\text{ dx}+\frac{8}{7}\int\frac{1}{\text{x}^2+25}\ \text{dx}$
$=\frac{-1}{7}\times\frac{1}{2}\tan^{-1}\frac{\text{x}}{2}+\frac{8}{7}\times\frac{1}{5}\tan^{-1}\frac{\text{x}}{5}+\text{C}$
$=\frac{-1}{14}\tan^{-1}\frac{\text{x}}{2}+\frac{8}{35}\tan^{-1}\frac{\text{x}}{5}+\text{C}$
View full question & answer→Question 2575 Marks
Evaluate the following integrals:
$\int\frac{1}{1+\sqrt{\text{x}}}\text{dx}$
View full question & answer→Question 2585 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{x}}}{\text{x}}\Big\{\text{x}(\log\text{x})^2+2\log\text{x}\Big\}\text{dx}$
AnswerWe have,
$\text{I}=\int\frac{\text{e}^{\text{x}}}{\text{x}}\Big\{\text{x}(\log\text{x})^2+2\log\text{x}\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}\Big\{(\log\text{x})^2+\frac{2}{\text{x}}\log\text{x}\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}(\log\text{x})^2+2\int\frac{\text{e}^{\text{x}}}{\text{x}}\log\text{x dx}$
Integrating by parts
$=\text{e}^{\text{x}}(\log\text{x})^2-\int\text{e}^{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})^2\text{dx}+2\int\text{e}^{\text{x}}\frac{1}{\text{x}}\log\text{x dx}$
$=\text{e}^{\text{x}}(\log\text{x})^2-\int\text{e}^{\text{x}}\frac{2\log\text{x}}{\text{x}}\text{dx}+2\int\text{e}^\text{x}\frac{\log\text{x}}{\text{x}}\text{dx}$
$=\text{e}^{\text{x}}(\log\text{x)}^2+\text{C}$
View full question & answer→Question 2595 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\sin\text{x}\cos\text{x }\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}\sin\text{x}\cos\text{x }\text{dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}2\sin\text{x}\cos\text{x dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}\sin2\text{x dx}$
We know that
$\int\text{e}^{2\text{x}}\sin\text{bx dx}=\frac{\text{e}^{2\text{x}}}{\text{a}^2+\text{b}^2}\{\text{a}\sin\text{bx}-\text{b}\cos\text{bx}\}+\text{C}$
$\Rightarrow\int\text{e}^{2\text{x}}\sin\text{2x dx}=\frac{\text{e}^{2\text{x}}}{8}\{2\sin2\text{x}-2\cos2\text{x}\}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}.\frac{\text{e}^{2\text{x}}}{8}\{2\sin2\text{x}-2\cos2\text{x}\}+\text{C}$
$\therefore\ \text{I}=\frac{\text{e}^{2\text{x}}}{8}\{\sin2\text{x}-\cos2\text{x}\}+\text{C}$
View full question & answer→Question 2605 Marks
Evaluate the following intregals:
$\int\frac{18}{(\text{x}+2)(\text{x}^2+4)}\text{ dx}$
AnswerLet $\frac{18}{(\text{x}+2)(\text{x}^2+4)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{Bx}+\text{C}}{\text{x}^2+4}$$\Rightarrow18=\text{A}(\text{x}^2+4)+(\text{Bx}+\text{C})(\text{x}+2)$
$18=(\text{A}+\text{B})\text{x}^2+(2\text{B}+\text{C})\text{x}+(4\text{A}+2\text{C})$
Equating similar terms, get,
$\text{A}+\text{B}=0,2\text{B}+\text{C}=0,4\text{A}+2\text{C}=18$
Solving, we get,
$\text{A}=\frac{9}{4},\text{B}=-\frac{9}{4},\text{C}=\frac{9}{2}$
Thus,
$\text{I}=\frac{9}{4}\int\frac{\text{dx}}{\text{x}+2}+\Big(-\frac{9}{4}\Big)\frac{\text{x}}{\text{x}^2+4}\ \text{dx}+\frac{9}{2}\int\frac{\text{dx}}{\text{x}^2+4}$
$\text{I}=\frac{9}{4}\log|\text{x}+2|-\frac{9}{8}\log|\text{x}^2+4|+\frac{9}{4}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$ $\Big[\because\int\frac{\text{dx}}{\text{x}^2+\text{a}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{9}\Big]$
View full question & answer→Question 2615 Marks
Evaluate the following intregals:
$\int\frac{2}{2+\sin^22\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{2}{2+\sin^22\text{x}}\text{ dx}$
$\text{I}=\int\frac{2}{2+2\sin\text{x}\cos\text{x}}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{\frac{1}{\cos^2\text{x}}+\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{\sec^2\text{x}+\tan\text{x}}\ \text{dx}$
$\text{I}=\int\frac{\sec^2\text{x}}{1+\tan^2\text{x}+\tan\text{x}}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2+\text{t}+1}$
$=\int\frac{\text{dt}}{\text{t}^2+2\text{t}\Big(\frac{1}{2}\Big)+\Big(\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2+1}$
$\text{I}=\int\frac{\text{dt}}{\Big(\text{t}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}$
$=\frac{1}{\frac{\sqrt{3}}{2}}\tan^{-1}\Bigg(\frac{\text{t}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg)+\text{C}$
$=\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{t}+1}{\sqrt{3}}\Big)+\text{C}$
$\text{I}=\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{2\tan\text{x}+1}{\sqrt{3}}\Big)+\text{C}$
View full question & answer→Question 2625 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^7}{(\text{a}^2-\text{x}^2)^5}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^7}{(\text{a}^2-\text{x}^2)^5}\text{ dx}$
Let $\text{x}=\text{a}\sin\theta$
On differentiating both sides, we get
$\text{dx}=\text{x}\cos\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{\text{a}^8\sin^{7}\theta\cos\theta}{(\text{a}^2-\text{a}^2\sin^{2}\theta)^5}\text{ d}\theta$
$=\int\frac{\text{a}^8\sin^{7}\theta\cos\theta}{\text{a}^{10}(1-\sin^2\theta)^5}\text{ d}\theta$
$=\int\frac{\sin^7\theta}{\text{a}^2\cos^9\theta}\text{ d}\theta$
$=\frac{1}{\text{a}^2}\int\tan^7\theta\sec^2\theta\text{ d}\theta$
Let $\tan\theta=\text{t}$
On differentiating both sides, we get
$\sec^2\theta\text{ d}\theta=\text{dt}$
$\therefore\ \text{I}=\frac{1}{\text{a}^2}\int\text{t}^7\text{dt}$
$=\frac{1}{\text{a}^2}\frac{\text{t}^8}{8}+\text{C}$
$=\frac{1}{8\text{a}^2}(\tan^8\theta)+\text{C}$
$=\frac{1}{8\text{a}^2}\Big(\tan\Big(\sin^{-1}\frac{\text{x}}{\text{a}}\Big)\Big)^8+\text{C}$
$=\frac{1}{8\text{a}^2}\Big(\tan\Big(\tan^{-1}\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big)\Big)^8+\text{C}$
$=\frac{1}{8\text{a}^2}\frac{\text{x}^8}{(\text{a}^2-\text{x}^2)^4}+\text{C}$
Hence, $\int\frac{\text{x}^7}{(\text{a}^2-\text{x}^2)^5}\text{ dx}=\frac{1}{8\text{a}^2}\frac{\text{x}^8}{(\text{a}^2-\text{x}^2)^4}+\text{C}$
View full question & answer→Question 2635 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2-1}{\text{x}^4+1}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2-1}{\text{x}^4+1}\ \text{dx}$
Dividing numerator and denominator bt $x^2$
$\therefore\text{I}=\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)}{\text{x}^2+\frac{1}{\text{x}^2}}\ \text{dx}$
$=\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)}{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2}$
Let $\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{t}$
$\Rightarrow\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2-2}$
$=\frac{1}{2\sqrt{2}}\log\Big|\frac{\text{t}-\sqrt{2}}{\text{t}+\sqrt{2}}\Big|+\text{C}$
So,
$\text{I}=\frac{1}{2\sqrt{2}}\log\Big|\frac{\text{x}^2+1-\sqrt{2}\text{x}}{\text{x}^2+1+\sqrt{2}\text{x}}\Big|+\text{C}$
View full question & answer→Question 2645 Marks
$\int\text{x}\sqrt{\text{x}+2}\ \text{dx}$
AnswerLet I $=\int\text{x}\sqrt{\text{x}+2}\text{ dx}.$ Then,
$\text{I}=\int\{(\text{x}+2)-2\}\text{x}+2\text{dx}\ \ \ [\because\text{x}=(\text{x}+2)-2]$
$\Rightarrow\text{I}=\int\Big\{(\text{x}+2)^\frac{3}{2}-2(\text{x}+2)^\frac{1}{2}\Big\}\text{dx}$
$\Rightarrow\text{I}=\frac{2}{5}(\text{x}+2)^\frac{5}{2}-\frac{4}{3}(\text{x}+2)^\frac{3}{2}+\text{C}$
View full question & answer→Question 2655 Marks
Evaluate the following integrals:
$\int\big\{\tan(\log\text{x})+\sec^2(\log\text{x})\big\}\text{dx}$
AnswerLet $\text{I}=\int\big\{\tan(\log\text{x})+\sec^2(\log\text{x})\big\}\text{dx}$
Let $\log\text{x}=\text{z}$
$\Rightarrow\text{x = e}^{\text{z}}$
$\Rightarrow\text{dx}=\text{e}^{\text{z}}\text{dz}$
$\therefore\text{I}=\int\big\{\tan\text{z}+\sec^2\text{z}\big\}\text{e}^{\text{z}}\text{dz}$
Here, $\text{f(z)}=\tan\text{z}$ and $\text{f}'\text{(z)}=\sec^2\text{z}$
And we know that
$\int\text{e}^{\text{ax}}(\text{af(x)}+\text{f}'(\text{x}))\text{dx}=\text{e}^{\text{ax}}\text{f(x) + C}$
$\therefore\int\text{e}^{\text{z}}\big\{\tan\text{z}+\sec^2\text{z}\big\}\text{dz}=\text{e}^{\text{z}}\tan\text{z + C}$
$\therefore\text{I}=\text{x}\tan(\log\text{x})+\text{C}$
View full question & answer→Question 2665 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2}{(\text{x}^2+4)(\text{x}^2+9)}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{(\text{x}^2+4)(\text{x}^2+9)}\ \text{dx}$
We express
$\frac{\text{x}^2}{(\text{x}^2+4)(\text{x}^2+9)}=\frac{\text{Ax}+\text{B}}{\text{x}^2+4}+\frac{\text{Cx}+\text{D}}{\text{x}^2+9}$
$\Rightarrow\text{x}^2=(\text{Ax}+\text{B})(\text{x}^2+9)+(\text{Cx}+\text{D})(\text{x}^2+4)$
Equating the coefficient of $x^3, x^2, x $ and constants, we get
$0 = A + C$ and $1 = B + D$ and $0 = 9A + 4C$
and $0 = 9B + 4D $ or $A = 0$ and or $A = 0$ and
$\text{B}=-\frac{4}{5}\text{ and }\text{C}=0,\text{ D }=\frac{9}{5}$
$\therefore\text{I}=\int\bigg(\frac{-\frac{4}{5}}{\text{x}^2+4}+\frac{\frac{9}{5}}{\text{x}^2+9}\bigg)\text{dx}$
$=-\frac{4}{5}\int\frac{1}{\text{x}^2+4}\text{ dx}+\frac{9}{5}\int\frac{1}{\text{x}^2+9}\ \text{dx}$
$=-\frac{4}{5}\times\frac{1}{2}\tan^{-1}\frac{\text{x}}{2}+\frac{9}{5}\times\frac{1}{3}\tan^{-1}\frac{\text{x}}{3}+\text{C}$
$=-\frac{2}{5}\tan^{-1}\frac{\text{x}}{2}+\frac{3}{5}\tan^{-1}\frac{\text{x}}{3}+\text{C}$
Hence, $\int\frac{\text{x}^2}{(\text{x}^2+4)(\text{x}^2+9)}\ \text{dx}=-\frac{2}{5}\tan^{-1}\frac{\text{x}}{2}+\frac{3}{5}\tan^{-1}\frac{\text{x}}{3}+\text{C}$
View full question & answer→Question 2675 Marks
Evaluate the following integrals:
$\int\frac{\text{x}\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{ dx}\ ....(1)$ Let $\sin^{-1}\text{x}^2=\text{t}$ then, $\text{d}\big(\sin^{-1}\text{x}^2\big)=\text{dt}$ $\Rightarrow2\text{x}\times\frac{1}{\sqrt{1-\text{x}^4}}\text{ dx}=\text{dt}$ $\Rightarrow\frac{\text{x}}{\sqrt{1-\text{x}^4}}\text{ dx}=\frac{\text{dt}}{2}$ Putting $\sin^{-1}\text{x}^2=\text{t}$ and $\frac{\text{x}}{\sqrt{1-\text{x}}^4}\text{ dx}=\frac{\text{dt}}{2}$ in equation (1),We get,
$\text{I}=\int\text{t}\frac{\text{dt}}{2}$ $=\frac{1}{2}\times\frac{\text{t}^2}{2}+\text{C}$ $=\frac{1}{4}\big(\sin^{-1}\text{x}^2\big)^2+\text{C}$ $\text{I}=\frac{1}{4}\big(\sin^{-1}\text{x}^2\big)^2+\text{C}$
View full question & answer→Question 2685 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$
AnswerWe have$\text{I}=\int\frac{(\text{x}^2+\text{x}+1)}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$
Let, $\int\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{C}}{\text{x}+2}$
$\Rightarrow\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}=\frac{\text{A}(\text{x}+1)(\text{x}+2)+\text{B}(\text{x}+2)+\text{C}(\text{x}+1)^2}{(\text{x}+1)^2(\text{x}+2)}$
$\Rightarrow\text{x}^2+\text{x}+1=\text{A}(\text{x}^2+\text{x}+2\text{x}+2)\\+\text{Bx}+2\text{B}+\text{C}(\text{x}^2+2\text{x}+1)$
$\Rightarrow\text{x}^2+\text{x}+1=(\text{A}+\text{C})\text{x}^2+(3\text{A}+\text{B}+2\text{C})\text{x}\\+(2\text{A}+2\text{B}+\text{C})$
Equating coefficient of like terms,
$\text{A}+\text{C}=1\ ...(1)$
$3\text{A}+\text{B}+2\text{C}=1\ ...(2)$
$2\text{A}+2\text{B}+\text{C}=1\ ...(3)$
Solving these three equation we get
$\text{A}=-2$
$\text{B}=1$
$\text{C}=3$
Hence, $\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}=\frac{-2}{\text{x}+1}+\frac{1}{(\text{x}+1)^2}+\frac{3}{\text{x}+2}$
$\therefore\text{I}=-2\int\frac{\text{dx}}{\text{x}+1}+\int\frac{\text{d}}{(\text{x}+1)^2}+3\int\frac{\text{dx}}{\text{x}+2}$
$=-2\log|\text{x}+1|-\frac{1}{\text{x}+1}+3\log|\text{x}+2|+\text{C}$
View full question & answer→Question 2695 Marks
Evaluate the following integrals:
$\int\cos^{-1}(4\text{x}^3-3\text{x})\text{dx}$
AnswerLet $\text{I}=\int\cos^{-1}(4\text{x}^3-3\text{x})\text{dx}$
Let $\text{x}=\cos\theta$
$\text{dx }=-\sin\theta\text{d}\theta$
$\text{I}=-\int\cos^{-1}(4\cos^3\theta-3\cos\theta)\sin\theta\text{d}\theta$
$=-\int\cos^{-1}(\cos3\theta)\sin\theta\text{d}\theta$
$=-\int3\theta\sin\theta\text{d}\theta$
$=-3[\theta\int\sin\theta\text{d}\theta-\int(1\int\sin\theta\text{d}\theta)\text{d}\theta]$
$=-3[-\theta\cos\theta+\int\cos\theta\text{d}\theta]$
$=3\theta\cos\theta-3\sin\theta+\text{C}$
$\text{I}=3\text{x}\cos^{-1}\text{x}-3\sqrt{1-\text{x}^2}+\text{C}$
View full question & answer→Question 2705 Marks
$\int\frac{\text{x}+1}{\sqrt{2\text{x}+3}}\text{dx}$
Answer$\int\Big(\frac{\text{x}+1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\frac{2\text{x}+2}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\frac{2\text{x}+3-1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\frac{2\text{x}+3}{\sqrt{2\text{x}+3}}-\frac{1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\int\Big(\sqrt{2\text{x}+3}-\frac{1}{\sqrt{2\text{x}+3}}\Big)\text{dx}$
$=\frac{1}{2}\Big[\int(2\text{x}+3)^\frac{1}{2}\text{dx}-\int(2\text{x}+3)^{-\frac{1}{2}}\text{dx}\Big]$
$=\frac{1}{2}\Bigg[\frac{(2\text{x}+3)^{\frac{1}{2}+1}}{2\big(\frac{1}{2}+1\Big)}-\frac{(2\text{x}+3)^{-\frac{1}{2}+1}}{2\big(-\frac{1}{2}+1\big)}+\text{C}\Bigg]$
$=\frac{1}{2}\Big[\frac{1}{3}(2\text{x}+3)^\frac{3}{2}-(2\text{x}+3)^\frac{1}{2}+\text{C}\Big]$
$=\frac{1}{6}(2\text{x}+3)^\frac{3}{2}-\frac{1}{2}(2\text{x}+3)^\frac{1}{2}+\text{C}$
View full question & answer→Question 2715 Marks
Write a value of $\int\text{e}^{\text{ax}}\cos\text{bx}\text{ dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{ax}}\cos\text{bx}\text{ dx}$
$=\cos\text{bx}\int\text{e}^{\text{ax}}\text{ dx}-\Big\{\frac{\text{d}}{\text{dx}}(\cos\text{bx})\int\text{e}^{\text{ax}}\text{ dx}\Big\}\text{dx}$
$=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}-\int-\sin\text{bx}\cdot\text{b}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}$
$=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}}{\text{a}}\int\text{e}^{\text{ax}}\cdot\sin\text{bx}\text{ dx}$
$=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}}{\text{a}}\text{I}_1\ ....(\text{i})$
$\therefore\ \text{I}_1=\int\text{e}^{\text{ax}}\cdot\sin\text{bx}\text{ dx}$
$=\sin\text{bx}\int\text{e}^{\text{ax}}\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\sin\text{bx})\int\text{e}^{\text{ax}}\text{dx}\Big\}\text{dx}$
$=\sin\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}-\int\text{b}\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}\text{ dx}$
$=\sin\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}-\frac{\text{b}}{\text{a}}\text{I}\ ....(\text{ii})$
From (i) & (ii)
$\therefore\ \text{I}=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}}{\text{a}}\Big\{\sin\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}-\frac{\text{b}}{\text{a}}\text{I}\Big\}$
$\text{I}=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}}{\text{a}^2}\sin\text{bx}\text{e}^{\text{ax}}-\frac{\text{b}^2}{\text{a}^2}\text{I}$
$\text{I}+\frac{\text{b}^2}{\text{a}^2}\text{I}=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}\sin\text{bx}\text{e}^{\text{ax}}}{\text{a}^2}$
$\big(\text{a}^2+\text{b}^2\big)\text{I}=(\text{a}\cos\text{bx}+\text{b}\sin\text{bx})\text{e}^{\text{ax}}$
$\text{I}=\frac{(\text{a}\cos\text{bx}+\text{b}\sin\text{bx})\text{e}^{\text{ax}}}{\text{a}^2+\text{b}^2}+\text{C}$
View full question & answer→Question 2725 Marks
$\int\frac{2\text{x}+1}{\sqrt{3\text{x}+2}}\text{dx}$
Answer$\text{Let I}=\int\frac{2\text{x}+1}{\sqrt{3\text{x}+2}}\text{dx}$
Let $2\text{x}+1=\lambda(3\text{x}+2)+\mu$ On equating the coefficients of like powers of x on both sides, we get
$3\lambda=2\text{ and }2\lambda+\mu=1$
$\Rightarrow\lambda=\frac{2}{3}\text{ and }2\times\frac{2}{3}+\mu=1$
$\Rightarrow\lambda=\frac{2}{3}\text{ and }\mu=\frac{-1}{3}$
$\therefore\text{I}=\int\frac{\lambda(3\text{x}+2)+\mu}{\sqrt{3\text{x}+2}}\text{dx}$
$=\lambda\int\frac{3\text{x}+2}{\sqrt{3\text{x}+2}}\text{dx}+\mu\int\frac{1}{\sqrt{3\text{x}+2}}\text{dx}$
$=\lambda\int(3\text{x}+2)^\frac{1}{2}\text{dx}+\mu\int(3\text{x}+2)^\frac{-1}{2}\text{dx}$
$=\lambda\times\frac{(3\text{x}+2)^\frac{3}{2}}{\frac{3}{2}\times3}+\mu\frac{(3\text{x}+2)^\frac{1}{2}}{\frac{1}{2}\times3}\text{c}$
$=\frac{2}{3}\times\frac{2}{9}\times(3\text{x}+2)^\frac{3}{2}-\frac{1}{3}\times\frac{2}{3}(3\text{x}+2)^\frac{1}{2}+\text{c}$
$=\frac{4}{27}\times(3\text{x}+2)^\frac{3}{2}-\frac{2}{9}\times(3\text{x}+2)^\frac{1}{2}+\text{c}$
$=\frac{2}{9}\times\sqrt{3\text{x}+2}\Big[\frac{2}{3}\times(3\text{x}+2)-1\Big]+\text{c}$
$=\frac{2}{9}\times\sqrt{3\text{x}+2}\Big[\frac{6\text{x}+4-3}{3}\Big]+\text{c}$
$=\frac{2}{27}\times\sqrt{3\text{x}+2}(6\text{x}+1)+\text{c}$
$\therefore\text{I}=\frac{2}{27}\times(6\text{x}+1)\sqrt{3\text{x}+2}+\text{c}$
View full question & answer→Question 2735 Marks
Evaluate the following integrals:$\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}$Let the first function be (log x) and second function be $\frac{1}{(\text{x}+1)^2}.$
First we find the integral of the second function, i.e, $\int\frac{1}{(\text{x}+1)^2}\text{dx}.$
Put t = (x + 1) Then dt = dx
Therefore,
$\int\frac{1}{(\text{x}+1)^2}\text{dx}=\int\text{t}^{-2}\text{dt}$
$=-\frac{1}{\text{t}}$
$=-\frac{1}{1+\text{x}}$
Hence, using integration by parts, we get
$\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}=(\log\text{x})\int\frac{1}{(\text{x}+1)^2}\text{dx}-\int\Big[\Big(\frac{\text{d}(\log\text{x})}{\text{dx}}\Big)\int\frac{1}{(\text{x}+1)^2}\text{dx}\Big]\text{dx}$
$=(\log\text{x})\Big(-\frac{1}{1+\text{x}}\Big)-\int\big(\frac{1}{\text{x}}\big)\Big(-\frac{1}{1+\text{x}}\Big)\text{dx}$
$=-\frac{\log\text{x}}{1+\text{x}}+\int\Big(\frac{1}{\text{x}^2+\text{x}}\Big)\text{dx}$
$=-\frac{\log\text{x}}{1+\text{}x}+\int\frac{1}{\text{x}^2+\text{x}+\frac{1}{4}-\frac{1}{4}}\text{dx}$
$=-\frac{\log\text{x}}{1+\text{x}}+\int\frac{1}{\big(\text{x}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\text{dx}$
$=-\frac{\log\text{x}}{1+\text{x}}+\frac{1}{2\times\frac{1}{2}}\log\Bigg|\frac{\text{x}+\frac{1}{2}-\frac{1}{2}}{\text{x}+\frac{1}{2}+\frac{1}{2}}\Bigg|+\text{C}$
$=-\frac{\log\text{x}}{1+\text{x}}+\log\Big|\frac{\text{x}}{\text{x}+1}\Big|+\text{C}$
Hence, $\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}=-\frac{\log\text{x}}{1+\text{x}}+\log\Big|\frac{\text{x}}{\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 2745 Marks
Evaluate the following intregals:
$\int\frac{2\text{x}+1}{\sqrt{\text{x}^2+2\text{x}-1}}\ \text{dx}$
AnswerLet $\text{I}\int\frac{2\text{x}+1}{\sqrt{\text{x}^2+2\text{x}-1}}\ \text{dx}$Let $2\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+2\text{x}-1)+\mu$
$=\lambda(2\text{x}+2)+\mu$
$2\text{x}+1=(2\lambda)\text{x}+2\lambda+\mu$
Comapring the coefficient of like powers of x,
$2\lambda=2\Rightarrow\lambda=1$
$2\lambda+\mu=1\Rightarrow2(1)+\mu=1$
$\mu=-1$
So, $\text{I}=\int\frac{(2\text{x}+2)-1}{\sqrt{\text{x}^2+2\text{x}-1}}\text{dx}$
$\text{I}=\int\frac{(2\text{x}+2)}{\sqrt{\text{x}^2+2\text{x}-1}}\text{dx}-\int\frac{1}{\sqrt{\text{x}^2+2\text{x}+(1)^2-(1)^2-1}}$
$\text{I}=\int\frac{2\text{x}+2}{\sqrt{\text{x}^2+2\text{x}-1}}\text{dx}-\int\frac{1}{\sqrt{(\text{x}+1)^2-(\sqrt{2}^2})}$
$\text{I}=(2\sqrt{\text{x}^2+2\text{x}-1})-\log\big|(\text{x}+1)+\sqrt{(\text{x}+1)^2-(\sqrt{2}})^2\big|+\text{C}$ $\Big[\text{since},\int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{x}^2-\text{a}^2}}\text{dx}=\log\big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\big|+\text{C}\Big]$
$\text{I}=2\sqrt{\text{x}^2+2\text{x}-1}-\log\big|\text{x}+1+\sqrt{\text{x}^2+2\text{x}-1}\big|+\text{c}$
View full question & answer→Question 2755 Marks
Write a value of $\int\text{e}^{\log\sin\text{x}}\cos\text{x}\text{ dx}$
Answer$\int\text{e}^{\log\sin\text{x}}\cos\text{x}\text{ dx}$
Let $\text{t}=\sin\text{x}\rightarrow\text{dt}=\cos\text{x dx}$
$\int\text{e}^{\log\sin\text{x}}\cos\text{x}\text{ dx}=\int\text{e}^{\log\text{t}}\text{dt}=\text{I}$
$\text{e}^{\log\text{t}}\int1\text{dt}-\Big(\int\frac{\text{de}^{\log\text{t}}}{\text{dt}}\big(\int1\text{dt}\big)\text{dt}\Big)$
$=\text{e}^{\log\text{t}}\text{t}-\Big(\int\text{e}^{\log\text{t}}\frac{1}{\text{t}}\text{t dt}\Big)$
$=\text{e}^{\log\text{t}}\text{t}-\big(\int\text{e}^{\log\text{t}}\text{dt}\big)=\text{I}$
$\rightarrow\text{e}^{\log\text{t}}\text{t}-\text{I}=\text{I}\rightarrow2\text{I}=\text{e}^{\log\text{t}}+\text{C}$
$\text{I}=\frac{1}{2}\Big[\text{te}^{\log\text{t}}\Big]+\text{C}$
Substitute back $\text{t}=\sin\text{x}$ in above expression
We get, $\text{I}=\frac{1}{2}\big[\sin{\text{x}}\text{e}^{\log\sin\text{x}}\big]+\text{C}$
$=\frac{\sin^2\text{x}}{2}+\text{C}$ $[\because\log$ with base 10 term can be changed to in (natural log) term along with a constant$]$
View full question & answer→Question 2765 Marks
Evaluate the following integrals:$\int\frac{\sqrt{1-\sin\text{x}}}{1+\cos\text{x}}\text{e}^{\frac{-\text{x}}{2}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sqrt{1-\sin\text{x}}}{1+\cos\text{x}}\text{e}^{\frac{-\text{x}}{2}}\text{dx}$
Put $=\frac{\text{x}}{2}=\text{t}$
$\Rightarrow\text{x}=2\text{t}$
$\text{dx}=2\text{dt}$
$\therefore\int\frac{\sqrt{1-\sin\text{x}}}{1+\cos\text{x}}\text{e}^{-\frac{\text{x}}{2}}\text{dx}$
$=2\int\frac{\sqrt{1-\sin2\text{t}}}{1+\cos2\text{t}}\text{e}^{-\text{t}}\text{dt}$ $\big[\because\sin^2\text{t}+\cos^2\text{t}=1\big]$
$=2\int\frac{\sqrt{\sin^2\text{t}+\cos^2\text{t}-2\sin\text{t}\cos\text{t}}}{1+\cos2\text{t}}\text{e}^{-\text{t}}\text{dt}$
$=2\int\frac{\sqrt{(\cos\text{t}-\sin\text{t})^2}}{2\cos^2\text{t}}\text{e}^{-\text{t}}\text{dt}$
$=2\int\frac{(\cos\text{t}-\sin\text{t})}{2\cos^2\text{t}}\text{e}^{-\text{t}}\text{dt}$
$=\int(\sec\text{t}-\tan\text{t}\sec\text{t})\text{e}^{-\text{t}}\text{dt}$
$=\int\sec\text{e}^{-\text{t}}\text{dt}-\int\tan\text{t}\sec\text{e}^{-\text{t}}\text{dt}$
Integrating by parts
$=\text{e}^{-\text{t}}\sec\text{t}+\int\text{e}^{-\text{t}\frac{\text{d}}{\text{dt}}}(\sec\text{t})\text{dt}-\int\tan\text{t}\sec\text{t}\text{ e}^{-\text{t}}\text{dt}$
$=-\text{e}^{-\text{t}}\sec\text{t}+\int\text{e}^{-\text{t}}\sec\text{t}\tan\text{t dt}-\int\sec\text{t}\tan\text{t}\text{ e}^{-\text{}t}\text{dt}$
$=-\text{e}^{-\text{t}}\sec\text{t+C}$
Putting the value of t
$=\text{-e}^{-\frac{\text{x}}{2}}\sec\frac{\text{x}}{2}+\text{C}$
View full question & answer→Question 2775 Marks
$\int\frac{\text{x}^2}{\sqrt{3\text{x}-4}}\text{dx}$
Answer$\int\frac{\text{x}^2}{\sqrt{3\text{x}-4}}\text{dx}$
Let $3\text{x}+4=\text{t}$
$\Rightarrow\text{x}=\frac{\text{t}^{-4}}{3}$
$\Rightarrow1=\frac{1}{3}\cdot\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{3}$
Now, $\int\frac{\text{x}^2}{\sqrt{3\text{x}-4}}$
$=\frac{1}{3}\int\frac{\Big(\frac{\text{t}^{-4}}{3}\Big)^2}{\sqrt{\text{t}}}\text{dt}$
$=\frac{1}{27}\int\Big(\frac{\text{t}^2}{\sqrt{\text{t}}}-\frac{8\text{t}}{\sqrt{\text{t}}}+\frac{16}{\sqrt{t}}\Big)\text{dt}$
$=\frac{1}{27}\int\Big(\text{t}^\frac{3}{2}-8\text{t}^\frac{1}{2}+16\text{t}^{-\frac{1}{2}}\Big)\text{dt}$
$=\frac{1}{27}\Bigg[\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{8\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{16\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=\frac{1}{27}\Big[\frac{2}{5}\text{t}^{\frac{5}{2}}-\frac{8\times2}{3}\text{t}^{\frac{3}{2}}+32\text{t}^{\frac{1}{2}}\Big]+\text{C}$
$=\frac{2}{135}(\text{t})^{\frac{5}{2}}-\frac{16}{81}\text{t}^{\frac{3}{2}}+\frac{32}{27}\text{t}^\frac{1}{2}+\text{C}$
$=\frac{2}{135}(3\text{x}+4)^{\frac{5}{2}}-\frac{16}{81}(3\text{x}+4)^{\frac{3}{2}}+\frac{32}{27}(3\text{x}+4)^{\frac{1}{2}}+\text{C}$
View full question & answer→Question 2785 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{(\text{x}-2)^2(\text{x}+3)}\ \text{dx}$
AnswerLet $\int\frac{\text{x}^2+1}{(\text{x}-2)^2(\text{x}+3)}=\frac{\text{A}}{\text{x}-2}+\frac{\text{B}}{(\text{x}-2)^2}+\frac{\text{C}}{\text{x}+3}$ $\Rightarrow\text{x}^2+1=\text{A}(\text{x}-2)(\text{x}+3)(\text{x}+3)+\text{B}(\text{x}+3)+\text{C}(\text{x}-2)^2$ $=(\text{A}+\text{C})\text{x}^2+(\text{A}+\text{B}-4\text{C})\text{x}+(-6\text{A}+3\text{B}+4\text{C})$ Equating similar terms, we get,A + C = 1, A + B - 4C = 0, -6A + 3B + 4C = 1
Solving we get, $\text{A}=\frac{3}{5},\text{B}=1,\text{C}=\frac{2}{5}$
Thus,
$\text{I}=\frac{3}{5}\int\frac{\text{dx}}{\text{x}-2}+\int\frac{\text{dx}}{(\text{x}-2)^2}+\frac{2}{5}\int\frac{\text{dx}}{\text{x}+3}$
$\text{I}=\frac{3}{5}\log\text{x}-2|-\frac{1}{(\text{x}-2)}+\frac{2}{5}\log|\text{x}+3|+\text{C}$
View full question & answer→Question 2795 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\log\text{x}+\frac{1}{\text{x}^2})\text{dx}$
AnswerWe have,
$\text{I}=\int\text{e}^{\text{x}}\Big(\log\text{x}+\frac{1}{\text{x} ^2}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big(\log\text{x}+\frac{1}{\text{x}}-\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)\text{dx}+\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}$
Integrating by parts
$=\text{e}^{\text{x}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)-\int\text{e}^{\text{x}}\frac{\text{d}}{\text{dx}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)\text{dx}+\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}$
$=\text{e}^{\text{x}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)-\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}+\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}$
$=\text{e}^{\text{x}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)+\text{C}$
View full question & answer→Question 2805 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\Big(\frac{\sin4\text{x}-4}{1-\cos4\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{\sin4\text{x}-4}{2\sin^22\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big\{\frac{2\sin2\text{x}\cos2\text{x}}{2\sin^22\text{x}}-\frac{4}{2\sin^{2}2\text{x}}\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}\big(\cot2\text{x}-2\text{cosec}^22\text{x}\big)\text{dx}$
$=\int\text{e}^{\text{x}}\cot2\text{x dx}-2\int\text{e}^{\text{x}}\text{cosec}^22\text{x dx}$
integrating by parts
$=\text{e}^{\text{x}}\cot2\text{x}-\int\text{e}^{\text{x}}\frac{\text{d}}{\text{dx}}(\cot2\text{x})\text{dx}-2\int\text{e}^{\text{x}}\text{cosec}^22\text{x dx}$
$=\text{e}^{\text{x}}\cot2\text{x}+2\int\text{e}^{\text{x}}\text{cosec}^22\text{x}-2\int\text{e}^{\text{x}}\text{cosec}^22\text{x dx}$
$=\text{e}^{\text{x}}\cot2\text{x}+\text{C}$
View full question & answer→Question 2815 Marks
Evaluate the following intregals:
$\int\frac{5}{(\text{x}^2+1)(\text{x}+2)}\text{ dx}$
AnswerWe have
$\text{I}=\int\frac{5}{(\text{x}^2+1)(\text{x}+2)}\text{ dx}$
Let $\frac{5}{(\text{x}^2+1)(\text{x}+2)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\frac{5}{(\text{x}^2+1)(\text{x}+2)}=\frac{\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}+2)}{(\text{x}+2)(\text{x}^2+1)}$
$\Rightarrow5=\text{A}(\text{x}^2+1)+\text{Bx}^2+2\text{Bx}+\text{Cx}+2\text{C}$
$\Rightarrow5=(\text{A}+\text{B})\text{x}^2+(2\text{B}+\text{C})\text{x}+(\text{A}+2\text{C})$
Equating coefficient of like terms
A + B = 0 ...(1)
2B + C = 0 ...(2)
A + 2C = 5 ...(3)
Solving (1), (2) and (3), we get
A = 1
B = -1
C = 2
$\therefore\frac{5}{(\text{x}+2)(\text{x}^2+1)}=\frac{1}{\text{x}+2}+\Big(\frac{-\text{x}+2}{\text{x}^2+1}\Big)$
$\Rightarrow\int\frac{5\text{dx}}{(\text{x}+2)(\text{x}^2+1)}+\int\frac{\text{dx}}{\text{x}+2}-\int\frac{\text{x dx}}{\text{x}^2+1}+2\int\frac{\text{dx}}{\text{x}^2+1}$
Let $\text{x}^2+1=\text{t}$
$\Rightarrow2\text{xdx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\int\frac{\text{dx}}{\text{x}+2}-\frac{1}{2}\int\frac{\text{dt}}{\text{t}}+2\int\frac{\text{dx}}{\text{x}^2+1^2}$
$=\log|\text{x}+2|-\frac12\log|\text{t}|+2\tan^{-1}\text{x}+\text{C}$
$=\log|\text{x}+2|-\frac{1}{2}\log|\text{x}^2+2|+2\tan^{-1}\text{x}+\text{C}$
View full question & answer→Question 2825 Marks
Evaluate the following intregals:
$\int\frac{1}{\text{x}\log\text{x}(2+\log\text{x})}\text{ dx}$
AnswerLet $\int\frac{1}{\text{x}\log\text{x}(2+\log\text{x})}=\frac{\text{A}}{\text{x}\log\text{x}}+\frac{\text{B}}{\text{x}(2+\log\text{x})}$
$\Rightarrow1=\text{A}(2+\log\text{x})+\text{B}\log\text{x}$
Put $x = 1$
$\Rightarrow1=2\text{A}\Rightarrow\text{A}=\frac{1}{2}$
Put $x = 10^{-2}$
$\Rightarrow1=-2\text{B}\Rightarrow\text{B}=-\frac{1}{2}$
Thus,
$\text{I}=\frac{1}{2}\int\frac{\text{dx}}{\text{x}\log\text{x}}+\Big(-\frac{1}{2}\Big)\int\frac{\text{dx}}{\text{x}(2+\log\text{x})}$
$=\frac{1}{2}\log|\log\text{x}|-\frac{1}{2}\log|2+\log\text{x}|+\text{C}$
$\text{I}=\frac{1}{2}\log\Big|\frac{\log\text{x}}{2+\log\text{x}}\Big|+\text{C}$
View full question & answer→Question 2835 Marks
Evaluate the following intregals:
$\int\frac{\cos\text{x}}{\cos3\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{\cos3\text{x}}\ \text{dx}$
$=\int\frac{\cos\text{x}}{(4\cos^3\text{x}-3\cos\text{x})}\ \text{dx}$ $\big[\cos3\text{A}=4\cos^3\text{A}-3\cos\text{A}\big]$
$=\int\frac{1}{4\cos^2\text{x}-3}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\Rightarrow\text{I}=\int\frac{\sec^2\text{x}}{4-3\sec^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{4-3(1+\tan^2\text{x})}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1-3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1-(\sqrt{3}\tan\text{x})^2}\ \text{dx}$
Let $\sqrt{3}\tan\text{x}=\text{t}$
$\Rightarrow\sqrt{3}\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\sec^2\text{x}\text{ dx}=\frac{\text{dt}}{\sqrt{3}}$
$\therefore\text{I}=\frac{1}{\sqrt{3}}\int\frac{\text{dt}}{1^2-\text{t}^2}$
$=\frac{1}{\sqrt{3}}\times\frac{1}{2}\ln\big|\frac{1+\text{t}}{1-\text{t}}\big|+\text{C}$
$=\frac{1}{2\sqrt{3}}\ln\Big|\frac{1+\sqrt{3}\tan\text{x}}{1-\sqrt{3}\tan\text{x}}\Big|+\text{C}$
View full question & answer→Question 2845 Marks
Evaluvate the following intregals:
$\int\frac{5\cos\text{x}+6}{2\cos\text{x}+\sin\text{x}+3}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{5\cos\text{x}+6}{2\cos\text{x}+\sin\text{x}+3}\ \text{dx}$
Let $(5\cos\text{x}+6)=\lambda\frac{\text{d}}{\text{dx}}(2\cos\text{x}+\sin\text{x}+3)+\mu(2\cos\text{x}+\sin\text{x}+3)+\text{v}$
$(5\cos\text{x}+6)=\lambda(-2\sin\text{x}+\cos\text{x})+\mu(2\cos\text{x}+\sin\text{x}+3)+\text{v}$
$(5\cos\text{x}+6)=(-2\lambda+\mu)\sin\text{x}(\lambda+2\mu)\cos\text{x}+(3\mu+\text{v})$
Comparing the coefficient of $\sin\text{x}\ \&\cos\text{x}$ on the both the sides,
$-2\lambda+\mu=0\dots\dots(1)$
$\lambda+2\mu=5\dots\dots(2)$
$3\mu+\text{v}=6\dots\dots(3)$
Solving equations (1), (2) and (3),
$\text{I}=\int\frac{(-2\sin\text{x}+\cos\text{x})}{(2\cos\text{x}+\sin\text{x}+3)}\text{dx}+2\int\text{dx}$
$\text{I}=\log|2\cos\text{x}+\sin\text{x}+3|+2\text{x}+\text{C}$
View full question & answer→Question 2855 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{(\text{x}^2+2\text{x}+2)\sqrt{\text{x}+1}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}}{(\text{x}^2+2\text{x}+2)\sqrt{\text{x}+1}}\text{ dx}$
Let $\text{x}+1=\text{t}^2$
$\text{dx}=2\text{t dt}$
$=2\int\frac{(\text{t}^2-1)\text{t dt}}{(\text{t}^4+1)\text{t}}$
$=2\int\frac{(\text{t}^2-1)\text{ dt}}{(\text{t}^4+1)}$
$=2\int\frac{\big(1-\frac{1}{\text{t}^2}\big)\text{ dt}}{\text{t}+\frac{1}{\text{t}^2}}$
$=2\int\frac{\big(1-\frac{1}{\text{t}^2}\big)}{\big(\text{t}+\frac{1}{\text{t}}\big)^2-2}$
Let $\text{t}+\frac{1}{\text{t}}=\text{y}$
$\Big(1-\frac{1}{\text{t}^2}\Big)\text{ dt}=\text{dy}$
$\therefore\ \text{I}=2\int\frac{\text{dy}}{\text{y}^2-2}$
$=\frac{2}{2\sqrt{2}}\log\bigg|\frac{\text{y}-\sqrt{2}}{\text{y}+\sqrt{2}}\bigg|+\text{C}$
Thus, $\text{I}=\frac{1}{\sqrt{2}}\log\bigg|\frac{\text{t}^2+1-\sqrt{2}\text{t}}{\text{t}^2+1+\sqrt{2}\text{t}}\bigg|+\text{C}$
Hence,
$\text{I}=\frac{1}{\sqrt{2}}\log\begin{vmatrix}\frac{\text{x}+2-\sqrt{2(\text{x}+1)}}{\text{x}+2\sqrt{2(\text{x}+1)}}\end{vmatrix}+\text{C}$
View full question & answer→Question 2865 Marks
Evaluate the following intregals:
$\int\frac{\text{x}}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}$
AnswerLet $\text{I}\int\frac{\text{x}}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}$
Let $\text{x}=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+6\text{x}+10)+\mu$
$=\lambda(2\text{x}+6)+\mu$
$\text{x}=(2\lambda)\text{x}+6\lambda+\mu$
Comparing the co-efficient of like powers of x.
$2\lambda=1\Rightarrow\lambda=\frac{1}{2}$
$6\lambda+\mu=0\Rightarrow6\Big(\frac{1}{2}\Big)+\mu=0$
$\mu=-3$
So, $\text{I}_1=\int\frac{\frac{1}{2}(2\text{x}+6)=3}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}$
$\frac{1}{2}\int\frac{{2\text{x}+6}=3}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}-3\text{ I }\frac{1}{\sqrt{\text{x}^2+2\text{x}(3)+(3)^2+10}}$
$\text{I}_1=\frac{1}{2}\int\frac{2\text{x}+6}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}-3\int\frac{1}{\sqrt{(\text{x}+3)^2+(1)^2}}\text{dx}$
$\text{I}=\sqrt{\text{x}^2+6\text{x}+10}-3\log\big|\text{x}+3+\sqrt{\text{x}^2+6\text{x}+10}\big|+\text{c}$
$\text{I}_1=\frac{1}{2}(2\sqrt{\text{x}^2+6\text{x}+10})-3\log\big|\text{x}+3+\sqrt{(\text{x}+3)^2+1}\big|+\text{c}$ $\Big[\text{since},\int\frac{1}{\sqrt{\text{x}}}\text{dx}-2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{x}^2+\text{a}^2}}\text{dx}-\log\big|\text{x}+\sqrt{\text{x}}^2+\text{a}^2\big|+\text{c}\Big]$
$\text{I}=\sqrt{\text{x}^2+6\text{x}+10}-3\log\big|\text{x}+3+\sqrt{\text{x}^2+6\text{x}+10}\big|+\text{c}$
View full question & answer→Question 2875 Marks
Evaluate the following intregals:
$\int\frac{1}{(\text{x}^2+1)(\text{x}^2+2)}\ \text{dx}$
AnswerLet $\text{x}^2=\text{y}$
Then, $\frac{1}{(\text{y}+1)(\text{y}+2)}=\frac{\text{A}}{\text{y}+1}+\frac{\text{B}}{\text{y}+2}$
$\Rightarrow1=\text{A}(\text{y}+2)+\text{B}(\text{y}+1)=(\text{A}+\text{B})\text{y}+(2\text{A}+\text{B})$
Equating similar terms, we get,
A + B = 0, and 2A + B = 1
Solving, we get,
Thus,
$\text{I}=\int\frac{\text{dx}}{\text{x}^2+1}-\int\frac{\text{dx}}{\text{x}^2+2}$
$\text{I}=\tan^{-1}\text{x}-\frac{1}{\sqrt{2}}\tan^{-1}\frac{\text{x}}{\sqrt{2}}+\text{C}$
View full question & answer→Question 2885 Marks
$\int\frac{\text{x}^2+3\text{x}-1}{(\text{x}+1)^2}\text{dx}$
Answer$\text{Let I}=\int\frac{\text{x}^2+3\text{x}-1}{(\text{x}+1)^2}\text{dx. Then}$$\text{I}=\int\frac{\text{x}^2+\text{x}+2\text{x}-1}{(\text{x}+1)^2}\text{dx}$
$=\int\frac{\text{x}(\text{x}+1)+2\text{x}-1}{(\text{x}+1)^2}\text{dx}$
$=\int\frac{\text{x}(\text{x}+1)}{(\text{x}+1)^2}\text{dx}+\int\frac{2\text{x}-1}{(\text{x}+1)^2}\text{dx}$
$=\int\frac{\text{x}}{\text{x}+1}\text{dx}+\int\frac{\sqrt{2\text{x}+2-2-1}}{(\text{x}+1)^2}\text{dx}$
$=\int\frac{\text{x}+1-1}{\text{x}+1}\text{dx}+\int\frac{2(\text{x}+1)-3}{(\text{x}+1)^2}\text{dx}$
$=\int\frac{\text{x}+1}{\text{x}+1}\text{dx}-\int\frac{1}{\text{x}+1}\text{dx}+\int\frac{2(\text{x}+1)}{(\text{x}+1)^2}\text{dx}-3\int\frac{1}{(\text{x}+1)^2}\text{dx}$
$=\int\text{dx}-\int\frac{1}{\text{x}+1}\text{dx}+2\int\frac{1}{\text{x}+1}\text{dx}-3\int(\text{x}+1)^{-2}\text{dx}$
$=\text{x}-\log|\text{x}+1|+2\log|\text{x}+1|+\frac{3}{\text{x}+1}+\text{C}$
$=\text{x}+\log|\text{x}+1|+\frac{3}{\text{x}+1}+\text{C}$
$\therefore\text{I}=\text{x}+\log|\text{x}+1|+\frac{3}{\text{x}+1}+\text{C}$
View full question & answer→Question 2895 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^5}{\sqrt{1+\text{x}^2}}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}^5}{\sqrt{1+\text{x}^2}}\text{ dx}\ ....(1)$ Let $1+\text{x}^3=\text{t}^2$ then, $\text{d}\big(1+\text{x}^3\big)=\text{d}\big(\text{t}^2\big)$ $\Rightarrow3\text{x}^2\text{dx}=\text{dt }2\text{t}$ $\Rightarrow\text{dx}=\frac{\text{dt}}{3\text{x}^2}\text{ 2}\text{t}$ Putting $1+\text{x}^3=\text{t}^2$ and $\text{dx}=\frac{2\text{t}}{3\text{x}^2}\text{ dt}$ in equation (1), we get,,$\text{I}=\int\frac{\text{x}^5}{\sqrt{{t}^2}}\times\frac{2\text{t}}{3\text{x}^2}\text{ dt}$
$=\int\frac{\text{x}^5}{\text{t}}\times\frac{2\text{t}}{3\text{x}^2}\text{ dt}$ $=\frac{2}{3}\int\text{x}^3\text{dt}$ $=\frac{2}{3}\int\big(\text{t}^2-1\big)\text{dt}$ $=\frac{2}{3}\times\frac{\text{t}^3}{3}-\frac{2}{3}\text{t}+\text{C}$ $\text{I}=\frac{2}{9}\big(1+\text{x}^3\big)^{\frac{3}{2}}-\frac{2}{3}\sqrt{1+\text{x}^3}+\text{C}$
View full question & answer→Question 2905 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{\text{x}(\text{x}^2-1)}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+1}{\text{x}(\text{x}^2-1)}\ \text{dx}=\int\frac{\text{x}^2+1}{\text{x}(\text{x}+1)(\text{x}-1)}\ \text{dx}$
Let $\frac{\text{x}^2+1}{\text{x}(\text{x}+1)(\text{x}-1)}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}+1}+\frac{\text{C}}{\text{x}-1}$
$\Rightarrow\text{x}^2+1=\text{A}(\text{x}+1)(\text{x}-1)+\text{Bx}(\text{x}-1)+\text{Cx}(\text{x}+1)$
Put x = 0
⇒ 1 = -A ⇒ A = -1
Put x = -1
⇒ 2 = 2B ⇒ B = 1
Put x = 1
⇒ 2 = 2C ⇒ C = 1
Thus,
$\text{I}=-\int\frac{\text{dx}}{\text{x}}+\int\frac{\text{dx}}{\text{x}+1}+\int\frac{\text{dx}}{\text{x}-1}$
$=-\log|\text{x}|+\log|\text{x}+1|+\log|\text{x}-1|+\text{C}$
$\text{I}=\log\Big|\frac{\text{x}^2-1}{\text{x}}\Big|+\text{C}$
View full question & answer→Question 2915 Marks
Evaluate the following integrals:
$\int\frac{1}{\sin\text{x}\cos^3\text{x}}\text{ dx}$
Answer$\frac{1}{\sin\text{x}\cos^3\text{x}}=\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin\text{x}\cos^3\text{x}}$
$=\frac{\sin\text{x}}{\cos^3\text{x}}+\frac{1}{\sin\text{x}\cos\text{x}}$
$=\tan\text{x}\sec^2\text{x}+\frac{\frac{1\cos^2\text{x}}{\sin\text{x}\cos\text{x}}}{\cos^2\text{x}}$
$=\tan\text{x}\sec^2\text{x}+\frac{\sec^2\text{x}}{\tan\text{x}}$
$\therefore\ \int\frac{1}{\sin\text{x}\cos^3\text{x}}\text{ dx}=\int\tan\text{x}\sec^2\text{x}\text{ dx}+\int\frac{\sec^2\text{x}}{\tan\text{x}}\text{ dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x}\text{ dx}=\text{dt}$
$\int\frac{1}{\sin\text{x}\cos^3\text{x}}\text{ dx}=\int\text{t}\text{ dt}+\int\frac{1}{\text{t}}\text{ dt}$
$=\frac{\text{t}^2}{2}+\log|\text{t}|+\text{C}$
$=\frac{1}{2}\tan^2\text{x}+\log|\tan\text{x}|+\text{C}$
View full question & answer→Question 2925 Marks
Evaluate the following integrals:$\int\text{cosec}^3\text{x dx}$
AnswerLet $\text{I}=\int\text{cosec}^3\text{dx}$
$=\int\text{cosec x}-\text{cosec}^2\text{x dx}$
using integration by parts,
$=\text{cosec x}\times\int\text{cosec}^2\text{x dx}+\int(\text{cosec x}\cot\text{x}\int\text{cosec}^2\text{x dx})\text{dx}$
$=\text{cosec x}\times(-\cot\text{x})+\int\text{cosec x}\cot\text{x}(-\cot\text{x})\text{dx}$
$=-\text{cosec x}\cot\text{x}-\int\text{cosec x}\cot^2\text{x dx}$
$=-\text{cosec x}\cot \text{x}-\int\text{cosec x}(\text{cosec}^2\text{x}-1)\text{dx}$
$=-\text{cosec x}\cot\text{x}-\int\text{cosec}^3\text{x dx}+\int\text{cosec x dx}$
$\text{I}=-\text{cosec x}\cot\text{x}-\text{I}+\log\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}_1$
$2\text{I}=-\text{cosec x}\cot\text{x}+\log\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}_1$
$\text{I}=-\frac{1}{2}\text{cosec x}\cot\text{x}+\frac{1}{2}\log\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 2935 Marks
Evaluate the following integrals:$\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{8-\sin2\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{8-\sin2\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{9-1-\sin2\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{9-\sin^2\text{x}-\cos^2\text{x}-2\sin\text{x}\cos\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{9-(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
Let $(\sin\text{x}+\cos\text{x})=\text{t}$
On differentiating both sides, we get
$(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{1}{(3)^2-(\text{t})^2}\text{ dt}$
$=\sin^{-1}\Big(\frac{\text{t}}{3}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\sin\text{x}-\cos\text{x}}{3}\Big)+\text{C}$
Hence, $\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{8-\sin2\text{x}}}\text{ dx}=\sin^{-1}\Big(\frac{\cos\text{x}-\sin\text{x}}{3}\Big)+\text{C}$
View full question & answer→Question 2945 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{3x}}}{4\text{e}^{6\text{x}}-9}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{3\text{x}}}{4\text{e}^{6\text{x}}-9}\text{dx}$
Let $\text{e}^{3\text{x}}=\text{t}$
$\Rightarrow3\text{e}^{3\text{x}}\text{ dx = dt}$
$\Rightarrow\text{e}^{3\text{x}} \text{dx}=\frac{\text{dt}}{3}$
$\text{I}=\frac{1}3{}\int\frac{\text{dt}}{4\text{t}^2-9}$
$=\frac{1}{12}\int\frac{\text{dt}}{\text{t}^2-\frac{9}{4}}$
$=\frac{1}{12}\int\frac{\text{dt}}{\text{t}^2-\big(\frac{3}{2}\big)^2}$
$=\frac{1}{12}\times\frac{1}{2\big(\frac{3}{2}\big)}\log\Bigg|\frac{\text{t}-\frac{3}{2}}{\text{t}+\frac{3}{2}}\Bigg|+\text{C}$ $\bigg[\text{Since,}\int\frac{1}{\text{x}^2+\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\bigg|\frac{\text{x}-\text{a}}{\text{x+a}}\bigg|+\text{C}\bigg]$
$\text{I}=\frac{1}{36}\log\bigg|\frac{2\text{t}-3}{2\text{t}+3}\bigg|+\text{C}$
$\text{I}=\frac{1}{36}\log\bigg|\frac{2\text{e}^{3\text{x}}-3}{2\text{e}^{3\text{x}}+3}\bigg|+\text{C}$
View full question & answer→Question 2955 Marks
Evaluate the following integrals:$\int\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
AnswerLet $\text{I}=-\int\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
Let $\text{x}=\tan\theta$
$\text{dx}=\sec^2\theta\text{d}\theta$
$\text{I}=\int\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)\sec^2\theta\text{d}\theta$
$=\int\tan^{-1}(\tan2\theta)\sec^2\theta\text{d}\theta$
$=\int2\theta\sec^2\theta\text{d}\theta$
$=2\Big[\theta\int\sec^2\theta\text{d}\theta-\int(1\int\sec^2\theta\text{d}\theta)\text{d}\theta\Big]$
$=2\big[\theta\tan\theta-\int\tan\theta\text{d}\theta\big]$
$=2\big[\theta\tan\theta-\log\sec\theta\big]+\text{c}$
$=2\Big[\text{x}\tan^{-1}\text{x}-\log\sqrt{1+\text{x}^2}\Big]+\text{C}$
$\text{I}=2\text{x}\tan^{-1}\text{x}-\log\big|1+\text{x}^2\big|+\text{C}$
View full question & answer→Question 2965 Marks
$\int\frac{\text{x}^2}{\sqrt{\text{x}-1}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\sqrt{\text{x}-1}}\text{dx}$
Substituting x - 1 = t and dx = dt we get
$\text{I}=\int\frac{(\text{t}+1)^2}{\sqrt{\text{t}}}\text{dx}$
$=\int\frac{\text{t}^2+1+2\text{t}}{\sqrt{\text{t}}}\text{dt}$
$=\int\Big(\text{t}^\frac{3}{2}+\text{t}^\frac{-1}{2}+2\text{t}^\frac{-1}{2}\Big)\text{dt}$
$=\frac{2}{5}\text{t}^\frac{5}{2}+2\text{t}^\frac{1}{2}+\frac{4}{3}\text{t}^\frac{3}{2}+\text{C}$
$=\frac{6\text{t}^\frac{5}{2}+30\text{t}^\frac{1}{2}+20\text{t}^\frac{3}{2}}{15}+\text{C}$
$=\frac{2}{5}\text{t}^\frac{1}{2}\big(3\text{t}^2+15+10\text{t}\big)+\text{C}$
$=\frac{2}{5}\sqrt{\text{x}-1}\Big(3\big(\text{x}-1\big)^2+15+10(\text{x}-1)\Big)+\text{C}$
$=\frac{2}{5}\sqrt{\text{x}-1}\Big(3\big(\text{x}^2+1-2\text{x}\big)+15+10\text{x}-10\Big)+\text{C}$
$=\frac{2}{5}\sqrt{\text{x}-1}\big(3\text{x}^2+4\text{x}+8\big)+\text{C}$
$\therefore\ \text{I}=\frac{2}{5}\big(3\text{x}^2+4\text{x}+8\big)+\sqrt{\text{x}-1}+\text{C}$
View full question & answer→Question 2975 Marks
Evaluate the following intregals:
$\int\frac{1}{1+3\sin^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{1+3\sin^2\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{\frac{1}{\cos^2\text{x}}+\frac{3\sin^2\text{x}}{\cos^2\text{x}}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{\sec^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1+\tan^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1+4\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1+(2\tan\text{x})}\ \text{dx}$
Let $2\tan\text{x}=\text{t}$
$2\sec^2\times\text{dx}=\text{dt}$
$\text{I}=\frac{1}{2}\int\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{1}{2}\tan^{-1}\text{t}+\text{c}$
$\text{I}=\frac{1}{2}\tan^{-1}(2\tan\text{x})+\text{C}$
View full question & answer→Question 2985 Marks
Evaluate the following integrals:
$\int\tan\text{x}\sec^2\text{x}\sqrt{1-\tan^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\tan\text{x}\sec^2\text{x}\sqrt{1-\tan^2\text{x}}\text{ dx}\ ....(1)$ Let $1-\tan^2\text{x}=\text{t}$ then, $\Rightarrow\text{d}\big(1-\tan^2\text{x}\big)\text{dt}$ $\Rightarrow-2\tan\text{x}\sec^2\text{x}\text{ dx}=\text{dt}$ $\Rightarrow\tan\text{x}\sec^2\text{x}\text{ dx}=\frac{-\text{dt}}{2}$Putting $1-\tan^2\text{x}=\text{t}$ and $\tan\text{x}\sec^2\text{x}\text{ dx}=-\frac{\text{dt}}{2}$ in equation (1),
We get
$\text{I}=\int\sqrt{\text{t}}\times\frac{-\text{dt}}{2}$
$=\frac{-1}{2}\int\text{t}^{\frac{1}{2}}\text{dt}$
$=-\frac{1}{2}\times\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\text{C}$
$=-\frac{1}{3}\text{t}^\frac{3}{2}+\text{C}$
$=-\frac{1}{3}\big[1-\tan^2\text{x}\big]^{\frac{3}{2}}+\text{C}$
View full question & answer→Question 2995 Marks
Evaluate the following integrals:
$\int\frac{\cos^3\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
Answer$\int\frac{\cos^3\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
$=\int\frac{\cos^2\text{x}\cos\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
$=\int\frac{(1-\sin^2\text{x})\cos\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
$\text{Let }\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\text{Now,}\int\frac{(1-\sin^2\text{x})\cos\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
$=\int\frac{(1-\text{t}^2)}{\sqrt{\text{t}}}\text{dt}$
$=\int\Big(\frac{1}{\sqrt{\text{t}}}-\text{t}^\frac{3}{2}\Big)\text{dt}$
$=\int\Big(\text{t}^{-\frac{1}{2}}-\text{t}^\frac{3}{2}\Big)\text{dt}$
$=\Bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}-\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]+\text{C}$
$=2\sqrt{\text{t}}-\frac{2}{5}\text{t}^\frac{5}{2}+\text{C}$
$=2\sqrt{\sin\text{x}}-\frac{2}{5}\sin^\frac{5}{2}\text{x}+\text{C}$
View full question & answer→Question 3005 Marks
Evaluate the following integrals:
$\int\text{x}\cos^3\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\cos^3\text{x dx}$
$=\int\text{x}\Big(\frac{3\cos\text{x}+\cos3\text{x}}{4}\Big)\text{dx}$
$\frac{1}{4}\int\text{x}(3\cos\text{x}+\cos3\text{x})\text{dx}$
Using integration by parts,
$\text{I}=\frac{1}{4}\big[\text{x}\int(3\cos\text{x}+\cos3\text{x})\text{dx}-\int(1\int(3\cos\text{x}+\cos3\text{x})\text{dx})\text{dx}\big]$
$=\frac{1}{4}\Big[\text{x}\Big(3\sin\text{x}+\frac{\sin3\text{x}}{3}\Big)-\int\Big(3\sin\text{x}+\frac{\sin3\text{x}}{3}\Big)\text{dx}\Big]$
$=\frac{1}{4}\Big[3\text{x}\sin\text{x}+\frac{\text{x}\sin3\text{x}}{3}+3\cos\text{x}+\frac{\cos3\text{x}}{9}\Big]+\text{C}$
$\text{I}=\frac{3\text{x}\sin\text{x}}{4}+\frac{\text{x}\sin3\text{x}}{12}+\frac{3\cos\text{x}}{4}+\frac{\cos3\text{x}}{36}+\text{C}$
View full question & answer→