Question 513 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 523 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 533 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 543 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 553 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 563 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 573 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 583 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 593 Marks
Evaluate the following integrals:$\int\frac{\sin8\text{x}}{\sqrt{9+\sin^44\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin8\text{x}}{\sqrt{9+(\sin4\text{x})^4}}\text{ dx}$
Let $\sin^24\text{x}=\text{t}$
$\Rightarrow2\sin4\text{x}.\cos4\text{x}(4)\text{dx}=\text{dt}$
$\Rightarrow4\sin8\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\sin8\text{x}\text{ dx}=\frac{\text{dt}}{4}$
$\text{I}=\frac{1}{4}\int\frac{\text{dt}}{\sqrt{(3)^2+\text{t}^2}}$
$\text{I}=\frac{1}{4}\log\Big|\text{t}+\sqrt{(3)^2+\text{t}^2}\Big|+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\log\Big|\text{x}+\sqrt{\text{a}^2+\text{x}^2}\Big|+\text{C}\Big]$
$\text{I}=\frac{1}{4}\log\Big|\sin^24\text{x}+\sqrt{9+\sin^44\text{x}}\Big|+\text{C}$
View full question & answer→Question 603 Marks
Evaluate the following intregals:
$\int\frac{1}{\sin^2\text{x}-\sin2\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sin^2\text{x}-\sin(2\text{x})}\ \text{dx}$
$=\int\frac{1}{\sin^2\text{x}+2\sin\text{x}\cos\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{\tan\text{x}+2\tan\text{x}}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+2\text{t}}$
$=\int\frac{\text{dt}}{\text{t}^2+2\text{t}+1-1}$
$=\ln\frac{\text{dt}}{(\text{t}+1)^2-(-1)^2}$
$=\frac{1}{2}\ln\Big|\frac{\text{t}}{\text{t}+2}\Big|+\text{C}$
$=\frac{1}{2}\ln\Big|\frac{\tan\text{x}}{\tan\text{x}+2}\Big|+\text{C}$
View full question & answer→Question 613 Marks
Evaluate the following integrals:$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
Answer$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
Let $\text{t}=\cos^2\text{x}\rightarrow-\text{dt}=2\cos\text{x}\sin\text{x}\text{ dx}$
$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
$=\int\frac{-1}{\sqrt{\text{t}^2-(1-\text{t})+2}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+1}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+\frac{1}{4}+\frac{3}{4}}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\big(\text{t}+\frac{1}{2}\big)^2+\frac{3}{4}}}\text{ dt}$
$=-\log\Big|\Big(\text{t}+\frac{1}{2}\Big)+\sqrt{\text{t}^2+\text{t}+1}\Big|$
$=-\log\Big|\Big(\cos^2\text{x}+\frac{1}{2}\Big)+\sqrt{\cos^4\text{x}+\cos^2\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 623 Marks
Evaluate the following integrals:$\int\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{dx}$
Answer$\int\Big(\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\Big[\frac{\text{x}}{1+\cos\text{x}}+\frac{\sin\text{x}}{1+\cos\text{x}}\Big]\text{dx}$
$=\int\bigg[\frac{\text{x}}{2\cos^{2}\frac{\text{x}}{2}}+\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\bigg]\text{dx}$
$=\frac{1}{2}\int\text{x}\cdot\sec^2\frac{\text{x}}{2}\text{dx}+\int\tan\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{2}\bigg[\text{x}\cdot\frac{\tan\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}-\int1\times2\tan\big(\frac{\text{x}}{2}\big)\text{dx}\bigg]+\frac{\log\big|\sec\frac{\text{x}}{2}\big|}{\frac{1}{2}}+\text{C}$
$=\text{x}\tan\big(\frac{\text{x}}{2}\big)-\frac{\log\big|\sec\frac{\text{x}}{2}\big|}{\frac{1}{2}}+\log\frac{\big|\sec\frac{\text{x}}{2}\big|}{\frac{1}{2}}+\text{C}$
$=\text{x}\tan\big(\frac{\text{x}}{2}\big)+\text{C}$
View full question & answer→Question 633 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 643 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 653 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+\text{x}+1}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{(\text{x}-1)\sqrt{\text{x}^2+\text{x}+1}}\text{ dx}$
Let $\text{x}+1=\frac{1}{\text{t}}$
$\text{dt}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=-\int\frac{\frac{1}{\text{t}^2}\text{ dt}}{\frac{1}{\text{t}}\sqrt{\Big(\frac{1}{\text{t}^2}+\frac{1}{\text{t}}-1\Big)}}$
$=-\int\frac{\text{dt}}{\sqrt{1+\text{t}-\text{t}^2}}$
$=-\int\frac{\text{dt}}{\sqrt{\frac{5}{4}-\big(\frac{1}{4}-\text{t}+\text{t}^2\big)}}$
$=-\int\frac{\text{dt}}{\sqrt{\frac{5}{4}-\big(\text{t}-\frac{1}{2}\big)^2}}$
$=-\sin^{-1}\Bigg(\frac{\text{t}-\frac{1}{2}}{\frac{\sqrt{5}}{2}}\Bigg)+\text{C}$
$\therefore\ \text{I}=-\sin^{-1}\Big(\frac{2\text{t}-1}{\sqrt{5}}\Big)+\text{C}$ $\Big[\text{When}\text{t}=\frac{1}{\text{x}+1}\Big]$
View full question & answer→Question 663 Marks
Integrate the following integrals:
$\int\cos3\text{x}\cos4\text{x dx}$
AnswerLet I $=\int\cos3\text{x}\cos4\text{x dx}.$ Then,
$\text{I}=\frac{1}{2}\int(2\cos3\text{x}\cos4\text{x})\text{dx}$
$=\frac{1}{2}\int(\cos7\text{x}+\cos(-\text{x}))\text{dx}$
$=\frac{1}{2}\int\cos7\text{x}+\frac{1}{2}\int\cos\text{dx}$ $[\because\cos(-0)=\cos0]$
$=\frac{\sin7\text{x}}{2\times7}+\frac{\sin\text{x}}{2}+\text{C}$
$=\frac{1}{14}\times\sin7\text{x}+\frac{1}{2}\sin\text{x}+\text{C}$
$\therefore\text{I}=\frac{1}{14}\times\sin7\text{x}+\frac{1}{2}\times\sin\text{x}+\text{C}$
View full question & answer→Question 673 Marks
Evaluate the following integrals:
$\int2\text{x}\sec^3\big(\text{x}^2+3\big)\tan\big(\text{x}^2+3\big)\text{dx}$
Answer $\int2\text{x}\sec^3\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)\text{dx}$ $=\int\sec^2\big(\text{x}^2+3\big).\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)2\text{x}\text{ dx}$ Let $\sec\big(\text{x}^2+3\big)=\text{t}$ $\Rightarrow\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big).2\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big).2\text{x}\text{ dx}=\text{dt}$Now, $\int\sec^2\big(\text{x}^2+3\big).\sec\big(\text{x}^2+3\big).\tan\big(\text{x}^2+3\big)2\text{x}\text{ dx}$
$=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^2}{3}+\text{C}$
$=\frac{\sec^2(\text{x}^2+3)}{3}+\text{C}$
View full question & answer→Question 683 Marks
Evaluate the following integrals:
$\int5^{5^{5^{\text{x}}}}5^{5^{\text{x}}}5^\text{x}\text{ dx}$
Answer$\int5^{5^{5^{\text{x}}}}5^{5^{\text{x}}}5^\text{x}\text{ dx}$
Let $5^\text{x}=\text{t}$
$\Rightarrow5^\text{x}\log5=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow5^\text{x}\text{dx}=\frac{\text{dt}}{\log5}$
Now, $\int5^{5^{5^{\text{x}}}}5^{5^{\text{x}}}5^\text{x}\text{ dx}$
$=5\int5^{5^{\text{t}}}.5^\text{t}.\frac{\text{dt}}{\log5}$
Again let $5^\text{t}=\text{p}$
$\Rightarrow5^\text{t}\log5=\frac{\text{dp}}{\text{dt}}$
$\Rightarrow5^\text{t}\text{dt}=\frac{\text{dp}}{\log5}$
Again $\int5^{5^\text{t}}.5^\text{x}.\frac{\text{dt}}{\log5}$
$=\int5^\text{p}.\frac{\text{dp}}{(\log5)^2}$
$=\frac{5^\text{p}}{(\log5)^3}+\text{C}$
$=\frac{5^{5^{5^{\text{x}}}}}{(\log5)^3}+\text{C}$
View full question & answer→Question 693 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}^3+8)(\text{x}-1)}{\text{x}^2-2\text{x}+4}\text{dx}$
Answer$\int\frac{(\text{x}^3+8)(\text{x}-1)}{(\text{x}^2-2\text{x}+4)}\text{dx}$
$=\int\frac{(\text{x}^3+2^3)(\text{x}-1)}{(\text{x}^2-2\text{x}+4)}\text{dx}$
$=\int\frac{(\text{x}+2)(\text{x}^2-2\text{x}+4)(\text{x}-1)}{(\text{x}^2-2\text{x}+4)}\text{dx}$ $\big[\therefore\ \text{a}^3+\text{b}^3=(\text{a + b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=\int(\text{x}+2)(\text{x}-1)\text{dx}$
$=\int(\text{x}^2-\text{x}+2\text{x}-2)\text{dx}$
$=(\text{x}^2+\text{x}-2)\text{dx}$
$=\int\text{x}^2\text{dx}+\int\text{x dx}-2\int1\text{dx}$
$=\frac{\text{x}^3}{3}+\frac{\text{x}^2}{2}-2\text{x}+\text{C}$
View full question & answer→Question 703 Marks
Evaluate the following integrals:
$\int\cos^7\text{x}\text{ dx}$
Answer$\int\cos^7\text{x}\text{ dx}$
$=\int\cos^6\text{x}\cdot\cos\text{x}\text{ dx}$
$=\int(\cos^2\text{x})^3\cos\text{x}\text{ dx}$
$=\int(1-\sin^2\text{x})^3\cdot\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int(1-\sin^2\text{x})^3\cdot\cos\text{x}\text{ dx}$
$=\int(1-\text{t}^2)^3\text{dt}$
$=\int\big(1-\text{t}^6-3\text{t}^2+3\text{t}^4)\text{dt}$
$=\Big[\text{t}-\frac{\text{t}^7}{\text{7}}-\frac{3\text{t}^3}{3}+\frac{3\text{t}^5}{5}\Big]+\text{C}$
$=\sin\text{x}-\frac{1}{7}\sin^7\text{x}-\sin^3\text{x}+\frac{3}{5}\sin^5\text{x}+\text{C}$
View full question & answer→Question 713 Marks
Evaluate the following integrals:
$\int\sqrt{16\text{x}^2+25}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{16\text{x}^2+25}\text{dx}$
$=\int\sqrt{(4\text{x})^2+5^2}\text{dx}$
$=4\int\sqrt{\text{x}^2+\Big(\frac{5}{4}\Big)^2}\text{dx}$
$=4\begin{Bmatrix}\frac{\text{x}}{2}\sqrt{\text{x}^2+\Big(\frac{5}{4}\Big)^2}+\frac{\big(\frac{5}{4}\big)^2}{2}\log\Bigg|\text{x}+\sqrt{\text{x}^2+\Big(\frac{5}{4}\Big)^2}\Bigg|+\text{C}\end{Bmatrix}$
$\therefore\ \text{I}=2\text{x}\sqrt{\text{x}^2+\frac{25}{16}}+\frac{25}{8}\log\bigg|\text{x}+\sqrt{\text{x}^2+\frac{25}{16}}\bigg|+\text{C}$
View full question & answer→Question 723 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{1+\cos\text{x}}}\text{dx}$
AnswerWe have,
$\int\frac{1}{\sqrt{1+\cos\text{x}}}\text{dx}$
$\int\frac{1}{\sqrt{2\cos^2\frac{\text{x}}{2}}}\text{dx}$
$=\int\frac{1}{\sqrt{2}\cos\frac{\text{x}}{2}}\text{dx}$
$=\frac{1}{\sqrt{2}}\int\sec\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{\sqrt{2}}\int\text{cosec}\Big(\frac{\pi}{2}+\frac{\text{x}}{2}\Big)\text{dx}$
$=\frac{2}{\sqrt{2}}\log\Big|\tan\Big(\frac{\pi}{4}+\frac{\text{x}}{4}\Big)\Big|+\text{C}$
$\because\int\frac{1}{\sqrt{1+\cos\text{x}}}\text{dx}=\sqrt{2}\log\Big|\tan\Big(\frac{\pi}{4}+\frac{\text{x}}{4}\Big)\Big|+\text{C}$
View full question & answer→Question 733 Marks
Evaluate the following integrals:
$\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Answer$\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$ Let $\sqrt{\text{x}}=\text{t}$ $\Rightarrow\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\frac{\text{dx}}{\sqrt{\text{x}}}=2\text{dt}$ Now, $\int\frac{\sec^2\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$$=2\int\sec^2\text{t dt}$
$=2\tan(\text{t})+\text{C}$
$=2\tan\big(\sqrt{\text{x}}\big)+\text{C}$
View full question & answer→Question 743 Marks
Evaluate the following integrals:
$\int (3\text{x}\sqrt{\text{x}}+4\sqrt{\text{x}}+5)\text{dx}$
Answer$\int(3\text{x}\sqrt{5}+4\sqrt{\text{x}}+5)\text{dx}$
$=\int3\text{x}\sqrt{5}\text{dx}+\int4\sqrt{\text{x}}\text{dx}+\int5\text{dx}$
$=\int3\text{x}^{\frac{3}{2}}\text{dx}+4\int\text{x}^{\frac{1}{2}}\text{dx}+5\int\text{dx}$
$=\frac{\text{x}\frac{3}{2}+1}{\frac{3}{2}+1}+\frac{4\text{x}^{\frac{1}{2}}}{\frac{1}{2}+1}+5\text{x}+\text{C}$
$=\frac{6}{5}\text{x}^{\frac{5}{2}}+\frac{8}{3}\text{x}^{\frac{3}{2}}+5\text{x}+\text{C}$
View full question & answer→Question 753 Marks
Evaluate the following integrals:
$\int\big\{\sqrt{\text{x}}\big(\text{ax}^2+\text{bx}+\text{c}\big)\big\}\text{dx}$
Answer$\int\sqrt{\text{x}}\Big(\text{ax}^2+\text{bx}+\text{c}\Big)\text{dx}$
$=\int\text{x}^{\frac{1}{2}}\Big(\text{ax}^2+\text{bx}+\text{c}\Big)\text{dx}$
$=\int\Big(\text{ax}^{2+\frac{1}{2}}+\text{bx}^{\frac{1}{2}+1}+\text{cx}^{\frac{1}{2}}\Big)\text{dx}$
$=\text{a}\int\text{x}^{\frac{5}{2}}\text{dx}+\text{b}\int\text{x}^{\frac{3}{2}}\text{dx}+\text{c}\int\text{x}^{\frac{1}{2}}\text{dx}$
$=\text{a}\begin{bmatrix}\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}\end{bmatrix}+\text{b}\begin{bmatrix}\frac{\text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\end{bmatrix}+\text{c}\begin{bmatrix}\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\end{bmatrix}+\text{C}$
$=\frac{2\text{a}}{7}\text{x}^{\frac{7}{2}}+\frac{2\text{b}}{5}\text{x}^{\frac{3}{2}}+\frac{2\text{c}}{3}\text{x}^{\frac{3}{2}}+\text{C}$
View full question & answer→Question 763 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}^2-2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\text{x}^2-2\text{x}}\text{dx}$
$\Rightarrow\text{I}=\int\sqrt{\text{x}^2-2\text{x}+1-1}\text{dx}$
$\Rightarrow\text{I}=\int\sqrt{(\text{x}-1)^2-1^2}\text{dx}$
$\because\ \int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2-\text{a}^2}-\frac{\text{a}^2}{2}\ln\big(|\text{x}+\sqrt{\text{x}^2-\text{a}^2}|\big)+\text{C}$
$\therefore\ \text{I}=\frac{(\text{x}-1)}{2}\sqrt{(\text{x}-1)^2-1}-\frac{1}{2}\ln\big|(\text{x}-1)+\sqrt{\text{x}^2-2\text{x}}\big|+\text{C}$
View full question & answer→Question 773 Marks
Evaluate the following integrals:
$\int\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}$
AnswerLet first function be $\sin^{-1}\text{x}$ and second dunction be $\frac{\text{x}}{\sqrt{1-\text{x}^2}}.$
First we find the intergral of the second function, i.e, $\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}.$
Put $\text{t}=1-\text{x}^{2}.$ Then $\text{dt}=-2\text{x dx}$
Therefore, $\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}=-\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}=-\sqrt{\text{t}}=-\sqrt{1-\text{x}^2}$
Hence, $\int\frac{\text{x}\sin^{-1}}{\sqrt{1-\text{x}^2}}\text{dx}=(\sin^{-1}\text{x})\Big(-\sqrt{1-\text{x}^2}\Big)-\int\frac{1}{\sqrt{1-\text{x}^2}}\Big(-\sqrt{1-\text{x}^2}\Big)\text{dx}$
$=-\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+\text{x + C}=\text{x}-\sqrt{1-\text{x}^2}\sin^{-1}\text{x+C}$
View full question & answer→Question 783 Marks
Evaluate the following integrals:
$\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
Answer$\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
$\text{Let, }\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\text{Now,}\int\tan^\frac{3}{2}\text{x}\sec^2\text{x dx}$
$=\int\text{t}^\frac{3}{2}\text{dt}$
$=\Bigg[\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]+\text{C}$
$=\frac{2}{5}\text{t}^\frac{5}{2}+\text{C}$
$=\frac{2}{5}\tan^\frac{5}{2}\text{x}+\text{C}$
View full question & answer→Question 793 Marks
Evaluate the following integrals:
$\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{dx}$
Answer$\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}=\frac{\cos\text{x}-\sin\text{x}}{(\sin^2\text{x}+\cos^2\text{x})+2\sin\text{x}\cos\text{x}}$
$[\sin^2\text{x}+\cos^2\text{x}=1;\ \sin2\text{x}=2\sin\text{x}\cos\text{x}]$
$=\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}$
$\sin\text{x}+\cos\text{x}=\text{t}$
$\therefore(\sin\text{x}+\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{dx}=\int\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}\text{dx}$
$=\int\frac{\text{dt}}{\text{t}^2} $
$=\int\text{t}^{-2}\text{dt}$
$=-\text{t}^{-1}+\text{C}$
$=-\frac{1}{\text{t}}+\text{C}$
$=\frac{-1}{\sin\text{x}+\cos\text{x}}+\text{C}$
View full question & answer→Question 803 Marks
$\int\frac{1}{1-\sin\frac{\text{x}}{2}}\text{dx}$
Answer$\int\frac{\text{dx}}{1-\sin\big(\frac{\text{x}}{2}\big)}$
$=\int\frac{\big(1+\sin\frac{\text{x}}{2}\big)}{\big(1-\sin\frac{\text{x}}{2}\big)\big(1+\sin\frac{\text{x}}{2}\big)}\text{dx}$
$=\int\bigg(\frac{1+\sin\frac{\text{x}}{2}}{1-\sin^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\bigg(\frac{1+\sin\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\big(\sec^2\frac{\text{x}}{2}+\sec\frac{\text{x}}{2}\tan\frac{\text{x}}{2}\big)\text{dx}$
$=\frac{\tan\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}+\frac{\sec\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}+\text{c}$
$=2\big(\tan\frac{\text{x}}{2}+\sec\frac{\text{x}}{2}\big)+\text{c}$
View full question & answer→Question 813 Marks
Evaluate the following integrals:
$\int\frac{(1+\sqrt{\text{x}})^2}{\sqrt{\text{x}}}\text{dx}$
Answer$\int\frac{(1+\sqrt{\text{x}})^2}{\sqrt{\text{x}}}\text{dx}$
$=\int\frac{1+\text{x}+2\sqrt{\text{x}}}{\text{x}^{\frac{1}{2}}}\text{dx}$
$\int\text{x}^{\frac{-1}{2}}+\int\text{x}^{\frac{1}{2}}\text{dx}+2\int\text{dx}$
$=2\sqrt{\text{x}}+\frac{2}{3}\text{x}^{\frac{3}{2}}+2\text{x}+\text{C}$
$\therefore\ \int\frac{(1+\sqrt{\text{x}})^2}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\frac{2}{3}\text{x}^{\frac{3}{2}}+2\text{x}+\text{C}$
View full question & answer→Question 823 Marks
Evaluate the following intregals: $\int\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}\ \text{dx}$
We express
$\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{C}}{\text{x}+3}$
$\Rightarrow3\text{x}-2=\text{A}(\text{x}+1)(\text{x}+3)+\text{B}(\text{x}+3)+\text{C}(\text{x}+1)^2$
Equating the coefficient of $x^2, x$ and constants, we get
$0 = A + C$ and $3 = 4A + B + 2C$ and $-2 = 3A + 3B + C$
$\text{or }\text{A}=\frac{11}{4}\text{ and }\text{B}=-\frac{5}{2}\text{ and }\text{C}=-\frac{11}{4}$
$\therefore\text{I}=\int\bigg(\frac{\frac{11}{4}}{\text{x}+1}+\frac{-\frac{5}{2}}{(\text{x}+1)^2}+\frac{-\frac{11}{4}}{\text{x}+3}\bigg)\ \text{dx}$
$=\frac{11}{4}\int\frac{1}{\text{x}+1}\text{ dx }-\frac{5}{2}\int\frac{1}{(\text{x}+1)^2}\text{ dx }-\frac{11}{4}\int\frac{1}{\text{x}+3}\text{ dx}$
$=\frac{11}{4}\log|\text{x}+1|+\frac{5}{2(\text{x}+1)}-\frac{11}{4}\log|\text{x}+3|+\text{C}$
Hence,
$\int\frac{3\text{x}-2}{(\text{x}+1)^2(\text{x}+3)}\ \text{dx}$
$=\frac{11}{4}\log|\text{x}+1|+\frac{5}{2(\text{x}+1)}-\frac{11}{4}\log|\text{x}+3|+\text{C}$
View full question & answer→Question 833 Marks
Evaluate the following integrals:
$\int \frac{\text{1}}{\sqrt{\text{x}} + \text{x}} \text{ dx}$
Answer$\int \frac{\text{dx}}{\sqrt{\text{x}} + \text{x}}\text{dx}$
$=\int \frac{\text{dx}}{\sqrt{\text{x}}\big(1 + \sqrt{\text{x}}\big)}$
Let $1 + \sqrt{\text{x}} = \text{t}$
$\Rightarrow \frac{1}{2\sqrt{\text{x}}} = \frac{\text{dt}}{\text{dx}}$
$\Rightarrow \frac{\text{dx}}{\sqrt{\text{x}}} = 2\text{dt}$
Now, $\int \frac{\text{dx}}{\sqrt{\text{x}}\big(1 + \sqrt{\text{x}}\big)}$
$= \int \frac{2\text{dt}}{\text{t}}$
$= 2\int\frac{\text{dt}}{\text{t}}$
$= 2\log|\text{t}| + \text{C}$
$= 2\log \big(1 + \sqrt{\text{x}}\big) + \text{C}$
View full question & answer→Question 843 Marks
Evalute the following integrals: $\int\frac{\text{a}}{\text{b}+\text{ce}^\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{a}}{\text{b}+\text{ce}^\text{x}}\text{dx}$
Dividing numerator and denomimator by $e^x$
$\Rightarrow\text{I}=\int\frac{\text{ae}^{-\text{x}}}{\text{be}^{-\text{x}}+\text{c}}\text{dx}$
Putting $e^{-x} = t$
$\Rightarrow-\text{e}^{-\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{-\text{x}}\text{dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{a}}{\text{bt}+\text{c}}\text{dt}$
$=\frac{-\text{a}}{\text{b}}\text{ ln}|\text{bt}+\text{c}|+\text{C}$
$\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\text{ ln}|\text{ax}+\text{b}|+\text{C}\Big]$
$=\frac{-\text{a}}{\text{b}}\text{ ln}|\text{be}^{-\text{x}}+\text{c}|+\text{C}\ \big[\because\text{t}=\text{e}^{-\text{x}}+\text{C}\big]$
View full question & answer→Question 853 Marks
Evalute the following integrals:
$\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{dx}$
AnswerNote: Here we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{dx}$
Putting $\log\sin\text{x}=\text{t}$
$\Rightarrow\cot\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cot\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\log\sin\text{x}|+\text{C}\ \big[\because\text{t}=\log\sin\text{x}\big]$
View full question & answer→Question 863 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}^2+\text{x}+1}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\text{x}^2+\text{x}+1}\text{dx}$
$=\int\sqrt{\text{x}^2+\text{x}+\frac{1}{4}+\frac{3}{4}}\text{dx}$
$=\int\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\text{dx}$
$=\frac{\big(\text{x}+\frac{1}{2}\big)}{2}\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}+\frac{\big(\frac{\sqrt{3}}{2}\big)^2}{2}\times\\\log\Bigg|\Big(\text{x}+\frac{1}{2}\Big)+\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\Bigg|+\text{C}$
$=\Big(\frac{2\text{x}+1}{4}\Big)\sqrt{\text{x}^2+\text{x}+1}+\frac{3}{8}\log\bigg|\Big(\frac{2\text{x}+1}{2}\Big)+\frac{1}{2}\sqrt{\text{x}^2+\text{x}+1}\bigg|+\text{C}$
$\therefore\ \text{I}=\Big(\frac{2\text{x}+1}{4}\Big)\sqrt{\text{x}^2+\text{x}+1}+\frac{3}{8}\log\bigg|2\text{x}+1+\sqrt{\text{x}^2+\text{x}+1}\bigg|+\text{C}$
View full question & answer→Question 873 Marks
Evalute the following integrals:
$\int\frac{2\cos2\text{x}+\sec^2\text{x}}{\sin2\text{x}+\tan\text{x}-5}\text{dx}$
AnswerLet $\text{I}=\int\frac{2\cos2\text{x}+\sec^2\text{x}}{\sin2\text{x}+\tan\text{x}-5}\text{dx}$
Putting $\sin2\text{x}+\tan\text{x}-5=\text{t}$
$\Rightarrow2\cos2\text{x}+\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(2\cos2\text{x}+\sec^2\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}|\sin2\text{x}+\tan\text{x}-5|+\text{C} $ $\big[\because\text{t}=\sin2\text{x}+\tan\text{x}-5\big]$
View full question & answer→Question 883 Marks
Evaluate the following integrals:
$\int\sqrt{9-{\text{x}^2}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{3^2-{\text{x}^2}}$
We know that,
$\int\sqrt{\text{a}^2-\text{x}^2}=\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
$\therefore\ \text{I}=\frac{\text{x}}{2}\sqrt{9-\text{x}^2}+\frac{9}{2}\sin^{-1}\frac{\text{x}}{3}+\text{C}$
View full question & answer→Question 893 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^{\frac{3}{2}}}\text{dx}$
Answer$\int\frac{1}{\text{x}^\frac{3}{2}}\text{dx}=\int\text{x}\frac{-3}{2}\text{dx}$
$=\int\text{x}^\frac{-3}{2}\text{dx}$
$=\frac{\text{x}^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+\text{c}$
$=\frac{\text{x}^\frac{-1}{2}}{\frac{-1}{2}}+\text{c}$
$=-2\text{x}\frac{1}{\sqrt{\text{x}}}+\text{c}$
$=\frac{-2}{\sqrt{\text{x}}}+\text{c}$
View full question & answer→Question 903 Marks
If $\int\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}=\text{f(x)}\text{e}^{\text{x}}+\text{C},$ then write the value of f(x).
Answer$\int\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}=\int\Big(\frac{\text{x}}{\text{x}^2}-\frac{1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{dx}$
$=\int\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$
Consider, $\text{f(x)}=\frac{1}{\text{x}},$ then $\text{f}'(\text{x})=-\frac{1}{\text{x}^2}$
Thus, the given integrand is of the from $\text{e}^{\text{x}}\big|\text{f}(\text{x})+\text{f}'(\text{x})\big|$
Therefore, $\int\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}=\frac{1}{\text{x}}\text{e}^{\text{x}}+\text{C}$
Hence, $\text{f(x)}=\frac{1}{\text{x}}$
View full question & answer→Question 913 Marks
$\int\frac{2\text{x}-1}{(\text{x}-1)^2}\text{dx}$
Answer$\int\bigg[\frac{2\text{x}-1}{(\text{x}-1)^2}\bigg]\text{dx}$
$=\int\bigg[\frac{2\text{x}-2+2-1}{(\text{x}-1)^2}\bigg]\text{dx}$
$=\int\bigg(\frac{2(\text{x}-1)}{(\text{x}-1)^2}+\frac{1}{(\text{x}-1)^2}\bigg)\text{dx}$
$=2\int\frac{\text{dx}}{\text{x}-1}+\int(\text{x}-1)^{-2}\text{dx}$
$=2\text{ ln}|\text{x}-1|+\frac{(\text{x}-1)^{-2+1}}{-2+1}+\text{C}$
$=2\text{ ln}|\text{x}-1|-\frac{1}{\text{x}-1}+\text{C}$
View full question & answer→Question 923 Marks
Evaluate the following integrals:
$\int\sqrt{2\text{ax}-\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{2\text{ax}-\text{x}^2}\text{dx}$
$=\int\sqrt{\text{a}^2+2\text{ax}-\text{x}^2-\text{a}^2}\text{dx}$
$=\int\sqrt{\text{a}^2-(\text{x}^2-2\text{ax}+\text{a}^2)}\text{dx}$
$=\int\sqrt{\text{a}^2-(\text{x}-\text{a})^2}\text{dx}$
$=\Big(\frac{\text{x}-\text{a}}{2}\Big)\sqrt{2\text{ax}-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\Big(\frac{\text{x}-\text{a}}{\text{a}}\Big)+\text{C}$
View full question & answer→Question 933 Marks
Evaluate the following integrals:
$\int\tan\text{x }\sec^4\text{x}\text{dx}$
AnswerLet $\text{I}=\int\tan\text{x }\sec^4\text{x}\text{dx}$ Then
$\text{I}=\int\tan\text{x }\sec^2\text{x}\sec^2\text{x}\text{dx}$
$=\int\tan\text{x}(1+\tan^2\text{x})\sec^2\text{x}\text{dx}$
$\text{I}=\int\big(\tan\text{x}+\tan^3\text{x}\big)\sec^2\text{x}\text{dx}$
Substituting $\tan\text{x}=\text{t}$ and $\sec^2\text{xdx}=\text{dt},$ we get
$\text{I}=\int(\text{t}+\text{t}^3)\text{dt}$
$=\frac{\text{t}^2}{2}+\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\tan^2\text{x}}{2}+\frac{\tan^4}{4}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\times\tan^2\text{x}+\frac{1}{4}\times\tan^4\text{x}+\text{C}$
View full question & answer→Question 943 Marks
Evaluate the following integrals:
$\int\text{x}^3\sin\text{x}^4\text{dx}$
Answer$\int\text{x}^3.\sin\text{x}^4\text{dx}$
Let $\text{x}^4=\text{t}$
$\Rightarrow4\text{x}^3=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{x}^3\text{dx}=\frac{\text{dt}}{4}$
Now, $\int\text{x}^3.\sin\text{x}^4\text{dx}$
$=\frac{1}{4}\int\sin\text{t}\text{ dt}$
$=\frac{1}{4}[-\cos(\text{t})]+\text{C}$
$=\frac{1}{4}\big[-\cos\text{x}^4\big]+\text{C}$
View full question & answer→Question 953 Marks
$\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\times\text{dx}.$ Then,
$\text{I}=\int\frac{2\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\times\text{dx}$
$=\int\frac{\sin^2\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}\times\text{dx}$
$=\int\tan^2\frac{\text{x}}{2}\text{dx}$
$=\int\Big(\sec^2\frac{\text{x}}{2}-1\Big)\text{dx}$
$=\frac{\tan\frac{\text{x}}{2}}{\frac{1}2{}}-\text{x+c}$
$=2\tan\frac{\text{x}}{2}-\text{x+c}$
View full question & answer→Question 963 Marks
Evaluate the following intregals:
$\int\frac{\sin2\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{2\tan\text{x}\sec^2\text{x}}{\tan^4\text{x}+1}\ \text{dx}$
Let $\tan^2\text{x}=\text{t}$
$2\tan\text{x}\sec^2\text{x}\text{ dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2+1}$
$=\tan^{-1}\text{t}+\text{C}$
$\text{I}=\tan^{-1}(\tan^2\text{x})+\text{C}$
View full question & answer→Question 973 Marks
Evaluate the following integrals:
$\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\text{dx}$
Answer$\int\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\frac{(1-\cos\text{x})^2}{1-\cos^2\text{x}}\text{dx}$
$=\int\frac{1+\cos^2\text{x}-2\cos\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\Big(\frac{1}{\sin^2\text{x}}+\frac{\cos^2\text{x}}{\sin^2\text{x}}-\frac{2\cos\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int(\text{cosec}^2\text{x}+\cot^2\text{x}-2\cot\text{x}\text{ cosec x})\text{dx}$
$=\int(\text{cosec}^2\text{x}+\text{cosec}^2\text{x}-1-2\cot\text{x cosec x})\text{dx}$
$=\int(2\text{cosec}^2\text{x}-1-2\cot\text{x cosec x})\text{dx}$
$=\int2\text{cosec}^2\text{x dx}-\int1\text{dx}-\int2\cot\text{x cosec x }\text{dx}$
$=-2\cot\text{x}-\text{x}+2\text{cosec x}+\text{C}$
$=2(\text{cosec x}-\cot\text{x})-\text{x + C}$
View full question & answer→Question 983 Marks
Evaluate the following integrals:
$\int\sqrt{4\text{x}^2-5}\text{dx}$
Answer$\text{I}=\int\sqrt{4\text{x}^2-5}\text{dx}$
$=\int\sqrt{4\Big(\text{x}^2-\frac{5}{4}\Big)}\text{dx}$
$=2\int\sqrt{\text{x}^2-\Big(\frac{\sqrt5}{2}}\Big)^2\text{dx}$
$=2\Big[\frac{\text{x}}{2}\sqrt{\text{x}^2-\frac{5}{4}}-\frac{5}{8}\int\Big|\text{x}+\sqrt{\text{x}^2-\frac{5}{4}}\Big|\Big]+\text{C}$
$\Big[\because\ \int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2-\text{a}^2}\frac{1}{2}\text{a}^2\int\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|+\text{C}\Big]$
$=\text{x}\sqrt{\text{x}^2-\frac{5}{4}}-\frac{5}{4}\int\bigg|\text{x}+\sqrt{\text{x}^2-\frac{5}{4}}\bigg|+\text{C}$
View full question & answer→Question 993 Marks
Write the anti-derivative of $\Big(3\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
AnswerLet $\text{I}=\int\Big(3\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)\text{dx}$
$\text{I}=3\sqrt{\text{x}}\text{ dx}+\int\frac{\text{dx}}{\sqrt{\text{x}}}$
$=3\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=3\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\text{C}$
$=2\times3\times\frac{\text{x}^\frac{3}{2}}{3}+2\times\frac{\text{x}^{\frac{1}{2}}}{1}+\text{C}$
$=2\Big(\text{x}^{\frac{3}{2}}+\text{x}^{\frac{1}{2}}\Big)+\text{C}$
View full question & answer→Question 1003 Marks
Evaluate the following integrals:
$\int\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\text{dx}$
Answer$\int\bigg(\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\bigg)\text{dx}$
$\text{Let }\sqrt{2\text{x}^2+3\text{x}+1}=\text{t}$
$\Rightarrow(4\text{x}+3)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(4\text{x}+3)\text{dx}=\text{dt}$
$\text{Now,}\int\bigg(\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\bigg)\text{dx}$
$=\int\frac{\text{dt}}{\sqrt{t}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\Bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{2\text{x}^2+3\text{x}+1}+\text{C}$
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