Question 2013 Marks
Evaluate the following integrals:
$\int\text{x}^2\sqrt{\text{a}^6-\text{x}^6}\text{dx}$
AnswerLet $\text{I}=\int\text{x}^2\sqrt{\text{a}^6-\text{x}^6}\text{dx}$
Let $\text{x}^3=\text{t}$
$\Rightarrow3\text{x}^2\text{dx}=\text{dt}$
$\therefore\ \text{I}=\frac{1}{3}\int\sqrt{\text{a}^6-\text{t}^2}\text{dt}$
$=\frac{1}{3}\begin{Bmatrix}\frac{\text{t}}{2}\sqrt{\text{a}^6-\text{t}^2}+\frac{\text{a}^6}{2}\sin^{-1}\Big(\frac{\text{t}}{\text{a}^3}\Big)\end{Bmatrix}+\text{C}$
$\therefore\ \text{I}=\frac{\text{x}^3}{6}\sqrt{\text{a}^6-\text{x}^6}+\frac{\text{a}^6}{6}\sin^{-1}\Big(\frac{\text{x}^3}{\text{a}^3}\Big)+\text{C}$
View full question & answer→Question 2023 Marks
Evaluate the following intregals:
$\int\frac{1}{\cos2\text{x}+3\sin^2\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos2\text{x}+3\sin^2\text{x}}\ \text{dx}$
$=\int\frac{1}{2\cos^2\text{x}-1+3\sin^2\text{x}}\ \text{dx}$
$\text{I}=\frac{1}{\sqrt{2}}\int\frac{1}{1+\text{t}^2}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{2-\sec^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{2-(1+\tan^2\text{x})^2+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{2-1-\tan^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\text{dt}}{1+2\tan^2\text{x}}$
Let $\sqrt{2}\tan\text{x}=\text{t}$
$\sqrt{2}\sec^2\text{x dx}=\text{dt}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\text{t}+\text{C}$
$\text{I}=\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan\text{x})+\text{C}$
View full question & answer→Question 2033 Marks
$\int\frac{1}{(7\text{x}-5)^2}+\frac{1}{\sqrt{5\text{x}-4}}\text{dx}$
AnswerLet $\text{I}=\int\bigg[\frac{1}{(7\text{x}-5)^3}+\frac{1}{\sqrt{5\text{x}-4}}\bigg]\text{dx}.$ Then,
$\text{I}=\int(7\text{x}-5)^{-3}\text{dx}+\int(5\text{x}-4)^{\frac{-1}{2}}\text{dx}$
$=\frac{(7\text{x}-5)^{-2}}{7\times(-2)}+\frac{(5\text{x}-4)^{\frac{1}{2}}}{5\times\frac{1}{2}}+\text{c}$
$=-\frac{(7\text{x}-5)^{6-2}}{14}+\frac{2}{5}\sqrt{(5\text{x}-4)}+\text{c}$
$\text{I}=\frac{-1}{14}(7\text{x}-5)^{-2}+\frac{2}{5}\times\sqrt{5\text{x}-4}+\text{c}.$
View full question & answer→Question 2043 Marks
Evaluate the following integrals:
$\int\Big(\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\tan^{-1}\text{x}+\frac{1}{1+\text{x}^2}\Big)\text{dx}$
Here, $\text{f(x)}=\tan^{-1}\text{x}$ and $\text{f}'\text{(x)}=\frac{1}{1+\text{x}^2}$
and we know thet,
$\int\text{e}^{\text{ax}}(\text{af}(\text{x})+\text{f}'(\text{x}))\text{dx}=\text{e}^{\text{ax}}\text{f(x)+C}$
$\therefore\int\text{e}^{\text{x}}\Big(\tan^{-1}\text{x}+\frac{1}{1+\text{x}^2}\Big)\text{dx}=\text{e}^{\text{x}}\tan^{-1}\text{x + C}$
Thus,
$\text{I}=\text{e}^{\text{x}}\tan^{-1}\text{x + C}$
View full question & answer→Question 2053 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{\tan^{-1}\text{x}}.(1+\text{x}^2)}\text{dx}$
Answer$\int\frac{\text{dx}}{\sqrt{\tan^{-1}\text{x}}(1+\text{x}^2)}$
$\text{Let }\tan^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{1+\text{x}^2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
$\text{Now,}\int\frac{\text{dx}}{\sqrt{\tan^{-1}\text{x}}(1+\text{x}^2)}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\tan^{-1}\text{x}}+\text{C}$
View full question & answer→Question 2063 Marks
Evaluate the following integrals:
$\int\cot^3\text{x }\text{cosec}^2\text{x}\text{ dx}$
Answer$\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}$
$\text{Let},\cot\text{x}=\text{t}$
$\Rightarrow-\text{cosec}^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{cosec}^2\text{x}\text{ dx}=-\text{dt}$
$\text{Now},\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}$
$=\int\text{t}^3(-\text{dt})$
$=\frac{-\text{t}^4}{4}+\text{C}$
$=\frac{-\cot^4\text{x}}{4}+\text{C}$
View full question & answer→Question 2073 Marks
Evaluate the following integrals:
$\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
Answer$\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
$\text{Let},\text{x}-\cos\text{x}=\text{t}$
$\Rightarrow(1+\sin\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\sin\text{x})\text{dx}=\text{dt}$
$\text{Now,}\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\text{x}-\cos\text{x}}+\text{C}$
View full question & answer→Question 2083 Marks
Evaluate the following integrals:
$\int\frac{\log\text{x}^2}{\text{x}}\text{dx}$
Answer$\int\frac{\log\text{x}^2\text{dx}}{\text{x}}$
$=\int\frac{2\log\text{x}}{\text{x}}\text{dx}$
$=2\int\frac{\log\text{x}}{\text{x}}\text{dx}$
$\text{Let }\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{Now, }2\int\frac{\log\text{x}}{\text{x}}\text{dx}$
$=2\int\text{t dt}$
$2\Big[\frac{\text{t}^2}{2}\Big]+\text{C}$
$=\text{t}^2+\text{C}$
$=(\log\text{x})^2+\text{C}$
View full question & answer→Question 2093 Marks
Write a value of $\int\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})\text{dx}$
$\because\ \int\text{e}^{\text{x}}\big(\text{f(x})+\text{f}'(\text{x})\big)\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}$
Here, $\text{f(x)}=\sin\text{x}$ and $\text{f}'(\text{x})=\cos\text{x}$
$\therefore\ \text{I}=\int\text{e}^{\text{x}}(\sin\text{x}+\cos\text{x})\text{dx}=\text{e}^{\text{x}}\sin\text{x}+\text{C}$
View full question & answer→Question 2103 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\Big(\frac{\sin\text{x}\cos\text{x}-1}{\sin^2\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{\sin\text{x}\cos\text{x}-1}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\big[\cot\text{x}-\text{cosec}^2\text{x}\big]\text{dx}$
Here, $\text{f(x)}=\cot\text{x}$
$\Rightarrow\text{f}'\text{(x)}=-\text{cosec}^2\text{x}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\cot\text{x}=\text{t}$
Diff both sides w.r.t x
$\text{e}^{\text{x}}(\cot\text{x}-\text{cosec}^2\text{x})\text{dx = dt}$
$\therefore\text{I}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\cot\text{x + C}$
View full question & answer→Question 2113 Marks
$\int(\text{e}^{\text{x}}+\frac{1}{\text{e}^{\text{x}}})^2\text{dx}$
Answer$\int(\text{e}^{\text{x}}+\frac{1}{\text{e}^{\text{x}}})^2\text{dx}$
$=\int(\text{e}^{2\text{x}}+\frac{1}{\text{e}^{2\text{x}}}2\text{e}^\text{x}\times\frac{1}{\text{e}^{\text{x}}})\text{dx}$
$=\int(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}+2)\text{dx}$
$=\frac{\text{e}^{2\text{x}}}{2}+\frac{\text{e}^{-2\text{x}}}{-2}+2\text{x}+\text{c}$
$=\frac{\text{e}^{2\text{x}}}{2}-\frac{\text{e}^{-2\text{x}}}{2}+2\text{x}+\text{c}$
View full question & answer→Question 2123 Marks
Evalute the following integrals:
$\int\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}\text{dx}$
Answer$\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}=\frac{2\cos\text{x}-3\sin\text{x}}{2(3\cos\text{x}+2\sin\text{x})}$
Let $3\cos\text{x}+2\sin\text{x}=\text{t}$
$(-3\sin\text{x}+2\cos\text{x})\text{dx}=\text{dt}$
$\int\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}\text{dx}=\int\frac{\text{dt}}{2\text{t}}$
$=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{2}\log|\text{t}|+\text{C}$
$=\frac{1}{2}\log|2\sin\text{x}+3\cos\text{x}|=\text{C}$
View full question & answer→Question 2133 Marks
Evaluate the following integrals:
$\int\frac{5\cos^3\text{x}+6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\text{dx}$
Answer$\int\bigg(\frac{5\cos^3\text{x}+6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int\bigg(\frac{5\cos^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}+\frac{6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int\Big(\frac{5}{2}\frac{\cos\text{x}}{\sin^2\text{x}}+3\frac{\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\frac{5}{2}\int\Big(\frac{\cos\text{x}}{\sin\text{x}}\times\frac{1}{\sin\text{x}}\Big)\text{dx}+3\int\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\text{dx}$
$=\frac{5}{2}\int(\text{cosec x}\cot\text{x})\text{dx}+3\int\sec\text{x}\tan\text{x dx}$
$=\frac{5}{2}(-\text{cosec x})+3\sec\text{x}+\text{C}$
$=-\frac{5}{2}\text{cosec x}+3\sec\text{x}+\text{C}$
View full question & answer→Question 2143 Marks
Evaluate the following integrals:
$\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sqrt{16+(\log\text{x})^2}}{\text{x}}\text{dx}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\sqrt{16+\text{t}^2}\text{dt}$
$=\int\sqrt{4^2+\text{t}^2}\text{dt}$
$=\frac{\text{t}}{2}\sqrt{16+\text{t}^2}+\frac{16}{2}\log\big|\text{t}+\sqrt{16+\text{t}^2}\big|+\text{C}$
$\therefore\ \text{I}=\frac{\log\text{x}}{2}\sqrt{16+(\log\text{x})^2}\\+8\log\Big|\log\text{x}+\sqrt{16+(\log\text{x})^2}\Big|+\text{C}$
View full question & answer→Question 2153 Marks
Evaluate the following integrals:
$\int\frac{5\text{x}^4+12\text{x}^3+7\text{x}^2}{\text{x}^2+\text{x}}\text{dx}$
Answer$\int\frac{5\text{x}^4+12\text{x}^3+7\text{x}^2}{\text{x}^2+\text{x}}\text{dx}$
$=\int\frac{5\text{x}^4+7\text{x}^3+5\text{x}^3+7\text{x}^2}{\text{x}^2+\text{x}}\text{dx}$
$=\int\frac{5\text{x}^3+7\text{x}^2+5\text{x}^2+7\text{x}}{\text{x}+1}\text{dx}$
$=\int\frac{5\text{x}^2(\text{x}+1)+7\text{x}(\text{x}+1)}{\text{x}+1}\text{dx}$
$=\int(5\text{x}^2+7\text{x})\text{dx}$
$=\frac{5\text{x}^3}{3}+\frac{7\text{x}^2}{2}+\text{C}$
View full question & answer→Question 2163 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2\text{dx}}{\text{x}^6-\text{a}^6}\text{dx}$
Answer$\int\frac{\text{x}^2\text{dx}}{\text{x}^6-\text{a}^6}$
Let $\text{x}^3=\text{t}$
$\Rightarrow3\text{x}^2\text{dx = dt}$
$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$
Now, $\int\frac{\text{x}^2\text{dx}}{\text{x}^6-\text{a}^6}$
$=\frac{1}{3}\int\frac{\text{dt}}{\text{t}^2-(\text{a}^3)^2}$
$=\frac{1}{3}\times\frac{1}{2\text{a}^3}\log\bigg|\frac{\text{t}-\text{a}^3}{\text{t}+\text{a}^3}\Big|+\text{C}$
$=\frac{1}{6\text{a}^3}\log\Big|\frac{\text{x}^3-\text{a}^3}{\text{x}^3+\text{a}^3}\Big|+\text{C}$
View full question & answer→Question 2173 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+1}}\text{ dx}$
AnswerWe have
$\text{I}=\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+1}}\text{ dx}$
Putting $\text{x}-1=\frac{1}{\text{t}}$
$\text{dx}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=\int\frac{-\frac{1}{\text{t}^2}\text{ dt}}{\big(\frac{1}{\text{t}}\big)\sqrt{\big(1+\frac{1}{\text{t}}}\big)^2+1}$
$=\int\frac{-\frac{1}{\text{t}}\text{ dt}}{\sqrt{1+\frac{1}{\text{t}^2}+\frac{2}{\text{t}}+1}}$
$=\int\frac{-\frac{1}{\text{t}}\text{ dt}}{\sqrt{\frac{\text{t}^2+1+2\text{t}+\text{t}^2}{\text{t}}}}$
$=\int\frac{-\text{dt}}{\sqrt{2\text{t}^2+2\text{t}+1}}$
$=-\frac{1}{\sqrt{2}}\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\frac{1}{2}}}$
$=-\frac{1}{\sqrt{2}}\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\frac{1}{4}-\frac{1}{4}+\frac{1}{2}}}$
$=-\frac{1}{\sqrt{2}}\int\frac{\text{dt}}{\big(\text{t}+\frac{1}{2}\big)^2+\big(\frac{1}{2}\big)^2}$
$=-\frac{1}{\sqrt{2}}\log\begin{vmatrix}\text{t}+\frac{1}{2}+\sqrt{\Big(\text{t}+\frac{1}{2}\Big)^2+\frac{1}{4}}\end{vmatrix}+\text{C}$ where $\text{t}=\frac{1}{\text{x}-1}$
View full question & answer→Question 2183 Marks
Evaluate the following integrals:
$\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$
Answer$\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$
Let $\text{e}^\text{x}-1=\text{t}^2$
$\Rightarrow\text{e}^\text{x}=\text{t}^2+1$
$\text{e}^\text{x}=2\text{t}\frac{\text{dt}}{\text{dx}}$
$\text{dx}=\frac{2\text{t dt}}{\text{e}^\text{x}}$
$\text{dx}=\frac{2\text{t dt}}{\text{t}^2+1}$
Now, $\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$
$=\int\frac{\text{t. 2t dt}}{\text{t}^2+1}$
$=2\int\frac{\text{t}^2\text{ dt}}{\text{t}^2+1}$
$=2\int\Big(\frac{\text{t}^2+1-1}{\text{t}^2+1}\Big)\text{ dt}$
$=2\int\text{dt}-2\int\frac{\text{dt}}{\text{t}^2+1}$
$=2\text{t}-2\tan^{-1}(\text{t})+\text{C}$
$=2\sqrt{\text{e}^\text{x}-1}-2\tan^{-1}\big(\sqrt{\text{e}^\text{x}-1}\big)+\text{C}$
View full question & answer→Question 2193 Marks
Write a value of $\int\frac{\log\text{x}^{\text{n}}}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\log\text{x}^{\text{n}}}{\text{x}}\text{ dx}$
Let $\log\text{x}^{\text{n}}=\text{t}$
$\frac{\text{nx}^{\text{n}-1}}{\text{x}^{\text{n}}}\text{ dx}=\text{dt}$
$\frac{\text{n}}{\text{x}}=\text{ dt}$
$\therefore\ \text{I}=\text{n}\int\text{t}\text{ dt}$
$=\text{n}\Big(\frac{\text{t}^2}{2}\Big)+\text{C}$
Putting the value of t
$\text{I}=\frac{\text{n}(\log\text{x}^{\text{n}})^2}{2}+\text{C}$
View full question & answer→Question 2203 Marks
Evaluate the following integrals:
$\int\sec^6\text{x }\tan\text{x}\text{ dx}$
Answer$\int\sec^6\text{x }\tan\text{x}\text{ dx}$
$\int\sec^6\text{x}.\sec\text{x}\tan\text{x}\text{ dx}$
Let $\sec\text{x}=\text{t}$
$\sec\text{x}\tan\text{x}\text{ dx}=\text{dt}$
Now, $\int\sec^6\text{x}.\sec\text{x}\tan\text{x}\text{ dx}$
$=\int\text{t}^6\text{dt}$
$=\frac{\text{t}^6}{6}+\text{C}$
$=\frac{\sec^6\text{x}}{6}+\text{C}$
View full question & answer→Question 2213 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
Answer$\int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1}{\sqrt{\text{x}}}+\frac{1}{\sqrt{\text{x}}\text{x}}\Big)\text{dx}$
$=\int\text{x}^{\frac{-1}{2}}+\int\text{x}^{\frac{-3}{2}}\text{dx}$
$=2\text{x}^{\frac{1}{2}}-2\text{x}^{\frac{-1}{2}}+\text{C}$
$=2\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}+\text{C}$
$\therefore\ \int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}=2\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}+\text{C}$
View full question & answer→Question 2223 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2(\text{xe}^\text{x})}\text{ dx}$
Answer$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2(\text{xe}^\text{x})}\text{ dx}$ Let $\text{xe}^\text{x}=\text{t}$ $\Rightarrow(1.\text{e}^\text{x}+\text{xe}^\text{x})=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow(\text{x}+1)\text{e}^\text{x}\text{dx}=\text{dt}$ Now, $\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2({\text{xe}^\text{x}})}=\text{dx}$$=\int\frac{\text{dt}}{\sin^2\text{t}}$
$=\int\text{cosec}^2\text{t}\text{ dt}$
$=-\cot(\text{t})+\text{C}$
$=-\cot(\text{xe}^\text{x})+\text{C}$
View full question & answer→Question 2233 Marks
Evaluate the following integrals:
$\int\text{x e}^{\text{x}^2}=\text{dx}$
AnswerLet $\text{I}=\int\text{x e}^{\text{x}^2}=\text{dx}\ ....(1)$ Let $\text{x}^2=\text{t}$ then, $\text{d}(\text{x}^2)=\text{dt}$ $\Rightarrow2\text{x dx}=\text{dt}$ $\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ Putting $\text{x}^2=\text{t}$ and $\text{x dx}=\frac{\text{dt}}{2}$ in equation (1),we get,
$\text{I}=\int\text{e}^\text{t}\frac{\text{dt}}{2}$
$=\frac{1}{2}\text{e}^\text{t}+\text{C}$
$=\frac{1}{2}\text{e}^{\text{x}^2}+\text{C}$
$\text{I}=\frac{1}{2}\text{e}^{\text{x}^2}+\text{C}$
View full question & answer→Question 2243 Marks
Evaluate the following integrals:
$\int\frac{(1+\text{x})^3}{\sqrt{\text{x}}}\text{dx}$
Answer$\int\frac{(1+\text{x})^3}{\sqrt{\text{x}}}\text{dx}$
$=\int\frac{1}{\sqrt{\text{x}}}\text{dx}+\int\frac{\text{x}^3}{\sqrt{\text{x}}}\text{dx}+\int\frac{3\text{x}^2}{\sqrt{\text{x}}}\text{dx}+\int\frac{3\text{x}}{\sqrt{\text{x}}}\text{dx}$
$=\int\text{x}^{\frac{-1}{2}}\text{dx}+\int\text{x}^{\frac{5}{2}}\text{dx}+3\int\text{x}^{\frac{3}{2}}\text{dx}+3\int\text{x}^{\frac{1}{2}}\text{dx}$
$=\frac{\text{x}^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}+\frac{3\text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{3\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{C}$
$=\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\frac{\text{x}^{\frac{7}{2}}}{\frac{7}{2}}+\frac{\text{3x}^{\frac{5}{2}}}{\frac{5}{2}}+\frac{\text{3x}^{\frac{3}{2}}}{\frac{3}{2}}+\text{C}$
$=2\text{x}^{\frac{1}{2}}+\frac{2}{7}\text{x}^{\frac{7}{2}}+\frac{6}{5}\text{x}^{\frac{5}{2}}+\frac{6}{3}\text{x}^{\frac{3}{2}}+\text{C}$
$=2\text{x}^{\frac{1}{2}}+\frac{2}{7}\text{x}^{\frac{7}{2}}+\frac{6}{5}\text{x}^{\frac{5}{2}}+2\text{x}^{\frac{3}{2}}+\text{C}$
View full question & answer→Question 2253 Marks
Write a value of $\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{dx}$
We know that,
$\int\text{e}^{\text{x}}\int\text{f}(\text{x})+\text{f}'(\text{x})=\text{e}^{\text{x}}\text{f}(\text{x})+\text{C}$
Hence, $\text{f}'(\text{x})=-\frac{1}{\text{x}^2}$
Then, $\int\text{e}^{\text{ax}}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{e}^{\text{x}}\cdot\frac{1}{\text{x}}+\text{C}$
$\therefore\ \text{I}=\text{e}^{\text{x}}\cdot\frac{1}{\text{x}}+\text{C}$
View full question & answer→Question 2263 Marks
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\sin\Big(\text{x}-\frac{\pi}{6}\Big)\sin\Big(\text{x}+\frac{\pi}{6}\Big)}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\sin\Big(\text{x}-\frac{\pi}{6}\Big)\sin\Big(\text{x}+\frac{\pi}{6}\Big)}\text{dx}$
$=\int\frac{\sin2\text{x}}{\sin^2\text{x}-\sin^2\frac{\pi}{6}}\text{dx}$
$\big[\because \sin(\text{A}+\text{B})\sin(\text{A}-\text{B})=\sin^2\text{A}-\sin^2\text{B}\big]$
$=\int\frac{\sin2\text{x}}{\sin^2\text{x}-\frac{1}{4}}\text{dx}$
Putting $\sin^2\text{x}-\frac{1}{4}=\text{t}$
$\Rightarrow2\sin\text{x}\cos\text{ x dx}=\text{dt}$
$\Rightarrow\sin2\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}\big|\sin^2\text{x}-\frac{1}{4}\big|+\text{C}\ \Big[\because\text{t}=\sin^2\text{x}-\frac{1}{4}\Big]$
View full question & answer→Question 2273 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^6+1}{\text{x}^2+1}\text{dx}$
Answer$\int\Big(\frac{\text{x}^6+1}{\text{x}^2+1}\Big)\text{dx}$
$=\int\bigg[\frac{(\text{x}^2)^3+1^3}{\text{x}^2+1}\bigg]\text{dx}$ $[\text{A}^3+\text{B}^3=(\text{A+B})(\text{A}^2-\text{AB}+\text{B}^2)]$
$=\int\frac{(\text{x}^2+1)(\text{x}^4-\text{x}^2+1)}{(\text{x}^2+1)}\text{dx}$
$=\int(\text{x}^4-\text{x}^2+1)\text{dx}$
$=\int\text{x}^4\text{dx}+\int\text{x}^2\text{dx}+\int1\text{dx}$
$=\frac{\text{x}^{4+1}}{4+1}-\frac{\text{x}^{2+1}}{2+1}+\text{x}+\text{C}$
$=\frac{\text{x}^5}{5}-\frac{\text{x}^3}{3}+\text{x}+\text{C}$
View full question & answer→Question 2283 Marks
Evaluate the following integrals:
$\int\sqrt{3+2\text{x}-\text{x}^2}\text{dx}$
Answer$\int\sqrt{3+2\text{x}-\text{x}^2}\text{dx}=\int\sqrt{4-(\text{x}-1)^2}\text{dx}$ Let X - 1 = t, so that dx = dt Thus,$\int\sqrt{3+2\text{x}-\text{x}^2}\text{dx}=\int\sqrt{4-\text{t}^2}\text{dt}$ $=\frac{1}{2}\text{t}\sqrt{4-\text{t}^2}+\frac{4}{2}\sin^{-1}\Big(\frac{\text{t}}{2}\Big)+\text{C}$$=\frac{1}{2}(\text{x}-1)\sqrt{3+2\text{x}-\text{x}^2}+2\sin^{-1}\Big(\frac{\text{x}-1}{2}\Big)+\text{C}$
View full question & answer→Question 2293 Marks
Integrate the following integrals:
$\int\sin4\text{x}\cos7\text{x dx}$
Answer$\int\sin4\text{x}\cos7\text{x dx}$
$=\frac{1}{2}\int2\cos7\text{x}\sin4\text{x dx}$
$=\frac{1}{2}\int\big[\sin(7\text{x}+4\text{x})-\sin(7\text{x}-4\text{x})\big]\text{dx}$ $[\therefore2\cos\text{A}\sin\text{B}=\sin(\text{A}+\text{B})-\sin(\text{A}-\text{B})\big]$
$=\frac{1}{2}\int\big(\sin(11\text{x})-\sin(3\text{x})\big)\text{dx}$
$=\frac{1}{2}\Big[-\frac{\cos(11\text{x})}{11}+\frac{\cos(3\text{x})}{3}\Big]+\text{c}$
$=-\frac{\cos(11\text{x})}{22}+\frac{\cos(3\text{x})}{6}$
View full question & answer→Question 2303 Marks
Write a value of $\int\frac{1}{\text{x}(\log\text{x})^{\text{n}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{x}(\log\text{x})^{\text{n}}}\text{ dx}$
Let $\log\text{x}=\text{t}$
$\frac{1}{\text{x}}\text{ dx}=\text{dt}$
$\text{dx}=\text{xdt}$
$\therefore\ \text{I}=\int\frac{1}{\text{t}^{\text{n}}}\text{ dt}$
$=\frac{\text{t}^{-\text{n}+1}}{-\text{n}+1}+\text{C}$
$\text{I}=\frac{(\log\text{x})^{1-\text{n}}}{1-\text{n}}+\text{C}$
View full question & answer→Question 2313 Marks
Evaluate the following integrals:
$\int\Big(\frac{\text{x}+1}{\text{x}}\Big)(\text{x}+\log\text{x})^2\text{dx}$
Answer $\int\Big(\frac{\text{x}+1}{\text{x}}\Big)(\text{x}+\log\text{x})^2\text{dx}$ Let $\text{x}+\log\text{x}=\text{t}$ $\Rightarrow\Big(1+\frac{1}{\text{x}}\Big)=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\Big(\frac{\text{x}+1}{\text{x}}\Big)\text{dx}=\text{dt}$Now, $\int\Big(\frac{\text{x}+1}{\text{x}}\Big)(\text{x}+\log\text{x})^2\text{dx}$
$=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^2}{3}+\text{C}$
$=\frac{(\text{x}+\log\text{x})^3}{3}+\text{C}$
View full question & answer→Question 2323 Marks
Evaluate the following integrals:
$\int\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{1+\cos4\text{x}}}\text{dx}$
Answer$\int\Big(\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{1+\cos4\text{x}}}\Big)\text{x}$
$=\int\frac{\cos(2\text{x})}{\sqrt{2\cos^2(2\text{x})}}\text{dx}$ $\Big[\therefore1+\cos\text{A}=2\cos^2\Big(\frac{\text{A}}{2}\Big)\text{ and }\cos^2\text{A}-\sin^2\text{A}=\cos\text{2A}\Big]$
$=\frac{1}{\sqrt2}\int\Big(\frac{\cos2\text{x}}{\cos\text{2x}}\Big)\text{dx}$
$=\frac{1}{\sqrt{2}}[\text{x}]+\text{C}$
$=\frac{\text{x}}{\sqrt{2}}+\text{C}$
View full question & answer→Question 2333 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
Here, $\text{f(x)}=\log\sin\text{x}$ Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{f}'\text{(x)}=\cot\text{x}$
let $\text{e}^{\text{x}}\log\sin\text{x = t}$
Diff. both sides w.r.t x
$\text{e}^{\text{x}}\log(\sin\text{x})+\text{e}^{\text{x}}\times\frac{1}{\sin\text{x}}\times\cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\big[\text{e}^{\text{x}}\log(\sin\text{x})+\text{e}^{\text{x}}\cot\text{x}\big]\text{dx = dt}$
$\Rightarrow\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx = dt}$
$\therefore \int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx} =\int\text{dt}$
$=\text{t + C}$
$=\text{e}^{\text{x}}\log\sin\text{x}+\text{C}$
View full question & answer→Question 2343 Marks
Evaluate the following integrals:
$\int\tan^32\text{x}\sec2\text{x dx}$
Answer$\int\tan^32\text{x}\sec2\text{x}=\tan^22\text{x}\tan2\text{x}\sec2\text{x}$
$=\big(\sec^22\text{x}-1\big)\tan2\text{x}\sec2\text{x}$
$=\sec^22\text{x}\tan2\text{x}\sec2\text{x}-\tan2\text{x}\sec2\text{x}$
$\therefore\ \int\tan^32\text{x}\sec2\text{x}\text{ dx}$
$=\int\sec^22\text{x}\tan2\text{x}\sec2\text{x dx}-\int\tan2\text{x}\sec2\text{x dx}$
$=\int\sec^22\text{x}\tan2\text{x}\sec2\text{x dx}-\frac{\sec2\text{x}}{2}+\text{C}$
Let $2\text{x}=\text{t}$
$\therefore\ 2\sec2\text{x}\tan2\text{x dx}=\text{dt}$
$\therefore\ \int\tan^32\text{x}\sec2\text{x dx}=\frac{1}{2}\int\text{t}^2\text{dt}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{\text{t}^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{(\sec2\text{x})^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
View full question & answer→Question 2353 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}}\Big(\text{x}^3-\frac{2}{\text{x}}\Big)\text{dx}$
Answer$\int\sqrt{\text{x}}\Big(\text{x}^3-\frac{2}{\text{x}}\Big)\text{dx}$
$=\int\Big(\text{x}^{\frac{7}{2}}-\frac{2}{\sqrt{\text{x}}}\Big)\text{dx}$
$=\int\Big(\text{x}^{\frac{7}{2}}-{2}{\text{x}^{-\frac{1}{2}}}\Big)\text{dx}$
$=\frac{\text{x}^{\frac{7}{2}+1}}{\frac{7}{2}+1}-2\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=\frac{2}{9}\text{x}^{\frac{9}{2}}-4\text{x}^{\frac{1}{2}}+\text{C}$
$=\frac{2}{9}\text{x}^{\frac{9}{2}}-4\sqrt{\text{x}}+\text{C}$
View full question & answer→Question 2363 Marks
Evaluate the following integrals:
$\int \frac{1}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
AnswerLet $\text{I}=\int \frac{1}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
Let $\text{x}+2=\text{t}^2$
$\therefore\ \text{I}=\int\frac{2\text{tdt}}{(\text{t}^2-3)\text{t}}$
$=2\int\frac{\text{dt}}{\text{t}^2-3}$
$=\frac{2}{\sqrt{3}}\log\Big|\frac{\text{t}-\sqrt{3}}{\text{t}+\sqrt{3}}\Big|+\text{C}$
Thus, $\text{I}=\frac{1}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}-2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}$
View full question & answer→Question 2373 Marks
Evaluate the following integrals:
$\int\text{x}^2\tan^{-1}\text{x dx} $
AnswerLet $\text{I}=\int\text{x}^2\tan^{-1}\text{x dx}$
$=\tan^{-1}\text{x}\int\text{x}^2\text{dx}-\int\Big(\frac{1}{1+\text{x}^2}\int\text{x}^2\text{dx}\Big)$
$=\tan^{-1}\text{x}\Big(\frac{\text{x}^3}{3}\Big)-\frac{1}{3}\frac{\text{x}^3}{1+\text{x}^2}\text{dx}$
$=\frac{1}3{\text{x}^3}\tan^{-1}\text{x}-\frac{1}{3}\int\Big(\text{x}-\frac{\text{x}}{1+\text{x}^2}\Big)\text{dx}$
$=\frac{1}{3}\text{x}^3\tan^{-1}\text{x}-\frac{1}{3}\times\frac{\text{x}^2}{2}+\frac{1}3{}\int\frac{\text{x}}{1+\text{x}^2}\text{dx}$
$\text{I}=\frac{1}{3}\text{x}^3\tan^{-1}\text{x}-\frac{1}{6}\text{x}^2+\frac{1}{6}\log\big|1+\text{x}^2\big|+\text{C}$
View full question & answer→Question 2383 Marks
Evalute the following integrals:
$\int\frac{1}{\text{x}\log\text{x}\log(\log\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.
Let $\text{I}=\int\frac{1}{\text{x}\log\text{x}\log(\log\text{x})}\text{dx}$
Putting $\log(\log\text{x})=\text{t}$
$\Rightarrow\frac{1}{\text{x}\log\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}\log\text{x}}\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\log\text{x})|+\text{C}\ \big[\because\text{t}=\log(\log\text{x})\big]$
View full question & answer→Question 2393 Marks
Evaluate the following integrals:
$\int\frac{\sin2\text{x}}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
Answer$\int\frac{\sin2\text{x}}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
$\text{Let a}+\text{b}\cos2\text{x}=\text{t}$
$\Rightarrow-\text{b}\sin(2\text{x})\text{dx}\times2=\text{dt}$
$\Rightarrow\sin(2\text{x})\text{dx}=\frac{-\text{dt}}{2\text{b}}$
$\text{Now,}\int\frac{\sin(2\text{x})}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
$=-\frac{1}{2\text{b}}\int\frac{\text{dt}}{\text{t}^2}$
$=\frac{-1}{2\text{b}}\int\text{t}^{-2}\text{dt}$
$=\frac{-1}{2\text{b}}\Big[\frac{\text{t}^{-2+1}}{-2+1}\Big]+\text{C}$
$=\frac{1}{2\text{b}}\times\frac{1}{\text{t}}+\text{C}$
$=\frac{1}{2\text{b}(\text{a}+\text{b}\cos2\text{x})}+\text{C}$
View full question & answer→Question 2403 Marks
Evaluate the following integrals:
$\int\text{x}\frac{\tan^{-1}\text{x}^2}{1+\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int\text{x}\frac{\tan^{-1}\text{x}^2}{1+\text{x}^4}\text{ dx}\ ....(1)$ Let $\tan^{-1}\text{x}^2=\text{t}$ then, $\text{d}\big(\tan^{-1}\text{x}^2\big)=\text{dt}$ $\Rightarrow\frac{1\times2\text{x}}{1+(\text{x}^2)^2}\text{ dx}=\text{dt}$ $\Rightarrow\frac{1\times\text{x}}{1+\text{x}^4}\text{ dx}=\frac{\text{dt}}{2}$ Putting, $\tan^{-1}\text{x}^2=\text{t}$ and $\frac{\text{x}}{1+\text{x}^4}\text{ dx}=\frac{\text{dt}}{2}$ in equation (1),we get,
$\text{I}=\int\text{t}\frac{\text{dx}}{2}$
$=\frac{1}{2}\int\text{t dt}$
$=\frac{1}{2}\times\frac{\text{t}^2}{2}+\text{C}$
$\text{I}=\frac{\text{t}^2}{4}+\text{C}$
$=\frac{(\tan^{-1}\text{x}^2)^2}{4}+\text{C}$
$\text{I}=\frac{1}{4}\big(\tan^{-1}\text{x}^2\big)^2+\text{C}$
View full question & answer→Question 2413 Marks
$\int\frac{2\text{x}+3}{(\text{x}-1)^2}\text{dx}$
Answer$\int\bigg(\frac{2\text{x}+3}{(\text{x}-1)^2}\bigg)\text{dx}$
$=\int\bigg[\frac{2\text{x}-2+2+3}{(\text{x}-1)^2}\bigg]\text{dx}$
$=\int\bigg[\frac{2(\text{x}-1)+5}{(\text{x}-1)^2}\bigg]\text{dx}$
$=2\int\frac{\text{dx}}{(\text{x}-1)}+5\int(\text{x}-1)^{-2}\text{dx}$
$=2\text{ln}|\text{x}-1|+5\bigg[\frac{(\text{x}-1)^{-2+1}}{-2+1}\bigg]+\text{C}$
$=2\text{ln}|\text{x}-1|-\frac{5}{\text{x}-1}+\text{c}$
View full question & answer→Question 2423 Marks
Evaluate the following integrals:
$\int\text{x}\cos^2\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\cos^2\text{x dx}$
Using integration by parts,
$\text{I}=\text{x}\int\cos^2\text{x dx}-\int(1\int\cos^2\text{x dx})\text{dx}$
$=\text{x}\int\Big(\frac{\cos2\text{x}+1}{2}\Big)\text{dx}-\int\bigg(\int\Big(\frac{1+\cos2\text{x}}{2}\Big)\text{dx}\bigg)\text{dx}$
$=\frac{\text{x}}{2}\Big[\frac{\sin2\text{x}}{2}+\text{x}\Big]-\frac{1}{2}\int\Big(\text{x}+\frac{\sin2\text{x}}{2}\Big)\text{dx}$
$=\frac{\text{x}}{4}\sin2\text{x}+\frac{\text{x}^2}{2}-\frac{1}{2}\times\frac{\text{x}^2}{2}-\frac{1}{4}\Big(-\frac{\cos2\text{x}}{2}\Big)+\text{C}$
$\text{I}=\frac{\text{x}}{4}\sin2\text{x}+\frac{\text{x}^2}{4}+\frac{1}{8}\cos2\text{x}+\text{C}$
View full question & answer→Question 2433 Marks
Evaluate the following integrals:$\int(\text{x}+1)\log\text{x dx}$
Answer$\int(\text{x}+1).\log\text{x dx}$
$=\log \text{x}\int(\text{x}+1)\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\log\text{x})\int(\text{x}+1)\text{dx}\Big\}\text{dx}$
$=\log\text{x}\Big[\frac{\text{x}^2}{2}+\text{x}\Big]-\int\frac{1}{\text{x}}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)-\int\big(\frac{\text{x}}{2}+1\big)\text{dx}$
$=\log\text{x}\Big(\frac{\text{x}^2}{2}+\text{x}\Big)-\Big(\frac{\text{x}^2}{4}+\text{x}\Big)+\text{C}$
View full question & answer→Question 2443 Marks
Write a value of $\int\tan^3\text{x}\sec^2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\tan^3\text{x}\sec^2\text{x}\text{ dx}$ Let $\tan\text{x}=\text{t}$ $\sec^2\text{x dx}=\text{dt}$ $\therefore\ \text{I}=\int\text{t}^3\text{ dt}$ $=\frac{\text{t}^4}{4}+\text{C}$$=\frac{\tan^{4}\text{x}}{4}+\text{C}$ $(\because\text{t}=\tan\text{x})$
View full question & answer→Question 2453 Marks
Evaluate the following integrals:
$\frac{\text{x}+1}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\frac{\text{x}+1}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{(\text{x}+1)(1+\text{xe}^\text{x}-\text{xe}^\text{x})}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{\int(\text{x}+1)(1+\text{xe}^\text{x})}{\text(1+\text{xe}^\text{x})}\ \text{dx}-\int\frac{(\text{x}+1)(\text{xe}^\text{x})}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{(\text{x}+1)}{\text{x}}\ \text{dx}-\int\frac{\text{e}^\text{x}(\text{x}+1)}{1+\text{xe}^\text{x}}\ \text{dx}$
$=\int\frac{(\text{x}+1)\text{e}^\text{x}}{\text{xe}^\text{x}}\ \text{dx}-\int\frac{\text{e}^\text{x}(\text{x}+1)}{1+\text{xe}^\text{x}}\ \text{dx}$
$=\log|\text{xe}^\text{x}|-\log|1+\text{xe}^\text{x}|+\text{C}$
$\therefore\text{I}=\log\Big|\frac{\text{xe}^\text{x}}{1+\text{xe}^\text{x}}\Big|+\text{C}$
View full question & answer→Question 2463 Marks
Evaluate the following integrals:
$\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$
Answer$\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$ Let $\text{x}+\tan^{-1}\text{x}=\text{t}$ $\Big(1+\frac{1}{1+\text{x}^2}\Big)=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\Big(\frac{\text{x}^2-1+1}{\text{x}^2+1}\Big)\text{dx}=\text{dt}$ $\Rightarrow\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}=\text{dt}$Now, $\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$
$=\int5^\text{t}\text{dt}$
$=\frac{5^\text{t}}{\log5}+\text{C}$
$=\frac{5^{\text{x}+\tan^{-1}\text{x}}}{\log5}+\text{C}$
View full question & answer→Question 2473 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{(\text{x}^2+4)\sqrt{\text{x}^2+9}}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}}{(\text{x}^2+4)\sqrt{\text{x}^2+9}}\text{ dx}$
Let $\text{x}^2+9=\text{u}^2$
$2\text{xdx}=2\text{udu}$
$\therefore\ \text{I}=\int\frac{\text{u}}{(\text{u}^2-5)\text{u}}\text{ du}$
$=\int\frac{\text{du}}{\text{u}^2-5}$
$=\frac{1}{2\sqrt{5}}\log\Big(\frac{\text{u}-\sqrt{5}}{\text{u}+\sqrt{5}}\Big)+\text{C}$
$=\frac{1}{2\sqrt{5}}\log\bigg(\frac{\sqrt{\text{x}^2+9}-\sqrt{5}}{\sqrt{\text{x}^2+9}+\sqrt{5}}\bigg)+\text{C}$
View full question & answer→Question 2483 Marks
Evalute the following integrals:
$\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\text{dx}$
AnswerLet $\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\text{dx}\ .....\text{(i)}$
Let $\tan\text{x}+2=\text{t}$ then,
$\text{d}(\tan\text{x}+2)=\text{dt}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{1}{\sec^2\text{x}}\text{dt}$
Putting $\tan\text{x}+2=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$ In equation (i), we get,
$\text{I}=\int\frac{\sec^2\text{x}}{\text{t}}\times\frac{1}{\sec^2\text{x}}\text{dt}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\tan\text{x}+2|+\text{C}$
$\Rightarrow\text{I}=\log|\tan\text{x}+2|+\text{C}$
View full question & answer→Question 2493 Marks
Evaluate the following integrals:
$\int\frac{\cos\text{x}}{\sin^2\text{x}+4\sin\text{x}+5}\text{dx}$
Answer$\int\frac{\cos\text{x dx}}{\sin^2\text{x}+4\sin\text{x}+5}$
Let $\sin\text{x = t}$
$\Rightarrow\cos\text{x dx = dt}$
Now, $\int\frac{\cos\text{x dx}}{\sin^2\text{x}+4\sin\text{x}+5}$
$=\int\frac{\text{dt}}{\text{t}^2+4\text{t}+5}$
$=\int\frac{\text{dt}}{\text{t}^2+2\times\text{t}\times2+4+1}$
$=\int\frac{\text{dt}}{(\text{t}+2)^2+1^2}$
$=\frac{1}{1}\tan^{-1}\Big(\frac{\text{t}+2}{1}\Big)+\text{C}$
$=\tan^{-1}(\sin\text{x}+2)+\text{C}$
View full question & answer→Question 2503 Marks
Evaluate the following integrals:
$\int\frac{\text{dx}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
Answer$\int\frac{\text{dx}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
$=\int\frac{\text{dx}}{\text{e}^{\text{x}}+\frac{1}{\text{e}^{\text{x}}}}$
$=\int\frac{\text{e}^{\text{x}}\text{ dx}}{\text{e}^{2\text{x}}+1}$
Let $\text{e}^{\text{x}}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\text{ dx = dt}$
Now, $\int\frac{\text{e}^{\text{x}}\text{ dx}}{\text{e}^{2\text{x}}+1}$
$=\int\frac{\text{dt}}{1+\text{t}^2}$
$=\tan^{-1}(\text{t})+\text{C}$
$=\tan^{-1}(\text{e}^{\text{x}})+\text{C}$
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