Question 2013 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 2023 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 2033 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 2043 Marks
Write a value of $\int\frac{1}{1+\text{e}^{\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{1+\text{e}^{\text{x}}}\text{dx}$
Dividing and multiplying by $e^x$
$=\frac{\text{e}^{-\text{x}}}{\text{e}^{-\text{x}}+1}\text{dx}$
Let $\text{e}^{-\text{x}}+1=\text{t}$
$-\text{e}^{-\text{x}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$\therefore\ \text{I}=-\log|1+\text{e}^{-\text{x}}|+\text{C}$
View full question & answer→Question 2053 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 2063 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 2073 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 2083 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 2093 Marks
Evaluate the following integrals:$\int\frac{\cos\text{x}}{\sqrt{\sin^2\text{x}-2\sin\text{x}-3}}\text{ dx}$
Answer$\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}-2\sin\text{x}-3}}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}-2\sin\text{x}-3}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2-2\text{t}-3}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2-2\text{t}+1-1-3}}$
$=\int\frac{\text{dt}}{\sqrt{(\text{t}-1)^2-2^2}}$
$=\log\Big|\text{t}-1+\sqrt{(\text{t}-1)^2-2^2}\Big|+\text{C}$
$=\log\Big|\text{t}-1+\sqrt{\text{t}^2-2\text{t}-3}\Big|+\text{C}$
$=\log\Big|\sin\text{x}-1+\sqrt{\sin^2\text{x}-2\sin\text{x}-3}\Big|+\text{C}$
View full question & answer→Question 2103 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 2113 Marks
Evaluate the following integrals:$\int\tan^{-1}\Big(\frac{3\text{x}-\text{x}^3}{1-3\text{x}^2}\Big)\text{dx}$
AnswerLet $\int\tan^{-1}\Big(\frac{3\text{x}-\text{x}^3}{1-3\text{x}^2}\Big)\text{dx}$$=\int3\tan^{-1}(\text{x})\text{dx}$
$=3\int\big[\tan^{-1}(\text{x})\times1\big]\text{dx}$
$=3\Big[\tan^{-1}\text{x}\times\text{x}-\int\frac{1}{1+\text{x}^2}\times\text{x dx}\Big]$
$=3\text{x}\tan^{-1}\text{x}-3\int\frac{\text{x}}{1+\text{x}^2}\text{dx}$
Let $1+\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx = dt}$
Then,
$\text{I}=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\int\frac{\text{dt}}{\text{t}}$
$=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\log|\text{t}|+\text{C}$
$=3\text{x}\tan^{-1}\text{x}-\frac{3}{2}\log|1+\text{x}^2|+\text{C}$
View full question & answer→Question 2123 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 2133 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 2143 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 2153 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 2163 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 2173 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 2183 Marks
Evaluate the following integrals:$\int\frac{\text{x}+5}{3\text{x}^2+13\text{x}-10}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}+5}{3\text{x}^2+13\text{x}-10}\text{ dx}$
$=\int\frac{\text{x}+5}{3\text{x}^2+15\text{x}-2\text{x}-10}\text{ dx}$
$=\int\frac{\text{x}+5}{3\text{x}(\text{x}+5)-2(\text{x}+5)}\text{ dx}$
$=\int\frac{\text{x}+5}{(3\text{x}-2)(\text{x}+5)}\text{ dx}$
$=\int\frac{1}{3\text{x}-2}\text{ dx}$
$\therefore\ \text{I}=\frac{1}{3}\int|3\text{x}-2|+\text{C}$
View full question & answer→Question 2193 Marks
Evaluate the following integrals:$\int\frac{\cos\text{x}}{\sqrt{4+\sin^2\text{x}}}\text{ dx}$
Answer$\int\frac{\cos\text{x}\text{ dx}}{\sqrt{4+\sin^2\text{x}}}$ Let $\sin\text{x}=\text{t}$ $\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$ Now, $\int\frac{\cos\text{x}\text{ dx}}{\sqrt{4+\sin^2\text{x}}}$$=\int\frac{\text{dt}}{\sqrt{2^2-\text{t}^2}}$
$=\log\Big|\text{t}+\sqrt{4+\text{t}^2}\Big|+\text{C}$$=\log\Big|\sin\text{x}+\sqrt{4+\sin^2\text{x}}\Big|+\text{C}$
View full question & answer→Question 2203 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 2213 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 2223 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 2233 Marks
Evaluate the following intregals:
$\int\frac{1}{\cos\text{x}(\sin\text{x}+2\cos\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos\text{x}(\sin\text{x}+2\cos\text{x})}\ \text{dx}$
$=\int\frac{1}{\sin\text{x}\cos\text{x}+2\cos^2\text{x}}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\ \text{dx}$
Let $2+\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$\text{I}=\log|2+\tan\text{x}|+\text{C}$
View full question & answer→Question 2243 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 2253 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 2263 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 2273 Marks
Evaluate the following integrals:$\int2\text{x}^3\text{e}^{\text{x}^{2}}\text{dx}$
Answer$\int2\text{x}^3\cdot\text{e}^{\text{x}^{2}}\text{dx}$
$=\int\text{x}^2\cdot\big(\text{e}^{\text{x}^2}\big)\cdot2\text{x dx}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx = dt}$
$=\int\text{t}\cdot\text{e}^{\text{t}}\text{dt}$
$=\text{t}\cdot\text{e}^{\text{t}}-\int1\cdot\text{e}^{\text{t}}\text{dt}$
$=\text{t e}^{\text{t}}-\text{e}^{\text{t}}+\text{C}$
$=\text{x}^2\text{e}^{\text{x}^{2}}-\text{e}^{\text{x}^{2}}+\text{C}$
$=\text{e}^{\text{x}^2}(\text{x}^2-1)+\text{C}$
View full question & answer→Question 2283 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 2293 Marks
Evaluate the following integrals:$\int\frac{\log(\log\text{x})}{\text{x}}\text{dx}$
Answer$\int\frac{\log(\log\text{x})}{\text{x}}\text{dx}$
Taking log log x as the first function and $\frac{1}{\text{x}}$ as the second function.
$=\log \log\text{x}\int\frac{1}{\text{x}}\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\log(\log\text{x})\int\frac{1}{\text{x}}\text{dx}\Big\}\text{dx}$
$=\log\text{x}.\log(\log\text{x})-\int\frac{1}{\text{x}\log\text{x}}(\log\text{x})\text{dx}$
$=\log\text{x}.\log(\log\text{x})-\int\frac{1}{\text{x}}\text{dx}$
$=\log\text{x}.\log(\log\text{x})-\log\text{x}+\text{C}$
$=\log\text{x}[\log(\log\text{x})-1]+\text{C}$
View full question & answer→Question 2303 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 2313 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 2323 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\sin\text{x }\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}\sin\text{x }\text{dx}$
Considering sin x as first function and $e^{2x}$ as second function
$\text{I}=\sin\text{x}\frac{\text{e}^{2\text{x}}}{2}-\int\cos\text{x}\frac{\text{e}^{2\text{x}}}{2}\text{dx}$
$\Rightarrow\text{I}=\sin\text{x}\frac{\text{e}^{2\text{x}}}{2}-\frac{1}{2}\int\cos\text{e}^{2\text{x}}\text{dx}$
$\Rightarrow\text{I}=\frac{\sin\text{x }\text{e}^{2\text{x}}}{2}-\frac{1}{2}\Big[\cos\text{x}\frac{\text{e}^{2\text{x}}}{2}-\int(-\sin\text{x})\frac{\text{e}^{2\text{x}}}{2}\text{dx}\Big]$
$\Rightarrow\text{I}=\frac{\sin\text{x }\text{e}^{2\text{x}}}{2}-\frac{\cos\text{x }\text{e}^{2\text{x}}}{4}-\frac{1}{2}\int\frac{\text{e}^{2\text{x}}\sin\text{x}}{2}\text{dx}$
$\text{I}=\frac{\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})}{4}-\frac{\text{I}}{4}$
$\Rightarrow5\text{I}=\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})$
$\Rightarrow\text{I}=\frac{\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})}{5}+\text{C}$
View full question & answer→Question 2333 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 2343 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 2353 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 2363 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 2373 Marks
Evaluate the following integrals:$\int\sec^{-1}\sqrt{\text{x}}\text{dx}$
Answer$\int1.\sec^{-1}\sqrt{\text{x}}\text{dx}$
$=\sec^{-1}\sqrt{\text{x}}\int1\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\Big(\sec^{-1}\sqrt{\text{x}}\Big)\int1\text{dx}\Big\}\text{dx}$
$=\sec^{-1}\sqrt{\text{x}}.\text{x}-\int\frac{1}{\sqrt{\text{x}}\sqrt{1-\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}\times\text{x dx}$
$=\text{x}\sec^{-1}\sqrt{\text{x}}-\frac{1}{2}\int(1-\text{x})^{-\frac{1}{2}\text{dx}}$
$=\text{x}\sec^{-1}\text{x}-\frac{1}{2}\Bigg[\frac{(1-\text{x})^{-\frac{1}{2}+1}}{\big(-\frac{1}{2}+1\big)(-1)}\Bigg]+\text{C}$
$=\text{x}\sec^{-1}\text{x}+(1-\text{x})^{\frac{1}{2}}+\text{C}$
View full question & answer→Question 2383 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 2393 Marks
Write a value of $\int\text{e}^{2\text{x}^2+\ln\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}^2+\ln\text{x}}\text{ dx}$
$=\int\text{e}^{2\text{x}^2}\cdot\text{e}^{\ln{\text{x}}}\text{dx}$
$=\int\text{x}\cdot\text{e}^{2\text{x}^2}\text{dx}$ $\big[\because\text{e}^{\ln\text{x}}=\text{x}\big]$
$=\int\text{x}\cdot\big(\text{e}^{\text{x}^2}\big)\text{dx}$
Let $\text{e}^{\text{x}^2}=\text{t}$
$\text{e}^{\text{x}^2}\cdot2\text{x dx}=\text{dt}$
$\therefore\ \frac{1}{2}\int\text{t dt}$
$=\frac{1}{2}\frac{\text{t}^2}{2}+\text{C}$
$=\frac{1}{4}\text{e}^{2\text{x}^2}+\text{C}$
$\therefore\ \text{I}=\frac{1}{4}\text{e}^{2\text{x}^2}+\text{C}$
View full question & answer→Question 2403 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 2413 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 2423 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 2433 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 2443 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}(\log\text{x}+\frac{1}{2})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\log\text{x}+\frac{1}{2})\text{dx}$
Here, $\text{f(x)}=\log\text{x}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\text{x}}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\log\text{x}=\text{t}$
Diff. both sides w.r.t x
$\text{e}^{\text{x}}\log\text{x}+\text{e}^{\text{x}}\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}(\log\text{x}+\frac{1}{\text{x}})\text{dx = dt}$
$\therefore\int\text{e}^{\text{x}}\big[\log\text{x}+\frac{1}{\text{x}}\big]\text{dx}=\int\text{dt}$
$\Rightarrow\text{I}=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\log\text{x}+\text{C}$
View full question & answer→Question 2453 Marks
$\int\frac{1}{\sqrt{\text{x+a}}+\sqrt{\text{x+b}}}\text{dx}$
AnswerLet $\text{l}=\int\frac{1}{\sqrt{\text{x+a}}+\sqrt{\text{x+b}}}\text{dx}. $ Then,
$\text{I}=\int\frac{1}{\sqrt{\text{x+a}}+\sqrt{\text{x+b}}}\times\frac{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}\times\text{dx}$
$=\int\frac{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}{\text{x+a}-\text{x-b}}\times\text{dx}$
$=\int\frac{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}{\text{a}-\text{b}}\times\text{dx}$
$=\frac{1}{\text{a}-\text{b}}\bigg[\frac{2}{3}(\text{x+a})^{\frac{3}{2}}-\frac{2}{3}(\text{x+b})^{\frac{3}{2}}\bigg]+\text{c}$
$=\frac{2}{3(\text{a}-\text{b})}\Big[(\text{x+a})^{\frac{3}{2}}-(\text{x+b})^{\frac{3}{2}}\Big]+\text{c}$
$ \text{I}=\frac{2}{3(\text{a}-\text{b})}\Big[(\text{x+a})^{\frac{3}{2}}-(\text{x+b})^{\frac{3}{2}}\Big]+\text{c}$
View full question & answer→Question 2463 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 2473 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 2483 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 2493 Marks
Evaluate the following integrals:$\int\sqrt{\text{cosec}\text{x}-1}\text{ dx}$
Answer$\int\sqrt{\text{cosec}\text{x}-1}\text{ dx}$
$=\int\sqrt{\frac{1}{\sin\text{x}}-1}\text{ dx}$
$=\int\frac{\sqrt{1-\sin\text{x}}}{\sqrt{\sin\text{x}}}\text{ dx}$
$=\int\frac{\sqrt{(1-\sin\text{x})(1+\sin\text{x})}}{\sqrt{\sin\text{x}(1+\sin\text{x})}}\text{ dx}$
$=\int\frac{\sqrt{1-\sin^2\text{x}}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
Now, $=\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}}$
$=\int\frac{\text{dt}}{\sqrt{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}}$
$=\log\Big|\Big(\text{t}+\frac{1}{2}\Big)+\sqrt{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\Big|+\text{C}$
$=\log\Big|\text{t}+\frac{1}{2}+\sqrt{\text{t}^2+\text{t}}\Big|+\text{C}$
$=\log\Big|\sin\text{x}+\frac{1}{2}\sqrt{\sin^2\text{x}+\sin\text{x}}\Big|+\text{C}$
View full question & answer→Question 2503 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{5-4\text{x}-2\text{x}^2}}\text{ dx}$
Answer$\int\frac{1}{\sqrt{5-4\text{x}-2\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{2\big[\frac{5}{2}-2\text{x}-\text{x}^2}\big]}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-2\text{x}-\text{x}^2}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}(\text{x}^2+2\text{x})}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-(\text{x}^2+2\text{x}+1-1)}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{5}{2}-(\text{x}+1)^2+1}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\frac{7}{2}-(\text{x}+1)^2}}$
$=\frac{1}{\sqrt2}\int\frac{\text{dx}}{\sqrt{\Big(\frac{\sqrt7}{\sqrt3}\Big)^2-(\text{x}+1)^2}}$
$=\frac{1}{\sqrt2}\sin^{-1}\Big(\frac{(\text{x}+1)\sqrt2}{\sqrt7}\Big)+\text{C}$
$=\frac{1}{\sqrt2}\sin^{-1}\Big(\sqrt{\frac{2}{7}}(\text{x}+1)\Big)+\text{C}$
View full question & answer→