Question 2515 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 2525 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 2535 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 2545 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 2555 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 2565 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 2575 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 2585 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 2595 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 2605 Marks
Evaluate the following intregals:
$\int\frac{5}{(\text{x}^2+1)(\text{x}+2)}\text{ dx}$
Answer We 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 2615 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 2625 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 2635 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 2645 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 2655 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 2665 Marks
Evaluate the following integrals:$\int\frac{\text{x}}{\text{x}^4-\text{x}^2+1}\text{dx}$
Answer$\int\frac{\text{x dx}}{\text{x}^4-\text{x}^2+1}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx = dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\text{x dx}}{\text{x}^4-\text{x}^2+1}$
$ =\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2-\text{t}+1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2-\text{t}+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2+1}$
$=\frac{1}{2}\int\frac{\text{dt}}{\big(\text{t}-\frac{1}{2}\big)^2+\frac{3}{4}}$
$=\frac{1}2{}\int\frac{\text{dt}}{\Big(\text{t}-\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}$
$=\frac{1}{2}\times\frac{2}{\sqrt{3}}\tan^{-1}\Bigg(\frac{\text{t}-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{t}-1}{\sqrt{3}}\Big)+\text{C}$
$=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{x}^2-1}{\sqrt{3}}\Big)+\text{C}$
View full question & answer→Question 2675 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 2685 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 2695 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 2705 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 2715 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→Question 2725 Marks
$\int\sin\text{x}\sqrt{1-\cos2\text{x}}\text{ dx}$
AnswerLet I $=\int\sin\text{x}\sqrt{1-\cos2\text{x}}\text{ dx}.$ Then,
$\text{I}=\int\sin\text{x}\times\sqrt{2\sin^2\text{x}}\times\text{dx}$
$=\int\sin\text{x}\times\sqrt{2}\times\sin\text{x dx}$
$=\sqrt{2}\int\sin^2\text{x}\times\text{dx}$
$=\frac{\sqrt{2}}{2}\int2\sin^2\text{x}\times\text{dx}$
$=\frac{\sqrt{2}}{2}\Big[\text{x}-\frac{\sin2\text{x}}{2}\Big]+\text{C}$
$=\frac{\sqrt{2}\text{x}}{2}-\frac{\sqrt{2}}{4}\times\sin2\text{x}+\text{C}$
$=\frac{1}{\sqrt{2}}\times\text{x}-\frac{\sin2\text{x}}{2\sqrt{2}}+\text{C}$
$\therefore\text{I}=\frac{1}{\sqrt{2}}\times\text{x}-\frac{\sin2\text{x}}{2\sqrt{2}}+\text{C}$
View full question & answer→Question 2735 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{m}\sin^{-1}\text{x}}}{\sqrt{1-\text{x}^2}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{\text{m}\sin^{-1}\text{x}}}{\sqrt{1-\text{x}^2}}\text{ dx}\ ....(1)$ Let $\text{m}\sin^{-1}\text{x}=\text{t}$ then, $\text{d}\big(\text{m}\sin^{-1}\text{x}\big)=\text{dt}$ $\Rightarrow\text{m}\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$ $\Rightarrow\frac{\text{dx}}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{m}}$Putting, $\text{m}\sin^{-1}\text{x}=\text{t}$ and $\frac{\text{dx}}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{m}}$ in equation (1),
We get,
$\text{I}=\int\text{e}^\text{t}\frac{\text{dt}}{\text{m}}$
$=\frac{1}{\text{m}}\text{e}^\text{t}+\text{C}$
$=\frac{1}{\text{m}}\text{e}^{\text{m}\sin^{-1}\text{x}}+\text{C}$
View full question & answer→Question 2745 Marks
Evaluate the following intregals: $\int\frac{1}{\sqrt{3}\sin\text{x}+\cos\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{3}\sin\text{x}+\cos\text{x}}\ \text{dx}$
Let $\sqrt{3}=\text{r}\cos\theta,\text{and }1=\text{r}\sin\theta$
$\tan\theta=\frac{1}{\sqrt{3}}$
$\theta=\frac{\pi}{6}$
$\text{r}=\sqrt{3+1}=2$
$\text{I}=\int\frac{1}{\text{r}\cos\theta\sin\text{x}+\text{r}\sin\theta\cos\text{x}}\ \text{dx}$
$=\frac{1}{\text{r}}\int\frac{1}{\sin(\text{x}+\theta)}\text{dx}$
$=\frac{1}{2}\int\text{cosec}(\text{x}+\theta)\text{dx}$
$=\frac{1}{2}\log\Big|\tan\Big(\frac{\text{x}}{2}+\frac{\theta}{2}\Big)\Big|+\text{c}$
$\text{I}=\frac{1}{2}\log\Big|\tan\Big(\frac{\text{x}}{2}+\frac{\pi}{12}\Big)\Big|+\text{C}$
View full question & answer→Question 2755 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
AnswerWE have,
$\text{I}=\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
$=\int\text{e}^{2\text{x}}\Big(\frac{1-2\sin\text{x}\cos\text{x}}{2\sin^2\text{x}}\Big)\text{dx}$
Put $\text{t}=2\text{x}.$ Then $\text{dt}=2\text{dx}$
Therefore,
$\text{I}=\frac{1}{2}\int\text{e}^{\text{t}}\bigg(\frac{1-2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{2\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}{4}\int\text{e}^{\text{t}}\bigg(\frac{1-2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}4{\int\text{e}^{\text{t}}}\bigg(\frac{1}{\sin^2\frac{\text{t}}{2}}-\frac{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}{4}\int\text{e}^{\text{t}}\big(\text{cosec}^2\frac{\text{t}}{2}-2\cot\frac{\text{t}}{2}\big)\text{dt}$
$=-\frac{1}{4}\int\text{e}^{\text{t}}\big(2\cot\frac{\text{t}}{2}-\text{cosec}^2\frac{\text{t}}{2}\big)\text{dt}$
Consider, $\text{f(x)}=2\cot\frac{\text{t}}{2},$ then $\text{f}'\text{(x)}=-\text{cosec}^2\frac{\text{t}}{2}$
Thus, the given integrand is of the from $\text{e}^{\text{x}}\big[\text{f(x)}+\text{f'}\text{(x)}\big].$
Therefore, $\text{I}=-\frac{1}{4}\big(2\cot\frac{\text{t}}{2}\big)\text{e}^{\text{t}}+\text{C}$
$=-\frac{1}{4}\big(2\cot\frac{2\text{x}}{2}\big)\text{e}^{2\text{x}}+\text{C}$
Hence, $\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}=-\frac{1}{2}(\cot\text{x})\text{e}^{2\text{x}}+\text{C}$
View full question & answer→Question 2765 Marks
$\int\frac{\text{x}-1}{\sqrt{\text{x}+4}}\ \text{dx}$
Answer$\text{Let I} =\int\Big(\frac{\text{x}-1}{\sqrt{\text{x}+4}}\Big)\text{dx}$
Putting x + 4 = t
Then, x = t - 4
Difference both sides
dx = dt
Now integral becomes,
$\text{I}=\int\Big(\frac{\text{t}-4-1}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\frac{\text{t}}{\sqrt{\text{t}}}-\frac{5}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\text{t}^{\frac{1}{2}}-5\text{t}^{-\frac{1}{2}}\Big)\text{dt}$
$=\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}-5\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=\frac{2}{3}\text{t}^\frac{3}{2}-10\sqrt{\text{t}}+\text{C}$
$=\frac{2}{3}(\text{x}+4)^\frac{3}{2}-10(\text{x}+4)^\frac{1}{2}+\text{C}$
View full question & answer→Question 2775 Marks
Evaluate the following integrals:
$\int\text{x}\sqrt{\text{x}^4+1}\text{dx}$
Answer$\text{I}=\int\text{x}\sqrt{\text{x}^4+1}\text{dx}$
$=\int\text{x}\sqrt{(\text{x}^2)^2+1}\text{dx}$
Putting $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\ \text{I}=\frac{1}{2}\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{1}{2}\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{1}{2}\Big[\frac{\text{t}}{2}\sqrt{\text{t}^2+1}+\frac{1^2}{2}\log\big|\text{t}+\sqrt{\text{t}^2+1}\big|\Big]+\text{C}$
$=\frac{1}{2}\Big[\frac{\text{x}^2}{2}\sqrt{\text{x}^4+1}+\frac{1}{2}\log\big|\text{x}^2+\sqrt{\text{x}^4+1}\big|\Big]+\text{C}$
$=\frac{\text{x}^2}{4}\sqrt{\text{x}^4+1}+\frac{1}{4}\log\big|\text{x}^2+\sqrt{\text{x}^4+1}+\text{C}$
View full question & answer→Question 2785 Marks
Evaluate the following integrals:
$\int\frac{\cot\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
AnswerLet I $=\int\frac{\cot\text{x}}{\sqrt{\sin\text{x}}}\text{dx}\ .....(1)$
Let $\sin\text{x}=\text{t}$ then,
$\text{d}(\sin\text{x})=\text{dt}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
$\text{Now,}\text{I}=\int\frac{\cot\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
$=\int\frac{\cot\text{x}}{\sin\text{x}\sqrt{\sin\text{x}}}\text{dx}$
$\int\frac{\cos\text{x}}{(\sin\text{x})^{\frac{3}{2}}}\text{dx}$
$\Rightarrow\ =\int\frac{\cos\text{x}}{(\sin\text{x})^\frac{3}{2}}\text{dx}\ ...(2)$
Putting $\sin\text{x}=\text{t}$ and $\cos\text{x}\text{ dx}=\text{dt}$ in equation (2), we get
$\text{I}=\int\frac{\text{dt}}{\text{t}^\frac{3}{2}}$
$=\int\text{t}^{-\frac{3}{2}}\text{dt}$
$=-2\text{t}^{-\frac{1}{2}}+\text{C}$
$=\frac{-2}{\sqrt{\text{t}}}+\text{C}$
$=\frac{-2}{\sqrt{\sin\text{x}}}+\text{C}$
$\therefore\text{I}=\frac{-2}{\sqrt{\sin\text{x}}}+\text{C}$
View full question & answer→Question 2795 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 2805 Marks
Integrate the following integrals:
$\int\sin2\text{x}\sin4\text{x}\sin6\text{x dx}$
Answer$\int\sin2\text{x}\sin4\text{x}\sin6\text{x dx}$
$=\frac{1}{2}\int(2\sin2\text{x}\sin4\text{x})\sin6\text{x dx}$
$=\frac{1}{2}\int\big[\cos(2\text{x}-4\text{x})-\cos(2\text{x}+4\text{x})\big]\sin6\text{x dx}$
$=\frac{1}{2}\int\big[\cos(2\text{x})-\cos(6\text{x})\big]\sin6\text{x dx}$
$=\frac{1}{2}\big[\int\cos(2\text{x})\sin(6\text{x})\text{dx}-\int\cos(6\text{x})\sin(6\text{x})\text{dx}\big]$
$=\frac{1}{4}\big[\int2\cos(2\text{x})\sin(6\text{x})\text{dx}-\int2\cos(6\text{x})\sin(6\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big\{\int\big[\sin(2\text{x}+6\text{x})-\sin(2\text{x}-6\text{x})\big]\text{dx}-\int\sin(12\text{x})\text{dx}\Big\}$
$=\frac{1}{4}\Big[\int\sin(8\text{x})\text{dx}+\int\sin(4\text{x})\text{dx}-\int\sin(12\text{x})\text{dx}\Big]$
$=\frac{1}{4}\Big[\frac{-\cos(8\text{x})}{8}+\frac{-\cos(4\text{x})}{4}+\frac{\cos(12\text{x})}{12}\Big]+\text{C}$
$=-\frac{\cos(8\text{x})}{32}-\frac{\cos(4\text{x})}{16}+\frac{\cos(12\text{x})}{48}+\text{C}$
View full question & answer→Question 2815 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 2825 Marks
Evaluate the following intregals:
$\int\frac{1}{4\sin^2\text{x}+5\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{4\sin^2\text{x}+5\cos^2\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{4\tan^2\text{x}+5}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{4\tan^2\text{x}+5}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x}\ \text{dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{4+9(\text{t})^2}$
$=\int\frac{\text{dt}}{4\text{t}^2+5}$
Let $2\text{t}=\text{u}$
$2\text{dt}=\text{du}$
$\text{I}=\frac{1}{2}\int\frac{\text{du}}{(4)^2+(\sqrt{5})^2}$
$=\frac{1}{2}\times\frac{1}{\sqrt{5}}\times\tan^{-1}\Big(\frac{\text{u}}{\sqrt{5}}\Big)+\text{C}$
$=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{2\text{t}}{\sqrt{5}}\Big)+\text{C}$
$\text{I}=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{2\tan\text{x}}{\sqrt{5}}\Big)+\text{C}$
View full question & answer→Question 2835 Marks
$\int\frac{\text{x}^2+3\text{x}+1}{(\text{x}+1)^2}\text{dx}$
Answer$\int\Big(\frac{\text{x}^2+3\text{x}+1}{(\text{x}+1)^2}\Big)\text{dx}$
Let $\text{x}+1=\text{t}$
$\Rightarrow\text{x}=\text{t}-1$
$\Rightarrow1=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=\text{dt}$
Now, $\int\Big(\frac{\text{x}^2+3\text{x}+1}{(\text{x}+1)^2}\Big)\text{dx}$
$=\int\Big[\frac{(\text{t}-1)^2+3(\text{t}-1)+1}{\text{t}^2}\Big]\text{dt}$
$=\int\Big(\frac{\text{t}^2-2\text{t}+1+3\text{t}-3+1}{\text{t}^2}\Big)\text{dt}$
$=\int\Big(\frac{\text{t}^2+\text{t}-1}{\text{t}^2}\Big)\text{dt}$
$=\int\Big(1+\frac{1}{\text{t}}-\text{t}^{-2}\Big)\text{dt}$
$=\text{t}+\log|\text{t}|-\frac{\text{t}^{-2+1}}{-2+1}+\text{C}$
$=\text{t}+\log|\text{t}|-\frac{\text{1}}{\text{t}}+\text{C}$
$=\text{x}+1+\log|\text{x+1}|+\frac{1}{\text{x}+1}+\text{C}$
Let $1+\text{C}=\text{C}'$
$=\text{x}+\log|\text{x+1}|+\frac{1}{\text{x}+1}+\text{C}'$
View full question & answer→Question 2845 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x}}{\log(\sec\text{x}+\tan\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{\sec\text{x}}{\log(\sec\text{x}+\tan\text{x})}\text{dx}$
Putting $\log(\sec\text{x}+\tan\text{x})=\text{t}$
$\Rightarrow\frac{\sec\text{x}\tan\text{x}+\sec^2\text{x}}{\sec\text{x}+\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x}\frac{(\sec\text{x}+\tan\text{x})}{\sec\text{x}+\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x }\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\sec\text{x}+\tan\text{x})|+\text{C}$
View full question & answer→Question 2855 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 2865 Marks
Evalute the following integrals:
$\int\frac{\cos\text{x}}{2+3\sin\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{2+3\sin\text{x}}\text{dx}\ .....\text{(i)}$
Let $2+3\sin\text{x}=\text{t}$ then,
$\text{d}(2+3\sin\text{x})=\text{dt}$
$\Rightarrow3\cos\text{x dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{3\cos\text{x}}$
Putting $2+3\sin\text{x}=\text{t and dx}=\frac{\text{dt}}{3\cos\text{x}}$ in equation (i), we get,
$\text{I}=\int\frac{\cos\text{x}}{\text{t}}\times\frac{\text{dt}}{3\cos\text{x}}$
$=\frac{1}{3}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{3}\log|\text{t}|+\text{C}$
$=\frac{1}{3}\log|2+3\sin\text{x}|+\text{C}$
View full question & answer→Question 2875 Marks
Evalute the following integrals:
$\int\frac{\sin(\text{x}-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin(\text{x}-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$ then,
$\text{I}=\int\frac{\sin(\text{x}-\alpha+\alpha-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\frac{\sin(\text{x}+\alpha-2\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\frac{\sin(\text{x}+\alpha)\cos2\alpha-\cos(\text{x}+\alpha)\sin2\alpha}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\Big[\frac{\sin(\text{x}+\alpha)\cos2\alpha}{\sin(\text{x}+\alpha)}-\frac{\cos(\text{x}+\alpha)\sin2\alpha}{\sin(\text{x}+\alpha)}\Big]\text{dx}$
$=\int\big(\cos2\alpha-\cot(\text{x}+\alpha)\sin2\alpha\big)\text{dx}$
$=\cos2\alpha\int\text{dx}-\sin2\alpha\int\cot(\text{x}+\alpha)\text{dx}$
$=\text{x}\cos2\alpha-\sin2\alpha\log|\sin(\text{x}+\alpha)|+\text{C}$
$\therefore\text{I}=\text{x}\cos2\alpha-\sin2\alpha\log|\sin(\text{x}+\alpha)|+\text{C}$
View full question & answer→Question 2885 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 2895 Marks
Evaluate the following integrals:
$\int\text{cosec x}\log({\text{cosec x}-\cot\text{x})}\text{dx}$
AnswerLet $\text{I}=\int\text{cosec x}\log(\text{cosec x}-\cot\text{x})\text{dx}\ ....(1)$
Let $\log(\text{cosec x}-\cot\text{x})=\text{t}$ then,
$\text{dx}[\log(\text{cosec x}-\cot\text{x})]=\text{dt}$
$\Rightarrow\text{cosec x}\text{ dx}=\text{dt}$ $\Big[\because\ \frac{\text{d}}{\text{dx}}(\log(\text{cosec x}-\cot\text{x}))=\text{cosec x}\Big]$
Putting $\log(\text{cosec x}-\cot\text{x})=\text{t}$ and $\text{cosec x}\text{ dx}=\text{dt}$ in equation (1), we get
$\text{I}=\int\text{t dt}$
$=\frac{\text{t}^2}{2}+\text{C}$
$\text{I}=\frac{1}{2}[\log(\text{cosec x}-\cot\text{x})]^2+\text{C}$
View full question & answer→Question 2905 Marks
Evaluate the following integrals:
$\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{dx}$
Answer$\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{dx}$
$=\int\frac{\sqrt{\tan\text{x}}}{\frac{\sin\text{x}}{\cos\text{x}}\times\cos^2\text{x}}\text{dx}$
$=\int\frac{\sqrt{\tan\text{x}}}{\tan\text{x}}\times\sec^2\text{x dx}$
$=\int\frac{1}{\sqrt{\tan\text{x}}}\times\sec^2\text{x dx}$
$=\int(\tan\text{x})^{-\frac{1}{2}}\sec^2\text{x dx}$
$\text{Let }\tan\text{x}=t$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\text{Now,}\int(\tan\text{x})^{-\frac{1}{2}}\sec^2\text{x dx}$
$=\int\text{t}^{{-\frac{1}{2}}}\text{dt}$
$=\Bigg[\frac{-{\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\tan\text{x}}+\text{C}$
View full question & answer→Question 2915 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 2925 Marks
Evaluate the following integrals:$\int\frac{\text{x}^3\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^3\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{dx}$
Let $\sin^{-1}\text{x}^2=\text{t}$
$\frac{1}{\sqrt{1-\text{x}^4}}(2\text{x})\text{dx}=\text{dt}$
$\text{I}=\int\frac{\text{x}^2\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^2}}\text{x dx}$
$=\int(\sin\text{t})\text{t}\frac{\text{dt}}{2}$
$=\frac{1}{2}\int\text{t}\sin\text{t dt}$
$=\frac{1}{2}\big[\text{t}\int\sin\text{t dt}-\int(1\int\sin\text{t dt})\text{dt}\big]$
$=\frac{1}{2}\big[\text{t}(-\cos\text{t})-\int(-\cos\text{t})\text{dt}\big]$
$=\frac{1}{2}\big[-\text{t}\cos\text{t}+\sin\text{t}\big]+\text{C}$
$\text{I}=\frac{1}{2}\Big[\text{x}^2-\sqrt{1-\text{x}^4}\sin^{-1}\text{x}^2\Big]+\text{C}$
View full question & answer→Question 2935 Marks
Evaluate the following intergrals:
$\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}$
$\Rightarrow-\text{e}^\text{ax}\frac{\cos(\text{bx}+\text{x})}{\text{b}}+\int\text{ae}^\text{ax}\frac{\cos(\text{bx}+\text{c})}{\text{b}}\text{dx}$
$=-\frac{1}{\text{b}}\text{e}^\text{ax}\cos(\text{bx}+\text{c})+\frac{\text{a}}{\text{b}}\int\text{e}^\text{ax}\cos(\text{bx}+\text{c})\text{dx}$
$=-\frac{1}{\text{b}}\text{e}^\text{ax}\cos(\text{bx}+\text{c})+\frac{\text{a}}{\text{b}}\Big[\int\text{e}^\text{ax}\frac{\sin(\text{bx}+\text{c})}{\text{b}}-\int\text{ae}^{\text{ax}}\frac{\sin(\text{bx}+\text{c})}{\text{b}}\text{dx}\Big]+\text{C}_1$
$=\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}\\-\frac{\text{a}^2}{\text{b}^2}\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}+\text{C}_1$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}\\-\frac{\text{a}^2}{\text{b}^2}\text{I}+\text{C}_1$
$\Rightarrow\text{I}=\Big\{\frac{\text{a}^2+\text{b}^2}{\text{b}^2}\Big\}-\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}+\text{C}_1$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{a}^2+\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}+\text{C}_1$
View full question & answer→Question 2945 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 2955 Marks
$\int\frac{\text{x}}{\sqrt{\text{x}+4}}\text{dx}$
Answer$\int\frac{\text{x}}{\sqrt{\text{x}+4}}\text{dx}$
$=\int\Big(\frac{\text{x}+4-4}{\sqrt{\text{x}+4}}\Big)\text{dx}$
$=\int\Big(\sqrt{\text{x}+4}-\frac{4}{\sqrt{\text{x}+4}}\Big)\text{dx}$
$=\int(\text{x}+4)^\frac{1}{2}\text{dx}-4\int(\text{x}+4)^{-\frac{1}{2}}\text{dx}$
$=\frac{(\text{x}+4)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-4\frac{[\text{x}+4]^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=\frac{2}{3}(\text{x}+4)^\frac{3}{2}-8(\text{x}+4)^\frac{1}{2}+\text{C}$
$=(\text{x}+4)^\frac{1}{2}\Big[\frac{2}{3}(\text{x}+4)-8\Big]+\text{C}$
$=(\text{x}+4)^\frac{1}{2}\Big[\frac{2\text{x}+8-24}{3}\Big]+\text{C}$
$=(\text{x}+4)^\frac{1}{2}\Big[\frac{2\text{x}-16}{3}\Big]+\text{C}$
$=\frac{2}{3}(\text{x}-8)\sqrt{\text{x}+4}+\text{C}$
View full question & answer→Question 2965 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 2975 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\cos(3\text{x}+4)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}\cos(3\text{x}+4)\text{dx}$
Integrating by parts
$\text{I}=\text{e}^{2\text{x}}\frac{\sin(3\text{x}+4)}{3}-\int2\text{e}^{2\text{x}}\frac{\sin(3\text{x}+4)}{3}\text{dx}$
$=\frac{1}{3}\text{e}^{2\text{x}}\sin(3\text{x}+4)-\frac{2}{3}\int\text{e}^{2\text{x}}\sin(3\text{x}+4)\text{dx}$
$=\frac{1}{3}\text{e}^{2\text{x}}\sin(3\text{x}+4)-\frac{2}{3}\Big[-\text{e}^{2\text{x}}\frac{\cos(3\text{x}+4)}{3}+\int2\text{e}^{2\text{x}}\frac{\cos(3\text{x}+4)}{3}\text{dx}\Big]+\text{C}$
$\text{I}=\frac{\text{e}^{2\text{x}}}{9}\{2\cos(3\text{x}+4)+3\sin(3\text{x}+4)\}+\text{C}$
Hence,
$\text{I}=\frac{\text{e}^{2\text{x}}}{13}\{2\cos(3\text{x}+4)+3\sin(3\text{x}+4)\}+\text{C}$
View full question & answer→Question 2985 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\cos^2\text{x }\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}\cos^2\text{x }\text{dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}2\cos^2\text{x dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}(1+\cos2\text{x})\text{dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}\text{dx}+\frac{1}{2}\int\text{e}^{2\text{x}}\cos2\text{x }\text{dx}$
$\because\ \int\text{e}^{2\text{x}}\cos\text{bx dx}=\frac{\text{e}^{2\text{x}}}{\text{a}^2+\text{b}^2}\{\text{a}\cos\text{bx}-\text{b}\sin\text{bx}\}+\text{C}$
$\therefore\ \text{I}=\frac{1}{4}\text{e}^{2\text{x}}+\frac{1}{2}\frac{\text{e}^{2\text{x}}}{8}\{2\cos2\text{x}+2\sin2\text{x}\}+\text{C}$
Hence,
$\text{I}=\frac{\text{e}^{2\text{x}}}{4}+\frac{\text{e}^{2\text{x}}}{16}\{2\cos2\text{x}+2\sin2\text{x}\}+\text{C}$
or
$\text{I}=\frac{\text{e}^{2\text{x}}}{4}+\frac{\text{e}^{2\text{x}}}{8}\{\cos2\text{x}+\sin2\text{x}\}+\text{C}$
View full question & answer→Question 2995 Marks
Evaluate the following integrals:
$\int(\text{e}^{\log\text{x}}+\sin\text{x})\cos\text{x dx}$
AnswerLet $\text{I}=\int(\text{e}^{\log\text{x}}+\sin\text{x})\cos\text{x dx}$
$=\int(\text{x}+\sin\text{x})\cos\text{x dx}$
$=\int\text{x}\cos\text{x dx}+\int\sin\text{x}\cos\text{x dx}$
$=\big[\text{x}\int\cos\text{x dx}-\int(1\int\cos\text{x dx})\text{dx}\big]+\frac{1}{2}\int\sin2\text{x dx}$
$=\big[\text{x}\sin\text{x}-\int\sin\text{x dx}\big]+\frac{1}{2}\Big(-\frac{\cos2\text{x}}{2}\Big)+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}-\frac{1}{4}\cos2\text{x}+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}-\frac{1}{4}\big[1-2\sin^2\text{x}\big]+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}-\frac{1}{4}+\frac{1}{2}\sin^2\text{x}+\text{C}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}+\frac{1}{2}\sin^2\text{x}+\text{C}-\frac{1}{4}$
$\text{I}=\text{x}\sin\text{x}+\cos\text{x}+\frac{1}{2}\sin^2\text{x}+\text{k},$ where $\text{k}=\text{C}-\frac{1}{4}$
View full question & answer→Question 3005 Marks
$\int\frac{2\text{x}-1}{(\text{x}-1)^2}\text{ dx}$
Answer$\int\Big[\frac{2\text{x}-1}{(\text{x}-1)^2}\text{ dx}\Big]$
Let $\text{x}-1=\text{t}$
$\Rightarrow\text{x}=1+\text{t}$
$\Rightarrow1=\frac{\text{dt}}{\text{dx}}$
Now, $\int\Big[\frac{2\text{x}-1}{(\text{x}-1)^2}\text{ dx}\Big]$
$=\int\Big[\frac{2(\text{t}+1)-\text{t}}{\text{t}^2}\Big]\text{dt}$
$=\int\Big(\frac{2\text{t}+1}{\text{t}^2}\Big)\text{dt}$
$=2\int\frac{\text{dt}}{\text{t}}+\int\text{t}^{-2}\text{dt}$
$=2\log|\text{t}|+\frac{\text{t}^{-2+1}}{-2+1}+\text{C}$
$=2\log(\text{x}-1)-\frac{1}{\text{x}-1}+\text{C}$
View full question & answer→