Question 1013 Marks
Evaluate the following integrals:
$\int\sin^3\text{x}\cos^6\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\sin^3\text{x}\cos^6\text{x}\text{ dx}$
Here, power of $\sin\text{x}$ is odd, so we substitute
$\cos\text{x}=\text{t}$
$-\sin\text{x}\text{dt}=\text{dt}$
$\therefore\ \text{I}=\int\sin^2\text{x}\text{ t}^6(-\text{dt})$
$=-\int\big(1-\cos^2\text{x}\big)\text{t}^6\text{dt}$
$=-\int\big(\text{t}^6-\text{t}^8\big)\text{dt}$
$=-\frac{\text{t}^7}{7}+\frac{\text{t}^9}{9}+\text{C}$
$\therefore\ \text{I}=\frac{1}{7}\cos^7\text{x}+\frac{1}{9}\cos^9\text{x}+\text{C}$
View full question & answer→Question 1023 Marks
Evalute the following integrals:
$\int\frac{1}{\cos3\text{x}-\cos\text{x}}\text{dx}$
Answer$\int\frac{1}{\cos3\text{x}-\cos\text{x}}\text{dx}$
$=\int\frac{1}{4\cos^2\text{x}-4\cos\text{x}}\text{dx}\big[\because\cos3\text{x}=\text{4}\cos^3\text{x}\cos\text{x}\big]$
$=\int\frac{1}{4\cos\text{x}(\cos^2\text{x}-1)}\text{dx}$
$=\frac{-1}{4}\int\frac{1}{\cos\text{x}\sin^2\text{x}}\text{dx}$
$=\frac{-1}{4}\int\Big(\frac{\sin^2\text{x}+\cos^2\text{x}}{\cos\text{x}\sin^2\text{x}}\text{dx}\Big)$
$=\frac{-1}{\text{4}}\Big[\int\sec\text{ x dx}+\int\cot\text{x cosec x dx}$
$=\frac{-1}{4}\big(\text{ln }|\sec\text{x}+\tan\text{x}|-\text{cosec x}\big)+\text{C}$
$=\frac{1}{4}\big(\text{cosec x}-\text{ln}|\sec\text{x}+\tan\text{x}|\big)+\text{C}$
View full question & answer→Question 1033 Marks
Evaluate the following integrals:
$\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$
Answer$\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$ Let $2+3\log\text{x}=\text{t}$ $\Rightarrow\frac{3}{\text{x}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\frac{\text{dx}}{\text{x}}=\frac{\text{dt}}{3}$ Now, $\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$$=\frac{1}{3}\int\sin\text{t dt}$
$=\frac{1}{3}[-\cos\text{t}]+\text{C}$
$=-\frac{1}{3}\cos(2+3\log\text{x})+\text{C}$
View full question & answer→Question 1043 Marks
Evaluate the following integrals:
$\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\Rightarrow\text{I}=\int\sqrt{4-\text{t}^2}\text{dt}$
$=\int\sqrt{2^2-\text{t}^2}\text{dt}$
$=\frac{\text{t}^2}{2}\sqrt{2^2-\text{t}^2}+\frac{4}{2}\sin^{-1}\frac{\text{t}}{2}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\sin\text{x}\sqrt{4-\sin^2\text{x}}+2\sin^{-1}\Big(\frac{\sin\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 1053 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
Answer$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{-(\text{x}^2-\text{x})}\text{dx}$
$=\int\sqrt{-\bigg\{\text{x}^2-\text{x}+\Big(\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2\bigg\}}\text{dx}$
$=\int\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$
$=\Big(\frac{\text{x}-\frac{1}{2}}{2}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}\bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\bigg)+\text{C}$
$\Big[\because\ \int\sqrt{\text{a}^2-\text{x}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{a}^2-\text{x}^2}+\frac{1}{2}\text{a}^2\sin^{-1}\frac{\text{x}}{\text{a}}=\text{C}\Big]$
$=\Big(\frac{2\text{x}-1}{4}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}(2\text{x}-1)+\text{C}$
View full question & answer→Question 1063 Marks
Evaluate the following integrals:
$\int\cot \text{x}. \text{log}\ \sin\text{x dx}$
Answer$\int\cot \text{x}. \text{log}\ \sin\text{x dx}$
$\text{Let}\ \text{log}\sin\text{x} =\text{t}$
$\Rightarrow\frac{1}{\sin\text{x}}\times \cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cot\text{x dx} =\text{dt}$
$\text{Now},\int\cot\text{x}. \log\sin\text{x dx} $
$=\int\text{t}.\text{dt}$
$=\frac{\text{t}^{2}}{2}+\text{C}$
$=\frac{(\text{log}\begin{vmatrix}\sin\text{x}\end{vmatrix})^{2}}{2}+\text{C}$
View full question & answer→Question 1073 Marks
Write a value of $\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$ $=\int\text{e}^{\text{x}}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}$ Let $\text{e}^{\text{x}}\sec\text{x}=\text{t}$ $(\text{e}^{\text{x}}\sec\text{x}+\text{e}^{\text{x}}\sec\text{x}\tan\text{x})\text{dx}=\text{dt}$ $\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}=\text{dt}$ $\therefore\text{I}=\int\text{dt}$ $=\text{t}+\text{C}$$=\text{e}^{\text{x}}\sec\text{x}+\text{C}$ $(\because\text{t}=\text{e}^{\text{x}}\sec\text{x})$
View full question & answer→Question 1083 Marks
Evalute the following integrals:
$\int\frac{1+\tan\text{x}}{1-\tan\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1+\tan\text{x}}{1-\tan\text{x}}\text{dx}$
$=\int\bigg(\frac{1+\frac{\sin\text{x}}{\cos\text{x}}}{1-\frac{\sin\text{x}}{\cos\text{x}}}\bigg)\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}-\sin\text{x}}\Big)\text{dx}$
Putting $\cos\text{x}-\sin\text{x}=\text{t}$
$\Rightarrow(-\sin\text{x}-\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow(\sin\text{x}+\cos\text{x})\text{dx}=-\text{dt}$
$\therefore\text{I}=-\int\frac{1}{\text{t}}\text{dt}$
$=-\text{ln}|\text{t}|+\text{C}$
$=-\text{ln}|\cos\text{x}-\sin\text{x}|+\text{C}\ \big[\because\text{t}\cos\text{x}-\sin\text{x}\big]$
View full question & answer→Question 1093 Marks
$\int\frac{1}{1+\cos3\text{x}}\text{dx}$
Answer$\int\frac{\text{dx}}{1+\cos3\text{x}}$
$=\int\frac{(1-\cos3\text{x})}{(1+\cos3\text{x})(1-\cos3\text{x})}\text{dx}$
$=\int\Big(\frac{1-\cos3\text{x}}{1-\cos^23\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1-\cos3\text{x}}{\sin^23\text{x}}\Big)\text{dx}$
$=\int\text{cosec}^23\text{x}\text{ dx}-\int\text{cosec}3\text{x}\cot3\text{x dx}$
$=-\frac{\cot3\text{x}}{3}+\frac{\text{cosec}3\text{x}}{3}+\text{c}$
$=\frac{1}{3}[\text{cosec}3\text{x}-\cot3\text{x}]+\text{c}$
$=\frac{1}{3}\bigg[\frac{1}{\sin3\text{x}}-\frac{\cos3\text{x}}{\sin3\text{x}}\bigg]+\text{c}$
$=\frac{1}{3}\bigg[\frac{1-\cos3\text{x}}{\sin3\text{x}}\bigg]+\text{c}$
View full question & answer→Question 1103 Marks
$\int(5\text{x}+3)\sqrt{2\text{x}-1}\text{ dx}$
Answer$\text{Let I}=\int(5\text{x}+3)\sqrt{2\text{x}-1}\text{ dx}$
$\text{Putting}\ \ 2\text{x}-1=\text{t}$
$\Rightarrow2\text{x}=\text{t}+1$
$\Rightarrow\text{x}=\frac{\text{t}+1}{2}$
$\&\ 2\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\int\Big[5\Big(\frac{\text{t}+1}{2}\Big)+3\Big]\times\sqrt{\text{t}}\times\frac{\text{dt}}{2}$
$=\int\Big(\frac{5\text{t}}{5}+\frac{5}{2}+3\Big)\times\frac{\sqrt{\text{t}}\text{ dt}}{2}$
$=\frac{1}{4}\int(5\text{t}+11)\text{t}^\frac{1}{2}\text{ dt}$
$=\frac{1}{4}\int(5\text{t}^\frac{3}{2}+11\text{t}^\frac{1}{2})\text{ dt}$
View full question & answer→Question 1113 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}^2}-\frac{2}{\text{x}^3}\Big)\text{dx}$
Answer$\text{I}=\int\text{e}^{\text{x}}\big(\text{x}^{-2}-2\text{x}^{-3}\big)\text{dx}$
$=\int\text{e}^{\text{x}}\text{x}^{-2}\text{dx}-2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}$
Integration by parts
$=\text{e}^{\text{x}}\text{x}^{-2}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\big(\text{x}^{-2}\big)\Big)\text{dx}-2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}$
$=\text{e}^{\text{x}}\text{x}^{-2}+2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}-2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}$
$=\frac{\text{e}^{\text{x}}}{\text{x}^2}+\text{C}$
$\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}^2}-\frac{2}{\text{x}^3}\Big)\text{dx}=\frac{\text{e}^\text{x}}{\text{x}^2}+\text{C}$
View full question & answer→Question 1123 Marks
If $\int\text{e}^{\text{x}}(\tan\text{x}+1)\sec\text{x dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C},$ then write the value of f(x).
AnswerSince, $\int\text{e}^{\text{x}}\big(\text{f(x)}+\text{f}(\text{x})\big)\text{dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C}$
We can write the expression as
$\int\text{e}^{\text{x}}(\tan\text{x}+1)\sec\text{x dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C}$
$\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{f}(\text{x})\text{e}^{\text{x}}+\text{C}$
Comparing with equation $\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{f}(\text{x})\text{e}^{\text{x}}+\text{C}$ with standard equation we have
$\text{f(x)}=\sec\text{x},\text{f}'(\text{x})=\sec\text{x}\tan\text{x}$
Therefore,
$\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{e}^{\text{x}}\sec\text{x}+\text{C}$
Thus, $\text{f(x)}=\sec\text{x}$
View full question & answer→Question 1133 Marks
Write a value of $\int\frac{\cos\text{x}}{3+2\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{3+2\sin\text{x}}\text{ dx}$
Let $3+2\sin\text{x}=\text{t}$
$2\cos\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{2}\log\text{t}+\text{C}$
$=\frac{1}{2}\log(3+2\sin\text{x})+\text{C}$
$\therefore\ \int\frac{\cos\text{x}}{3+2\sin\text{x}}\text{ dx}=\frac{1}{2}\log(3+2\sin\text{x})+\text{C}$
View full question & answer→Question 1143 Marks
Write a value of $\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}+\cos\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
Let $\sin\text{x}+\cos\text{x}=\text{t}$
$(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}}$
$=-\log\text{t}+\text{C}$
$\text{I}=\log|\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 1153 Marks
Evaluate the following integrals:
$\int\log(\text{x}+1)\text{dx}$
AnswerLet $\text{I}=\int\log(\text{x}+1)\text{dx}$
$=\int1\times\log(\text{x}+1)\text{dx}$
Using integration by parts,
$\text{I}=\log(\text{x}+1)\int1\text{dx}-\int\Big(\frac{1}{\text{x}+1}\times\int1\text{dx}\Big)\text{dx+C}$
$=\text{x}\log(\text{x}+1)-\int\Big(\frac{\text{x}}{\text{x}+1}\Big)\text{dx+C}$
$=\text{x}\log(\text{x}+1)-\int\Big(1-\frac{1}{\text{x}+1}\Big)\text{dx+C}$
$\text{I}=\text{x}\log(\text{x}+1)-\text{x}+\log(\text{x}+1)+\text{C}$
View full question & answer→Question 1163 Marks
Evaluate the following integrals:
$=\int\frac{\sin\text{x}}{(1+\cos\text{x})^2}\text{dx}$
AnswerLet I $=\int\frac{\sin\text{x}}{(1+\cos\text{x})^2}\text{dx}\ ....(1)$
Let $1+\cos\text{x}=\text{t}$ then,
$\text{d}(1+\cos\text{x})=\text{dt}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
$\Rightarrow\sin\text{x dx}=-\text{dt}$
Putting $1+\cos\text{x}=\text{t}$ and $\sin\text{dx}=-\text{dt}$ in equation (2), we get
$\text{I}=\int\frac{-\text{dt}}{\text{t}^2}$
$=-\int\text{t}^{-2}\text{dt}$
$=-\big(-1\text{t}^{-1}\big)+\text{C}$
$=\frac{1}{\text{t}}+\text{C}$
$=\frac{1}{1+\cos\text{x}}+\text{C}$
$\therefore\text{I}=\frac{1}{1+\cos\text{x}}+\text{C}$
View full question & answer→Question 1173 Marks
Evaluate the following integrals:
$\int\big(2-3\text{x}\big)\big(3+2\text{x}\big)\big(1-2\text{x}\big)\text{dx}$
Answer$\int\big(2-3\text{x}\big)\big(3+2\text{x}\big)\big(1-2\text{x}\big)\text{dx}$
$=\int\big(6+4\text{x}-9\text{x}-6\text{x}^2\big)\big(1-2\text{x}\big)\text{dx}$
$=\int\big(-6\text{x}^2-5\text{x}+6\big)\big(1-2\text{x}\big)\text{dx}$
$=\int\big(-6\text{x}^2+12\text{x}^3-5\text{x}+10\text{x}^2+6-12\text{x}\big)\text{dx}$
$=\int\big(4\text{x}^2+12\text{x}^3-17\text{x}+6\big)\text{dx}$
$=\int\big(12\text{x}^3+4\text{x}^2-17\text{x}+6\big)\text{dx}$
$=\frac{12}{4}\text{x}^4+\frac{4}{3}\text{x}^3-\frac{17}{2}\text{x}^2+6\text{x}+\text{C}$
$=3\text{x}^4+\frac{4}{3}\text{x}^3-\frac{17}{2}\text{x}^2+6\text{x}+\text{C}$
View full question & answer→Question 1183 Marks
Evalute the following integrals: $\int\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{10^\text{x}+\text{x}^{10}}\text{dx}$
AnswerLet $\text{I}=\int\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{10^\text{x}+\text{x}^{10}}\text{dx}\ .....(\text{i})$
Let $10^\text{x}+\text{x}^{10}=\text{t}$ then,
$\text{d}(10^\text{x}+\text{x}^{10})=\text{dt}$
$\Rightarrow(10^\text{x}\log_\text{e}10+10\text{x}^9)\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{(10\text{x}^9+10^\text{x}\log_\text{e}10)}$
Putting $10^x + x^{10} = t$ and $\text{dx}=\frac{\text{dt}}{10\text{x}^9+10^\text{x}\log_\text{e}10}$ in equation $(i),$ we get,
$\text{I}=\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{\text{t}}\times\frac{\text{dt}}{10\text{x}^9+10^\text{x}\log_\text{e}10}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|10^\text{x}+\text{x}^{10}|+\text{C}$
$\therefore\text{I}=\log|10^\text{x}+\text{x}^{10}|+\text{C}$
View full question & answer→Question 1193 Marks
Evaluate the following integrals:
$\int\cot^{\text{n}}\text{cosec}^2\text{x}\text{ dx},\text{ n}\neq-1$
Answer$\int\cot^{\text{n}}\text{cosec}^2\text{x}\text{ dx},\text{ n}\neq-1$
Let $\cot\text{x}=\text{t}$
$-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$
$\text{cosec}^2\text{x}\text{ dx}=-\text{dt}$
Now, $\int\cot^{\text{n}}\text{x }\text{cosec}^2\text{x}\text{ dx}$
$=-\int\text{t}^{\text{n}}\text{dt}$
$=\frac{-\text{t}^{\text{n}+1}}{\text{n}+1}+\text{C}$
$=-\frac{\cot^{\text{n}+1}}{\text{n}+1}+\text{C}$
View full question & answer→Question 1203 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x cosec x}}{\log(\tan\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.
Let $\text{I}=\int\frac{\sec\text{x cosec x}}{\log(\tan\text{x})}\text{dx}$
Putting $\log\tan\text{x}=\text{t}$
$\Rightarrow\frac{\sec^2\text{x}}{\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x cosec x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\tan\text{x})|+\text{C}$
View full question & answer→Question 1213 Marks
Evaluate the following integrals:$\int\frac{\text{e}^\text{x}}{(1+\text{e}^{\text{x}})(2+\text{e}^\text{x})}\text{dx}$
AnswerTo evaluate the following integral follow tha steps:
Let $\text{e}^\text{x}=\text{t}$ therefore $\text{e}^\text{x}\text{dx = dt}$
Now
$\int\frac{\text{e}^{\text{x}}}{(1+\text{e}^\text{x})(2+\text{e}^\text{x})}\text{dx}=\int\frac{\text{dt}}{(1+\text{t})(2+\text{t})}$
$=\int\frac{\text{dt}}{(1+\text{t})}-\int\frac{\text{dt}}{(2+\text{t})}$
$=\ln|1+\text{t}|-\ln|2+\text{t}|+\text{C}$
$=\ln\bigg|\frac{1+\text{t}}{2+\text{t}}\bigg|+\text{C}$
$=\ln\bigg|\frac{1+\text{e}^{\text{x}}}{2+\text{e}^{x}}\bigg|+\text{C}$
View full question & answer→Question 1223 Marks
$\int\frac{2\text{x}}{(2\text{x}+1)^2}\text{dx}$
Answer$\int\frac{2\text{x}}{(2\text{x}+1)^2}\text{dx}$
$=\int\bigg(\frac{2\text{x}+1-1}{(2\text{x}+1)^2}\bigg)\text{dx}$
$=\int\bigg[\frac{2\text{x}+1}{(2\text{x}+1)^2}-\frac{1}{(2\text{x}+1)^2}\bigg]\text{dx}$
$=\int\frac{\text{dx}}{2\text{x}+1}-\int(2\text{x}+1)^{-2}\text{dx}$
$=\frac{\log(2\text{x}+1)}{2}-\bigg[\frac{(2\text{x}+1)^{-2+1}}{2(-2+1)}\bigg]+\text{c}$
$=\frac{\log(2\text{x}+1)}{2}+\frac{(2\text{x}+1)^{-1}}{2}+\text{c}$
$=\frac{\log(2\text{x}+1)}{2}+\frac{1}{2(2\text{x}+1)}+\text{c}$
View full question & answer→Question 1233 Marks
Write a value of $\int\frac{1+\log\text{x}}{3+\text{x}\log\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1+\log\text{x}}{3+\text{x}\log\text{x}}\text{ dx}$
Let $3+\text{x}\log\text{x}=\text{t}$
$\Big(\log\text{x}+\text{x}\cdot\frac{1}{\text{x}}\Big)\text{dx}=\text{at}$
$(1+\log\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log\text{t}+\text{C}$
$\text{I}=\log(3+\text{x}\log\text{x})+\text{C}$
View full question & answer→Question 1243 Marks
Evaluate the following integrals: $\int\text{x}^2\cos\text{x dx}$
Answer$\int\text{x}^2\cos\text{x dx}$
Taking $x^2$ as the first function and $\cos x$ as the second function.
$=\text{x}^2\int\cos\text{x dx}-\int\big(\frac{\text{d}}{\text{dx}}\text{x}^2\int\cos\text{x dx}\big)\text{dx}$
$=\text{x}^2\sin\text{x}-\int2\text{x}\sin\text{x dx}$
$=\text{x}^2\sin\text{x}-2\big[\text{x}\int\sin\text{x}-\int\big\{\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{x dx}\big\}\text{dx}\big]$
$=\text{x}^2\sin\text{x}-2[-\text{x}\cos\text{x}+\int\cos\text{x dx}]$
$=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x+C}$
View full question & answer→Question 1253 Marks
$\int\frac{1}{\sqrt{\text{x}+3}-\sqrt{\text{x}+2}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{\text{x}+3}-\sqrt{\text{x}+2}}\text{dx}.$ Then,
$\text{I}=\int\frac{1}{\sqrt{\text{x}+3}-\sqrt{\text{x}+2}}\times\frac{\sqrt{\text{x}+3}+\sqrt{\text{x}+2}}{\sqrt{\text{x}+3}+\sqrt{\text{x}+2}}\text{dx}$
$=\int\frac{\sqrt{\text{x}+3}+\sqrt{\text{x}+2}}{\text{x}+3-\text{x}-2}\text{dx}$
$=\int\Big[(\text{x}+3)^{\frac{1}{2}}+(\text{x}+2)^{\frac{1}{2}}\Big]\text{dx}$
$=\frac{(\text{x}+3)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{(\text{x}+2)^{\frac{3}{2}}}{\frac{3}{2}}+\text{c}$
$=\frac{2}{3}\times(\text{x}+3)^{\frac{3}{2}}+\frac{2}{3}(\text{x}+2)^{\frac{3}{2}}+\text{c}$
$=\frac{2}{3}\Big\{(\text{x}+3)^{\frac{3}{2}}+(\text{x}+2)^{\frac{3}{2}}\Big\}+\text{c}$
$\text{I}=\frac{2}{3}\Big\{(\text{x}+3)^{\frac{3}{2}}+(\text{x}+2)^{\frac{3}{2}}\Big\}+\text{c}$
View full question & answer→Question 1263 Marks
$\int \text{(2x} - 3)^{5} + \sqrt{3\text{x + 2}}\text{ dx}$
Answer$\int\big[(2\text{x}-3)^5+\sqrt{3\text{x}+2}\big]\text{dx}$
$=\int(2\text{x}-3)^5\text{dx}+\int{(3\text{x}+2)^{\frac{1}{2}}}\text{dx}$
$=\frac{(2\text{x}-3)^{5+1}}{2(5+1)}+\frac{(3\text{x}+2)^{\frac{1}{2}{+1}}}{3\Big(\frac{1}{2}+1\Big)}+\text{c}$
$=\frac{(2\text{x}-3)^6}{12}+\frac{2}{9}(3\text{x}+2)^{\frac{3}{2}}+\text{c}$
View full question & answer→Question 1273 Marks
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\sin5\text{x}\sin3\text{x}}\text{dx}$
Answer$\int\frac{\sin2\text{x}}{\sin5\text{x}\sin3\text{x}}\text{dx}$
$=\int\frac{\sin(5\text{x}-3\text{x})}{\sin5\text{x}\sin3\text{x}}\text{dx}$
$=\int\frac{5\text{x}\cos3\text{x}-\cos5\text{x}\sin3\text{x}}{\sin5\text{x}\sin3\text{x}}$
$=\int\frac{\sin5\text{x}\cos3\text{x}}{\sin5\text{x}\sin3\text{x}}-\frac{\cos5\text{x}\sin3\text{x}}{\sin5\text{x}\sin3\text{x}}\text{dx}$
$=\int\big[\cot3\text{x}-\cot5\text{x}\big]\text{dx}$
$=\int\cot3\text{x dx}-\int\cot5\text{x dx}$
$=\frac{1}{3}\text{ln}|\sin3\text{x}|-\frac{1}{5}\text{ln}|\sin5\text{x}|+\text{C}$
View full question & answer→Question 1283 Marks
Evalute the following integrals:
$\int\sqrt{\frac{1-\sin2\text{x}}{1+\sin2\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\frac{1-\sin2\text{x}}{1+\sin2\text{x}}}\text{dx}$ then,
$=\int\sqrt{\frac{1-\cos\Big(\frac{\pi}{2}-2\text{x}\Big)}{1+\cos\Big(\frac{\pi}{2}-2\text{x}\Big)}}\text{dx}$
$=\int\sqrt{\frac{2\sin^2\Big(\frac{\pi}{4}-\text{x}\Big)}{2\cos^2\Big(\frac{\pi}{4}-\text{x}\Big)}}\text{dx}$
$=\int\sqrt{\tan^2\Big(\frac{\pi}{4}-\text{x}\Big)}\text{dx}$
$=\int\tan\Big(\frac{\pi}{4}-\text{x}\Big)\text{dx}$
$=\log\Big|\cos\Big(\frac{\pi}{4}-\text{x}\Big)\Big|+\text{C}$
View full question & answer→Question 1293 Marks
Write a value of $\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{ dx}$
Let $\text{x}+\log\sin\text{x}=\text{t}$
$(1+\cot\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
Hence,
$\text{I}=\log|\text{x}+\log\sin\text{x}|+\text{C}$
View full question & answer→Question 1303 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}\sqrt{\text{x}^4-1}}\text{ dx}$
Answer$\text{I}=\int\frac{1}{\text{x}\sqrt{\text{x}^4-1}}\text{ dx}\ ....(1)$
Let $\text{x}^2=\text{t}$ then,
$\text{d}\big(\text{x}^2\big)=\text{dt}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{2\text{x}}$
Putting $\text{x}^2=\text{t}$ and $\text{dx}=\frac{\text{dt}}{2\text{x}}$ in equation (1), we get,
$\text{I}=\int\frac{1}{\text{x}\sqrt{\text{t}^2-1}}\times\frac{\text{dt}}{2\text{x}}$
$=\frac{1}{2}\int\frac{1}{\text{x}^2\sqrt{\text{t}^2-1}}\text{ dt}$
$=\frac{1}{2}\int\frac{1}{\text{t}\sqrt{\text{t}^2-1}}\text{ dt}$
$=\frac{1}{2}\sec^{-1}\text{t}+\text{C}$
$=\frac{1}{2}\sec^{-1}\text{x}^2+\text{C}$
$\text{I}=\frac{1}{2}\sec^{-1}\big(\text{x}^2\big)+\text{C}$
View full question & answer→Question 1313 Marks
Evaluate the following integrals:
$\int\text{x}^3\log\text{x dx}$
AnswerLet $\text{I}=\int\text{x}^3\log\text{x dx}$
Using integration by parts,
$\text{I}=\log\text{x}\int\text{x}^3\text{dx}-\int\Big(\frac{1}{\text{x}}\times\int\text{x}^3\text{dx}\Big)\text{dx+C}$
$=\frac{\text{x}^4}{4}\log\text{x}-\int\frac{\text{x}^4}{4\text{x}}\text{dx+C}$
$=\frac{\text{x}^4}{4}\log\text{x}-\frac{1}{4}\int\text{x}^3\text{dx+C}$
$=\frac{\text{x}^4}{4}\log\text{x}-\frac{1}{4}\int\frac{\text{x}^4}{4}\text{dx+C}$
$\text{I}=\frac{\text{x}^4}{4}\log\text{x}-\frac{1}{16}\text{x}^4+\text{C}$
View full question & answer→Question 1323 Marks
Write a value of $\int\text{e}^{3\log\text{x}}\text{x}^{4}\text{ dx}.$
AnswerLet $\text{I}=\int\text{e}^{3\log\text{x}}\text{x}^{4}\text{ dx}$
$=\int\text{e}^{\log\text{x}^3}\cdot\text{x}^{4}\text{ dx}$
$=\int\text{x}^3\cdot\text{x}^4\text{ dx}$ $\big[\because\text{e}^{\log\text{x}}=\text{x}\big]$
$=\int\text{x}^{7}\text{ dx}$
$\therefore\ \text{I}=\frac{\text{x}^{8}}{8}+\text{C}$
View full question & answer→Question 1333 Marks
Evaluate the following integrals:
$\int\tan^{-1}\Big(\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big)\text{dx}$
Answer$\int\tan^{-1}\Big[\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big]\text{dx}$
$=\int\tan^{-1}\Big[\frac{2\sin\text{x}\cos\text{x}}{2\cos^2\text{x}}\Big]\text{dx}$
$=\int\tan^{-1}\Big[\frac{\sin\text{x}}{\cos\text{x}}\Big]\text{dx}$
$=\int\tan^{-1}(\tan\text{x})\text{dx}$
$=\int\text{x dx}$
$=\frac{\text{x}^2}{2}+\text{C}$
$\therefore\ \int\tan^{-1}\Big[\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big]\text{dx}=\frac{\text{x}^2}{2}+\text{C}$
View full question & answer→Question 1343 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^{\frac{-1}{3}}+\sqrt{\text{x}}+2}{\sqrt[3]{\text{x}}}\text{dx}$
Answer$\int\Bigg(\int\frac{\text{x}^{-\frac{1}{3}}+\sqrt{\text{x}}+2}{\text{x}^{\frac{1}{3}}}\Bigg)\text{dx}$
$=\int\Bigg(\frac{\text{x}^{-\frac{1}{3}}}{\text{x}^{\frac{1}{3}}}+\frac{\text{x}^{\frac{1}{2}}}{\text{x}^{\frac{1}{3}}}+\frac{2}{\text{x}^{\frac{1}{3}}}\Bigg)\text{dx}$
$=\int\Big(\text{x}^{-\frac{2}{3}}+\text{x}^{\frac{1}{6}}+2\text{x}^{-\frac{1}{3}}\Big)\text{dx}$
$=\Bigg[\frac{\text{x}^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}+\frac{\text{x}^{\frac{1}{6}+1}}{\frac{1}{6}+1}+2\frac{\text{x}^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}\Bigg]$
$=\Bigg[\frac{\text{x}^{\frac{1}{3}}}{\frac{1}{3}}+\frac{\text{x}^{\frac{7}{6}}}{\frac{7}{6}}+3\text{x}^{\frac{2}{3}}\Bigg]+\text{C}$
$=3\text{x}^{\frac{1}{3}}+\frac{6}{7}\text{x}^{\frac{7}{6}}+3\text{x}^{\frac{2}{3}}+\text{C}$
View full question & answer→Question 1353 Marks
$\int\frac{1}{2-3\text{x}}+\frac{1}{\sqrt{3\text{x}-2}}\text{dx}$
Answer$\int\Big(\frac{1}{2-3\text{x}}+\frac{1}{\sqrt{3\text{x}-2}}\Big)\text{dx}$
$=\int\frac{\text{dx}}{2-3\text{x}}+\int(3\text{x}-2)^{-\frac{1}{2}}\text{dx}$
$=\frac{\ln(2-3\text{x})}{-3}+\Bigg[\frac{(3\text{x}-2)^{-\frac1{2}+1}}{3\big(-\frac{1}{2}+1\big)}\Bigg]+\text{c}$
$=\frac{\ln(2-3\text{x})}{-3}+\frac{2}{3}(3\text{x}-2)^{\frac{1}{2}}+\text{c}$
$=-\frac{1}{3}\ln(2-3\text{x})+\frac{2}{3}\sqrt{3\text{x}-2}+\text{c}$
View full question & answer→Question 1363 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{1-\text{x}^2}(2+3\sin^{-1}\text{x})}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{1-\text{x}^2}(2+3\sin^{-1}\text{x})}\text{dx}$
Putting $\sin^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}=\text{dt}$
$\therefore\text{I}=\frac{1}{2+3\text{t}}\text{dt}$
$=\frac{1}{3}\text{ln}|2+3\text{t}|+\text{C}$
$=\frac{1}{3}\text{ln}|2+3\sin^{-1}\text{tx}|+\text{C }\big[\because\text{t}=\sin^{-1}\text{x}\big]$
View full question & answer→Question 1373 Marks
Evalute the following integrals:
$\int\big\{1+\tan\text{x}\tan(\text{x}+\theta)\big\}\text{dx}$
AnswerSince,
$\tan(\text{A}-\text{B})=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}$
$\therefore\tan(\text{x}+\theta-\text{x})=\frac{\tan(\text{x}+\theta)-\tan\text{x}}{1+\tan(\text{x}+\theta)\tan\text{x}}$
$\Rightarrow 1+\tan(\text{x}+\theta)\tan\text{x}=\frac{\tan(\text{x}+\theta)-\tan\text{x}}{\tan\theta}$
$\Rightarrow\int1+\tan(\text{x}+\theta)\tan\text{x dx}$
$=\frac{1}{\tan\theta}\big[\int\tan(\text{x}+\theta)\text{dx}-\int\tan\text{x dx}\big]$
$=\frac{1}{\tan\theta}\big[-\log|\cos(\text{x}+\theta)++\log|\cos\text{x}|\big]+\text{C}$
$=\frac{1}{\tan\theta}\big[\log|\cos\text{x}|-\log|\cos(\text{x}+\theta)|\big]+\text{C}$
$=\frac{1}{\tan\theta}\log\Big|\frac{\cos\text{x}}{\cos(\text{x}+\theta)}\Big|+\text{C}$
View full question & answer→Question 1383 Marks
Write a value of $\int\tan^6\text{x}\sec^2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\tan^6\text{x}\sec^2\text{x}\text{ dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\text{t}^6\text{dt}$
$=\frac{\text{t}^7}{7}+\text{C}$
Thus, $\text{I}=\frac{\tan^7\text{x}}{7}+\text{C}$
View full question & answer→Question 1393 Marks
Evaluate the following integrals:
$\int\sin^3\text{x}\cos^5\text{x}\text{ dx}$
Answer$\int\sin^3\text{x}\cos^5\text{x}\text{ dx}$
$=\int\sin^2\text{x}\cdot\cos^5\text{x}\cdot\sin\text{x}\text{ dx}$
$=\int(1-\cos^2\text{x})\cdot\cos^5\text{x}\sin\text{x}\text{ dx}$
Let $\cos\text{x}=\text{t}$
$-\sin\text{x}\text{ dx}=\text{dt}$
$\sin\text{x}\text{ dx}=-\text{dt}$
Now, $\int(1-\cos^2\text{x})\cdot\cos^5\text{x}\sin\text{x}\text{ dx}$
$=-\int(1-\text{t}^2)\text{t}^5\text{dt}$
$=-(\text{t}^5-\text{t}^7)\text{dt}$
$=-\int(\text{t}^7-\text{t}^5)\text{dt}$
$=\frac{\text{t}^8}{8}-\frac{\text{t}^6}{6}+\text{C}$
$=\frac{\cos^8\text{x}}{8}-\frac{\cos^6\text{6}\text{x}}{6}+\text{C}$
View full question & answer→Question 1403 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{\text{x}^2-1}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+1}{\text{x}^2-1}\ \text{dx}$
$=\int1+\frac{2}{\text{x}^2-1}\ \text{dx}$
$=\int\text{dx}+\int\frac{2\text{dx}}{(\text{x}+1)(\text{x}-1)}$
$=\int\text{dx}+\int\frac{-1}{\text{x}+1}+\frac{1}{\text{x}-1}\ \text{dx}$
$=\text{x}-\log|\text{x}+1|+\log|\text{x}-1|+\text{C}$
$\text{I}=\text{x}+\log\Big|\frac{\text{x}-1}{\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 1413 Marks
Evaluate the following integrals:
$\int\cot^5\text{x}\text{ cosec}^4\text{x}\text{ dx}$
Answer$\int\cot^5\text{x}\text{ cosec}^4\text{x}\text{ dx}$
$=\int\cot^5\text{x}\text{ cosec}^2\text{ x}.\text{ cosec}^2\text{x}\text{ dx}$
$=\int\cot^5\text{x}.(1+\cot^2\text{x}).\text{ cosec}^2\text{x}\text{ dx}$
Let $\cot\text{x}=\text{t}$
$=-\text{ cosec}^2\text{x}\text{ dx}=\text{dt}$
$=\text{ cosec}^2\text{x}\text{ dx}=-\text{dt}$
Now, $\int\cot^5\text{x}\text{ cosec}^4\text{x}\text{ dx}$
$=\int\text{t}^5(1+\text{t}^2)\text{dt}$
$=\int(\text{t}^5+\text{t}^7)\text{dt}$
$=-\Big[\frac{\text{t}^6}{6}+\frac{\text{t}^8}{8}\Big]+\text{C}$
$=-\Big[\frac{\cot^6\text{x}}{6}+\frac{\cot^8\text{x}}{8}\Big]+\text{C}$
View full question & answer→Question 1423 Marks
Evaluate the following integrals:
$\int\frac{\text{x}+\sqrt{\text{x}+1}}{\text{x}+2}\text{ dx}$
AnswerWe have,
$\text{I}=\int\frac{\text{x}+\sqrt{\text{x}+1}}{\text{x}+2}\text{ dx}$
Let $\text{x}+1=\text{t}^2$
Differentiating both sides we get
$\text{dx}=2\text{t dt}$
Now, integration becomes
$\text{I}=\int\frac{(\text{t}^2-1+\text{t})}{\text{t}^2+1}2\text{t dt}$
$=2\int\frac{\text{t}^3+\text{t}^2-\text{t}}{\text{t}^2+1}\text{ dt}$
$=2\int\frac{\text{t}^3+\text{t}-\text{t}+\text{t}^2+1-1-\text{t}}{\text{t}^2+1}\text{ dt}$
$=\int\frac{\text{t}^3+\text{t}+\text{t}^2+1-\text{t}-\text{t}-1}{\text{t}^2+1}\text{ dt}$
$=2\int\frac{\text{t}^3+\text{t}}{\text{t}^2+1}+2\int\frac{\text{t}^2+1}{\text{t}^2+1}+2\int\frac{-2\text{t}-1}{\text{t}^2+1}\text{ dt}$
$=2\int\text{t dt}+2\int\text{dt}-2\int\frac{2\text{t}}{\text{t}^2+1}\text{ dt}-2\int\frac{1}{\text{t}^2+1}\text{ dt}$
$=\text{t}^2+2\text{t}-2\log\big|\text{t}^2+1\big|-2\tan^{-1}\text{t}+\text{C}$
$=(\text{x}+1)+2\sqrt{\text{x}+1}-2\log|\text{x}+2|-2\tan^{-1}\sqrt{\text{x}+1}+\text{C}$
View full question & answer→Question 1433 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\cos\text{x}-\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\cos\text{x}-\sin\text{x})\text{dx}$
$=\int\text{e}^{\text{x}}\cos\text{x dx}-\int\text{e}^{\text{x}}\sin\text{x dx}$
Integrating by parts
$=\text{e}^{\text{x}}\cos\text{x}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\cos\text{x}\Big)\text{dx}-\int\text{e}^{\text{x}}\sin\text{x dx}$
$=\text{e}^{\text{x}}\cos\text{x}+\int\text{e}^{\text{x}}\sin\text{x dx}-\int\text{e}^{\text{x}}\sin\text{x dx}$
$=\text{e}^{\text{x}}\cos\text{x}+\text{C}$
$\therefore\int\text{e}^{\text{x}}(\cos\text{x}-\sin\text{x})\text{dx}=\text{e}^{\text{x}}\cos\text{x}+\text{C}$
View full question & answer→Question 1443 Marks
Evaluate the following integrals:
$\int\frac{\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
Answer$\int\frac{\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
$=\int\frac{\tan\text{x}}{(\sec\text{x}+\tan\text{x})}\times\Big(\frac{\sec\text{x}-\tan\text{x}}{\sec\text{x}-\tan\text{x}}\Big)\text{dx}$
$=\int\frac{\tan\text{x}(\sec\text{x}-\tan\text{x})}{(\sec^2\text{x}-\tan^2\text{x})}\text{dx}$
$=\int\Big(\frac{\sec\text{x}\tan\text{x}-\tan^2\text{x}}{1}\Big)\text{dx}$
$=\int\sec\text{x}\tan\text{x dx}-\int(\sec^2\text{x}-1)\text{dx}$
$=\sec\text{x}-\tan\text{x}+\text{x}+\text{C}$
View full question & answer→Question 1453 Marks
Evaluate $\int\frac{1}{\text{x}(1+\log\text{x})}\text{ dx}$
Answer$\text{I}=\int\frac{1}{\text{x}(1+\log\text{x})}\text{ dx}$
Let $(1+\log\text{x})=\text{t}$
Or, $\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$\text{I}=\log|\text{t}|+\text{C}$
$\therefore\ \text{I}=\log|1+\log\text{x}|+\text{C}$
View full question & answer→Question 1463 Marks
Evalute the following integrals:
$\int\frac{\cos2\text{x}+\text{x}+1}{\text{x}^2+\sin2\text{x}+2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos2\text{x}+\text{x}+1}{\text{x}^2+\sin2\text{x}+2\text{x}}\text{dx}$
Putting $\text{x}^2+\sin2\text{x}+2\text{x}=\text{t}$
$\Rightarrow2\text{x}+2\cos2\text{x}+2=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(\text{x}+\cos2\text{x}+1)\text{dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{2}\text{ln}|\text{t}|+\text{C}$
$=\frac{1}{2}\text{ln}|\text{x}^2+\sin2\text{x}+2\text{x}|+\text{C}$
$\big[\because\text{t}=\text{x}^2+\sin2\text{x}+2\text{x}\big]$
View full question & answer→Question 1473 Marks
$\int\sin^3(2\text{x}+1)\text{dx}$
AnswerWe need to evaluate $\int\sin^3(2\text{x}+1)\text{dx}$
By using the formula
$\sin3\theta=-4\sin^3\theta+3\sin\theta$
$\therefore\sin^3(2\text{x}+1)=\frac{3\sin(2\text{x}+1)-\sin3(2\text{x}+1)}{4}$
$\int\sin^3(2\text{x}+1)\text{dx}$
$=\int\frac{3\sin(2\text{x}+1)-\sin3(2\text{x}+1)}{4}\text{dx}$
$=-\frac{3}{8}\cos(2\text{x}+1)+\frac{1}{24}\cos3(2\text{x}+1)+\text{C}$
View full question & answer→Question 1483 Marks
$\int\frac{1+\cos4\text{x}}{\cot\text{x}-\tan\text{x}}\text{dx}$
Answer$\int\Big(\frac{1+\cos4\text{x}}{\cot\text{x}-\tan\text{x}}\Big)\text{dx}$
$=\int\frac{(1+\cos4\text{x})}{\big(\frac{\cos\text{x}}{\sin\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}\big)}\text{dx}$
$=\int\frac{2\cos^22\text{x}\times\sin\text{x}\cos\text{x}}{(\cos^2\text{x}-\sin^2\text{x})}\text{dx}$
$=\int\frac{\cos^22\text{x}\times2\sin\text{x}\cos\text{x}}{\cos2\text{x}}\text{dx}$
$=\int\cos2\text{x}\sin2\text{x}\text{ dx}$
$=\frac{1}{2}\int2\sin2\text{x}\cos2\text{x dx}$
$=\frac{1}{2}\int\sin4\text{x dx}$
$=\frac{1}{2}\Big[-\frac{\cos4\text{x}}{4}\Big]+\text{c}$
$=-\frac{1}{8}\cos4\text{x}+\text{c}$
View full question & answer→Question 1493 Marks
Evaluate the following intregals:
$\int\frac{1}{(\text{x}-1)(\text{x}+1)(\text{x}+2)}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{(\text{x}-1)(\text{x}+1)(\text{x}+2)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}+1}+\frac{\text{C}}{\text{x}+2}$
$\Rightarrow1=\text{A}(\text{x}+1)(\text{x}+2)+\text{B}(\text{x}-1)(\text{x}+2)+\text{C}(\text{x}^2-1)$
Put x = 1
$\Rightarrow1=6\text{A}\Rightarrow\text{A}=\frac{1}{6}$
put = -1
$\Rightarrow1=-2\text{B}\Rightarrow\text{B}=-\frac{1}{2}$
put = -2
$\Rightarrow1=3\text{C}\Rightarrow\text{C}=\frac{1}{3}$
So,
$\text{I}=\frac{1}{6}\int\frac{\text{dx}}{\text{x}-1}-\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}+\frac{1}{3}\int\frac{\text{dx}}{\text{x}+2}$
$\text{I}=\frac{1}{6}\log|\text{x}-1|-\frac{1}{2}\log|\text{x}+1|+\frac{1}{3}\log|\text{x}+2|+\text{C}$
View full question & answer→Question 1503 Marks
Evalute the following integrals: $\int\frac{\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}}{\text{e}^\text{x}+\text{x}^\text{e}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}}{\text{e}^\text{x}+\text{x}^\text{e}}\text{dx}$
Putting $e^x + x^e = t$
$\Rightarrow\text{e}^\text{x}+\text{ex}^{\text{e}-1}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}\big(\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}\big)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\big(\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}\big)\text{dx}=\frac{\text{dt}}{\text{e}}$
$\therefore\text{I}=\frac{1}{\text{e}}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{\text{e}}\text{ ln}|\text{t}|+\text{C}$
$=\frac{1}{\text{e}}\text{ ln}\Big|\text{e}^\text{x}+\text{x}^\text{e}\Big|+\text{C}\big[\because\text{t}=\text{e}^\text{x}+\text{x}^\text{e}\big]$
View full question & answer→