Question 513 Marks
Evaluate the following integrals:$\int\frac{\text{x}}{\sqrt{\text{x}^4+\text{a}^4}}\text{ dx}$
Answer$\int\frac{\text{x}\text{ dx}}{\sqrt{\text{x}^4+\text{a}^4}}$ $\int\frac{\text{x}\text{ dx}}{\sqrt{(\text{x}^2)^2+(\text{a}^2)^2}}$ Let $\text{x}^2=\text{t}$ $\Rightarrow2\text{x}\text{ dx}=\text{dt}$ $\Rightarrow\text{x}\text{ dx}=\frac{\text{dt}}{2}$Now, $\int\frac{\text{x}\text{ dx}}{\sqrt{(\text{x}^2)^2+(\text{a}^2)^2}}$
$=\frac{1}{2}\int\frac{\text{x}\text{ dx}}{\sqrt{{\text{t}^2+(\text{a}^2)^2}}}$
$=\frac{1}{2}\log\Big|\text{t}+\sqrt{\text{t}^2+\text{a}^4}\Big|+\text{C}$
$=\frac{1}{2}\log\Big|\text{x}^2+\sqrt{\text{x}^4+\text{a}^4}\Big|+\text{C}$
View full question & answer→Question 523 Marks
Evaluate the following integrals:$\int\sin^{-1}(3\text{x}-4\text{x}^3)\text{dx}$
AnswerLet $\text{I}=\int\sin^{-1}(3\text{x}-4\text{x}^3)\text{dx}$
Let $\text{x}=\sin\theta$
$\text{dx}=\cos\theta \text{d}\theta$
$=\int\sin^{-1}(3\sin\theta-4\sin^3\theta)\cos\theta\text{d}\theta$
$=\int\sin^{-1}(\sin3\theta)\cos\theta\text{d}\theta$
$=\int3\theta\cos\theta\text{d}\theta$
$=3[\theta\int\cos\theta\text{d}\theta-\int(1\int\cos\theta\text{d}\theta)\text{d}\theta]$
$=3[\theta\sin\theta-\int\sin\theta\text{d}\theta]$
$=3[\theta\sin\theta+\cos\theta]+\text{C}$
$\text{I}=3\Big[\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}\Big]+\text{C}$
View full question & answer→Question 533 Marks
$\text{Show that}\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{g}\text{(x)}\text{dx}=2\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx}\text,$if and are defined as $\text{f(x)}=\text{f(a}-\text{x)}$ and $\text{g(x)}+\text{g(a}-\text{x)}=4$
Answer$\text{Here}\ \text{f}\text{(x)}=\text{f}\text{(a}-\text{x)}\ .....(\text{i})\text{and}\ \ \text{g}\text{(x)}+\text{g}\text{(a}-\text{x)}=4 \ ......\text{(ii)}$$\text{Let}\ \text{I}=\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{g}\text{(x)}\text{dx}\ ....\text{(iii)}$
$\therefore\ \ \text{I}=\int^{\text{a}}\limits_{0}\text{(a}-\text{x)}\text{g}\text{(a}-\text{x)}\text{dx}=\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{g}\text{(a}-\text{x)}\text{dx}\ \ \ [\text{from eq.(i)}]\ ...\text{(iv)}$
Adding eq. (iii) and (iv)
$21=\int^{\text{a}}\limits_{0}(\text{f}\text{(x)}\text{g}\text{(x)}+\text{f}\text{(x)}\text{g}\text{(a}-\text{x)}\text{dx}=\int^{\text{a}}\limits_{0}\text{f}\text{(x)}(\text{g}\text{(x)}+\text{g}\text{(a}-\text{x)})\text{dx}$
$\Rightarrow\ \ \ \ 21\int^{\text{a}}_{0}\text{f}\text{(x)}(4)\text{dx}.......[\text{from eq. (ii)}]$
$\Rightarrow\ \ \ \ 21=4\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx}\ \Rightarrow\ \text{I}=2\int^{\text{a}}\limits_{0}\text{f}\text{(x)}$ Hence proved.
View full question & answer→Question 543 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 553 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 563 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 573 Marks
Verify the following:
$\int\frac{2\text{x}-1}{2\text{x}+3}\text{dx}=\text{x}-\log\big|(2\text{x}+3)^3\big|+\text{c}$
AnswerLet $\int\frac{2\text{x}-1}{2\text{x}+3}\text{dx}$
$=\int\frac{2\text{x}+3-2}{2\text{x}+3}\text{dx}$
$=\int\Big(\frac{2\text{x}+3}{2\text{x}+3}-\frac{2}{2\text{x}+3}\Big)\text{dx}$
$=\int1\text{dx}-4\int\frac{1}{2\text{x}+3}\text{dx}$
$=\text{x}-\frac{4}{2}\log\big|(2\text{x}+3)\big|+\text{C}$ $\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\log|\text{ax}+\text{b}|\Big]$
$=\text{x}-2\log|(2\text{x}+3)|+\text{C}$
$=\text{x}-\log|(2\text{x}+3)^2|+\text{C}$ $\big[\because\ \text{a}\log\text{b}=\log\text{b}^\text{a}\big]$
View full question & answer→Question 583 Marks
Evaluate the following integrals:
$\int\text{x}^2\text{e}^{\text{x}^3}\cos\big(\text{e}^{\text{x}^3}\big)\text{dx}$
AnswerLet $\text{I}=\int\text{x}^2\text{e}^{\text{x}^3}\cos\big(\text{e}^{\text{x}^3}\big)\text{dx}\ ....(1)$ Let $\text{e}^{\text{x}^3}=\text{t}$ then, $\text{d}\big(\text{e}^{\text{x}^3}\big)=\text{dt}$ $\Rightarrow3\text{x}^2\text{e}^{\text{x}^3}\text{dx}=\text{dt}$ $\Rightarrow\text{x}^2\text{e}^{\text{x}^3}\text{dx}=\frac{\text{dt}}{3}$Putting $\text{e}^{\text{x}^3}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{3}$ in equation (1), we get
$\text{I}=\int\cos\text{t}\frac{\text{dt}}{3}$ $=\frac{\sin\text{t}}{3}+\text{C}$ $=\frac{\sin\big(\text{e}^{\text{x}^3}\big)}{3}+\text{C}$ $\text{I}=\frac{1}{3}\sin\big(\text{e}^{\text{x}^3}\big)+\text{C}$
View full question & answer→Question 593 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 603 Marks
Evaluate the following integrals:
$\int^\limits4_1\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}7\text{x}+3,&\text{if }\ 1\leq\text{x}\leq3\\8\text{x},&\text{if }\ 3\leq\text{x}\leq9\end{cases}$
AnswerWe have,
$\int^\limits4_1\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}7\text{x}+3,&\text{if }\ 1\leq\text{x}\leq3\\8\text{x},&\text{if }\ 3\leq\text{x}\leq9\end{cases}$
$\text{I}=\int^\limits4_1\text{f(x)}\text{dx}$
$\Rightarrow\text{I}=\int^\limits{3}_1\text{f(x)}\text{dx}+\int^\limits4_3\text{f(x)}\text{dx}$ [Additive property]
$\Rightarrow\text{I}=\int^\limits{3}_1(7\text{x}+3)\text{dx}+\int^\limits{4}_38\text{x dx}$
$\Rightarrow\text{I}=\Big[\frac{7\text{x}^2}{2}+3\text{x}\big]^3_1+\big[4\text{x}^2\big]^4_2$
$\Rightarrow\text{I}=\frac{63}{2}+9-\frac{7}{2}-3+64-36$
$\Rightarrow\text{I}=\frac{56}{2}+34$
$\Rightarrow\text{I}=62$
View full question & answer→Question 613 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 623 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 633 Marks
Prove the following Exercise:
$\int^{\frac{\pi}{2}}_{0}2\tan^{3}\text{x dx}=1-\log2$
Answer$\text{Let I}=\int^{\frac{\pi}{4}}\limits_{0}2\tan^{3}\text{x dx}$
$\text{I}=\int^{\frac{\pi}{4}}\limits_{0}\tan^{2}\text{x}\ \tan\text{x dx}=2\int^{\frac{\pi}{4}}\limits_{0}(\sec^{2}\text{x}-1)\tan\text{x dx}$
$=2\int^{\frac{\pi}{4}}\limits_{0}\sec^{2}\text{x}\ \tan\text{x dx}-2\int^{\frac{\pi}{4}}\limits_{0}\tan\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_0\sec\text{x}.\sec\text{x}\tan\text{x dx}-2\int\limits^{\frac{\pi}{4}}_0\tan\text{x dx}$
$\text{Now,} \sec\text{x}=\text{t}$
$=2\Big[\frac{\text{t}^2}{2}\Big]^{\sqrt{2}}_1-2\int\limits^{\frac{\pi}{4}}_0\tan\text{x dx}$
$=2\Big[\frac{2}{2}-\frac{1}{2}\Big]-2\int\limits^{\frac{\pi}{4}}_0\tan\text{x dx}$
$=1-2\int\limits^{\frac{\pi}{4}}_0\tan\text{x dx}$
Hence, the given result is Proved
View full question & answer→Question 643 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 653 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 663 Marks
Evaluate the definite integral in Exercise:
$\int\limits_\frac{\pi}{2}^{\pi}\text{e}^{\text{x}}\bigg(\frac{1-\sin\text{x}}{1+\cos\text{x}}\bigg)\text{dx}$
Answer$\text{I}=\int^{\pi}\limits_{\frac{\pi}{2}}\text{e}^{\text{x}}\bigg(\frac{1-\sin\text{x}}{1+\cos\text{x}}\bigg)\text{dx}$ $\text{I}=\int^{\pi}\limits_{\frac{\pi}{2}}\text{e}^{\text{x}}\Bigg[\frac{1-2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\sin^{2}\frac{\text{x}}{2}}\Bigg]\text{dx}$ $\text{I}=\int^{\pi}\limits_{\frac{\pi}{2}}\text{e}^{\text{x}}\Bigg[\frac{\text{cosec}^{2}\frac{\text{x}}{2}}{2}-\cot\frac{\text{x}}{2}\Bigg]\text{dx}$ $\text{Let f (x)}=-\cot\frac{\text{x}}{2}$ $\Rightarrow\text{f (x)}=-\bigg(-\frac{1}{2}\text{cosec}^{2}\frac{\text{x}}{2}\bigg)=\frac{1}{2}\text{cosec}^{2}\frac{\text{x}}{2}$ $\therefore\ \text{I}=\int^{\pi}\limits_{\frac{\pi}{2}}\text{e}^{\text{x}}\Big[\text{f(x)}+\text{f}'\text{(x)}\Big]\text{dx}$$=\Big[\text{e}^{\text{x}}.\text{f (x) dx}\Big]_{\frac{\pi}{2}}^{\pi}$
$=-\bigg[\text{e}^\text{x}.\cot\frac{\text{x}}{2}\bigg]^{\pi}_{\frac{\pi}{2}}$ $=-\bigg[\text{e}^{\text{x}}\times\cot\frac{\pi}{2}-\text{e}^{\frac{\pi}{2}}\times\cot\frac{\pi}{4}\bigg]$ $=-\bigg[\text{e}^{\text{x}}\times0-\text{e}^{\frac{\pi}{2}}\times1\bigg]$$=\text{e}^{\frac{\pi}{2}}$
View full question & answer→Question 673 Marks
Evaluate the following definite integrals:
$\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{4}}\text{cosec}\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{4}}\text{cosec}\text{x}\text{ dx}$
$\int\text{cosec}\text{x dx}=\log|\text{cosecx}-\cot\text{x}|=\text{F}(\text{x})$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}\Big(\frac{\pi}{4}\Big)-\text{F}\Big(\frac{\pi}{6}\Big)$
$=\log\Big|\text{cosec}\frac{\pi}{4}-\cot\frac{\pi}{4}\Big|-\log\Big|\text{cosec}\frac{\pi}{6}-\cot\frac{\pi}{6}\Big|$
$=\log\big|\sqrt{2}-1\big|-\log\big|2-\sqrt{3}\big|$
$=\log\bigg(\frac{\sqrt{2}-1}{2-\sqrt{3}}\bigg)$
View full question & answer→Question 683 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 693 Marks
Find the integrals of the functions in Exercises:
$\sin4\text{x}\sin8\text{x}$
AnswerIt is known that, $\sin\text{A}\sin\text{B}=\frac{1}{2}\cos(\text{A}-\text{B})-\cos(\text{A}+\text{B})$ $=\int\bigg\{\frac{1}{2}\cos(4\text{x}-8\text{x})-\cos(4\text{x}+8\text{x})\bigg\}\text{ dx}$ $=\frac{1}{2}\int(\cos(-4\text{x})-\cos12\text{x})\text{dx}$ $=\frac{1}{2}\int\big\{(\cos4\text{x}-\cos12\text{x}\big\}\text{dx}$ $=\frac{1}{2}\bigg[\frac{\sin4\text{x}}{4}-\frac{\sin12\text{x}}{12}\bigg]+\text{C}$
View full question & answer→Question 703 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 713 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 723 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 733 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 743 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 753 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
Answer$\int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1}{\sqrt{\text{x}}}+\frac{1}{\sqrt{\text{x}}\text{x}}\Big)\text{dx}$
$=\int\text{x}^{\frac{-1}{2}}+\int\text{x}^{\frac{-3}{2}}\text{dx}$
$=2\text{x}^{\frac{1}{2}}-2\text{x}^{\frac{-1}{2}}+\text{C}$
$=2\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}+\text{C}$
$\therefore\ \int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}=2\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}+\text{C}$
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Evaluate the integral in Exercise:$\int^{2}_{0}\text{x}\sqrt{\text{x}+2}\ (\text{put}\ \text{x}+2=\text{t}^{2})$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{2}\text{x}\sqrt{\text{x}+2}\ \text{dx}$
$\text{putting}\sqrt{\text{x}+2}=\text{t}\ \Rightarrow\ \text{x}+2=\text{t}^{2}\ \Rightarrow\frac{\text{dx}}{\text{dt}}=2\text{t}\ \Rightarrow\text{dx}=2\text{t}\ \text{dt}$
$\text{Limits of integration}\ \text{when}\ \text{x}=0,\text{t}=\sqrt{2}\ \text{and}\ \text{when}\text{x}=2,\text{t}=\sqrt{4}=2$
$\therefore\text{from}\ \text{eq}.\text{(i)},\ \text{I}=\int\limits_{\sqrt{2}}^{2}\text{t}^{2}(\text{t}^{2}-2)\text{t}.2\text{t}\ \text{dt}=2\int\limits_{\sqrt{2}}^{2}\text{t}^{2}(\text{t}^{2}-2)\text{dt}=2\int\limits_{\sqrt{2}}^{2}(\text{t}^{2}-2\text{t}^{2})\text{dt} $
$=2\Bigg[\bigg(\frac{\text{t}^{5}}{5}\bigg)^{2}_{\sqrt{2}}-2\bigg(\frac{\text{t}^{3}}{3}\bigg)^{2}_{\sqrt{2}}\Bigg]=2\bigg[\frac{1}{5}\Big(2^{5}-\big(\sqrt{2}\big)^{5}\Big)-\frac{2}{3}\Big(2^{3}-\big(\sqrt{2}\big)^{3}\Big)\bigg]$
$=2\bigg[\frac{1}{5}\big(32-4\sqrt{2}\big)-\frac{2}{3}\big(8-2\sqrt{2}\big)\bigg]=2\bigg[\frac{32}{5}-\frac{4\sqrt{2}}{5}-\frac{16}{3}+\frac{4\sqrt{2}}{3}\bigg]=2\bigg[\frac{96-12\sqrt{2}-80+20\sqrt{2}}{15}\bigg]$
$=\frac{2}{15}\big(16+8\sqrt{2}\big)=\frac{16\sqrt{2}}{15}\big(\sqrt{2}+1\big)$
View full question & answer→Question 773 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int^{5}_{-5}|\text{x}+2|\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{-5}^{5}|\text{x}+2|\ \text{dx}$
$\text{putting}\ \ \text{x}+2=0\ \ \Rightarrow=-2\in(-5,5)$
$\therefore\ \ \text{From eq. (i)},\ \text{I}=\int\limits_{-5}^{-2}|\text{x}+2|\ \text{dx}+\int\limits_{-2}^{5}|\text{x}+2|\ \text{dx}=\int\limits_{-5}^{-2}-\text{(x}+2 )\ \text{dx}+\int\limits_{-2}^{5}\text{(x}+2)\ \text{dx}$
$=-\bigg(\frac{\text{x}^{2}}{2}+2\text{x}\bigg)^{-2}_{-5}+\bigg(\frac{\text{x}^{2}}{2}+2\text{x}\bigg)^{5}_{-2}=-\bigg[\bigg(\frac{4}{2}-4\bigg)-\bigg(\frac{25}{2}-10\bigg)\bigg]+\bigg[\bigg(\frac{25}{2}-10\bigg)-\bigg(\frac{4}{2}-4\bigg)\bigg]$
$=-\bigg(-2-\frac{5}{2}\bigg)+\bigg(\frac{45}{2}+2\bigg)=2+\frac{5}{2}+\frac{45}{2}+2=4+25=29$
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Prove the following Exercise:
$\int^{\frac{\pi}{2}}_{0}\sin^{3}\text{x dx}=\frac{2}{3}$
Answer$\text{Let I}=\int^{\frac{\pi}{2}}_{0}\sin^{3}\text{x dx}$
$\text{I}=\int^{\frac{\pi}{2}}_{0}\sin^{2}\text{x.}\sin\text{x dx}$
$=\int^{\frac{\pi}{2}}_{0}\Big(1-\cos^{2}\text{x}\Big)\sin\text{x dx}$
$=\int^{\frac{\pi}{2}}_{0}\sin\text{x dx}-\int^{\frac{\pi}{2}}_{0}\cos^{2}\text{x}.\sin\text{x dx}$
$=\Big[-\cos\text{x}\Big]^{\frac{\pi}{0}}_{0}-\Big[\frac{\cos^{3}\text{x}}{3}\Big]^{\frac{\pi}{2}}_{0}$
$=1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3}$
Hence, the given result is Proved.
View full question & answer→Question 793 Marks
Find the integrals of the functions in Exercises:
$\tan^4\text{x}$
Answer$\tan^4\text{x}$
$=\tan^2\text{x}\tan^2\text{x}$
$=\big(\sec^2\text{x}-1\big)\tan^2\text{x}$
$=\sec^2\text{x}\tan^2\text{x}-\tan^2\text{x}$
$=\sec^2\text{x}\tan^2\text{x}-\big(\sec^2\text{x}-1\big)$
$=\sec^2\text{x }\tan^2\text{x}-\sec^2\text{x}+1$
$\therefore\int\tan^4\text{x}\text{ dx}=\int\sec^2\text{x }\tan^2\text{x}\text{ dx}-\int\sec^2\text{x}\text{ dx}+\int1\cdot\text{dx}$
$=\int\sec^2\text{x }\tan^2\text{x}\text{ dx}-\tan\text{x}+\text{x}+\text{C} \ \ \ \ ...\text{(1)}$
Consider $\int\sec^2\text{x }\tan^2\text{x}\text{ dx}$
$\text{Let }\tan\text{x}=\text{t}\Rightarrow\sec^2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\int\sec^2\text{x }\tan^2\text{x }\text{dx}=\int\text{t}^2\text{dt}=\frac{\text{t}^3}{3}=\frac{\tan^3\text{x}}{3}$
From equation(1),we obtain
$\int\tan^4\text{x}\text{ dx}=\frac{1}{3}\tan^3\text{x}-\tan\text{x}+\text{x}+\text{C}$
View full question & answer→Question 803 Marks
Evaluate the following:
$\int\frac{\text{dx}}{\text{x}\sqrt{\text{x}^4}-1}$
Hint: Put $\text{x}^2=\sec\theta$
AnswerLet $\text{I}=\int\frac{\text{dx}}{\text{x}\sqrt{\text{x}^4}-1}$
Put $\text{x}^2=\sec\theta\Rightarrow\theta=\sec^{-1}\text{x}^2$
$\Rightarrow 2\text{xdx} =\sec\theta.\tan\theta \text{d}\theta$
$\therefore\ \text{I}=\frac{1}{2}\int\frac{\sec\theta\cdot\tan\theta}{\sec\theta\cdot\tan\theta}\text{d}\theta$ $=\frac{1}{2}\int\text{d}\theta=\frac{1}{2}\theta+\text{C}$
$=\frac{1}{2}\sec^{-1}(\text{x}^2)+\text{C}$
View full question & answer→Question 813 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 823 Marks
Evaluate the following definite integrals:
$\int_{1}^\limits{\text{e}}\frac{\log\text{x}}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{\text{e}}\frac{\log\text{x}}{\text{x}}\text{ dx}$
Let $\log\text{x}=\text{u}$
$\Rightarrow\frac{1}{\text{x}}=\text{dx}=\text{du}$
$\therefore\text{ I}=\int\text{u}\text{ du}$
$\Rightarrow\text{I}=\Big[\frac{\text{u}^2}{2}\Big]$
$\Rightarrow\text{I}=\Big[\frac{(\log\text{x})}{2}\Big]^{\text{e}}_1$
$\Rightarrow\text{I}=\frac{1}{2}-0$
$\Rightarrow\text{I}=\frac{1}{2}$
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Evaluate the following integrals:
$\int\limits^{\sqrt{2}}_0\big[\text{x}^2\big]\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{\sqrt{2}}_0\big[\text{x}^2\big]\text{dx}$
$=\int\limits^{1}_0\big[\text{x}^2\big]\text{dx}+\int\limits^{\sqrt{2}}_1\big[\text{x}^2\big]\text{dx}$
$=\int\limits^{1}_0(0)\text{dx}+\int\limits^{\sqrt{2}}_1(1)\text{dx}$ $\begin{pmatrix}\because\big[\text{x}\big]^2=\begin{cases}0,&0<\text{x}<1\\1,&1<\text{x}<\sqrt{2}\end{cases}\end{pmatrix}$
$=0+\big[\text{x}\big]^{\sqrt{2}}_1$
$=\sqrt{2}-1$
View full question & answer→Question 843 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
Let $\text{f}(\text{x})=\text{x}\cos^2\text{x}$
$\Rightarrow \text{f}(-\text{x})= (-\text{x})\cos^2(-\text{x})$
$= - \text{x}\cos^2\text{x}$
$\therefore \text{f}(-\text{x})=-\text{f}(\text{x})$
i.e., f(x) is odd function.
We know that $\int\limits^\text{a}_{-\text{a}}\text{f}(\text{x})\text{dx} = 0, $ if f(x) is odd function.
$\therefore\ \text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}=0$
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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 863 Marks
Evaluate the following:
$\int\frac{\sqrt{\text{x}}}{\sqrt{\text{a}^3-\text{x}^3}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sqrt{\text{x}}}{\sqrt{\text{a}^3-\text{x}^3}}\text{dx}$ $=\int\frac{\sqrt{\text{x}}}{\sqrt{\Big(\text{a}^{\frac{3}{2}}\Big)^2-\Big(\text{x}^{\frac{3}{2}}\Big)^2}}$
Put $=\text{x}^{\frac{3}{2}}=\text{t}\Rightarrow\frac{3}{2}\text{x}^{\frac{1}{2}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\frac{2}{3}\int\frac{\text{dt}}{\sqrt{\Big(\text{a}^{\frac{3}{2}}\Big)^2}-\text{t}^2}$ $=\frac{2}{3}\sin^{-1}\frac{\text{t}}{\text{a}^{\frac{3}{2}}}+\text{C}$
$=\frac{2}{3}\sin^{-1}\frac{\text{x}^{\frac{3}{2}}}{\text{a}^{\frac{3}{2}}}+\text{C}$ $=\frac{2}{3}\sin^{-1}\sqrt{\frac{\text{x}^3}{\text{a}^3}}+\text{C}$
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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 883 Marks
Evaluate the following:
$\int\limits^1_0\frac{\text{dx}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\limits^1_0\frac{\text{dx}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$
$=\int\limits^1_0\frac{\text{dx}}{\text{e}^\text{x}+\frac{1}{\text{e}^\text{x}}}$
$=\int\limits^0_1\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{dx}$
Put $\text{e}^\text{x}=\text{t}$
$\text{e}^\text{x}\text{dx}=\text{dt}$
Substituting ex = t and ex dx = dt
$\therefore\ \text{I}=\int\limits^\text{e}_1\frac{\text{dt}}{1+\text{t}^2}$
$=\big[\tan^{-1}\text{t}\big]^\text{e}_\text{1}$ $\Big[\because\int\frac{1}{1+\text{x}^2}\text{dx}=\tan^{-1}\text{x}+\text{C}\Big]$
$=\tan^{-1}\text{e}-\tan^{-1}1$
$=\tan^{-1}\text{e}-\frac{\pi}{4}$ $\Big[\because\tan^{-1}1=\frac{\pi}{4}\Big]$
View full question & answer→Question 893 Marks
Evaluate the following:
$\int\sqrt{2\text{ax}-\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{2\text{ax}-\text{x}^2}\text{dx}$ $=\int\sqrt{-(\text{x}^2-2\text{ax})\text{dx}}$
$=\int\sqrt{-(\text{x}^2-2\text{ax}+\text{a}^2-\text{a}^2)}\text{dx}$ $=\int\sqrt{-(\text{x}-\text{a})^2-\text{a}^2}\text{dx}$
$=\int\sqrt{\text{a}^2-(\text{x}-\text{a})^2}\text{dx}$
$=\frac{\text{x}-\text{a}}{2}\sqrt{\text{a}^2-(\text{x}-\text{a})^2}+\frac{\text{a}^2}{2}\sin^{-1}\Big(\frac{\text{x}-\text{a}}{\text{x}}\Big)+\text{C}$
$=\frac{\text{x}-\text{a}}{2}\sqrt{2\text{ax}-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\Big(\frac{\text{x}-\text{a}}{\text{a}}\Big)+\text{C}$
View full question & answer→Question 903 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 913 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 923 Marks
Evaluate the following:
$\int\sqrt{1+\sin\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{1+\sin\text{x}}\text{dx}$
$=\int\sqrt{\sin^2\frac{\text{x}}{2}+\cos^2\frac{\text{x}}{2}+2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}\text{dx}$ $\Big[\because\ \sin^2\frac{\text{x}}{2}+\cos^2\frac{\text{x}}{2}=1\Big]$
$=\int\sqrt{\big(\sin\frac{\text{x}}{2}+\cos\frac{\text{x}}{2}\big)^2}\text{dx}$
$=\int\Big(\sin\frac{\text{x}}{2}+\cos\frac{\text{x}}{2}\Big)\text{dx}$
$=-2\cos\frac{\text{x}}{2}+2\sin\frac{\text{x}}{2}+\text{C}$
View full question & answer→Question 933 Marks
Evaluate the following:
$\int\frac{\text{dx}}{1+\cos\text{x}}$
AnswerLet $\int\frac{\text{dx}}{1+\cos\text{x}}$
$=\int\frac{\text{x}}{2\cos^2\frac{\text{x}}{2}}$ $\Big[\because\ 1+\cos\text{A}=2\cos^2\frac{\text{A}}{2}\Big]$
$=\frac{1}{2}\int\frac{1}{\cos^2\frac{\text{x}}{2}}\text{dx}$
$=\frac{1}{2}\int\sec^2\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{2}\cdot\tan\frac{\text{x}}{2}\cdot2+\text{C}$ $\big[\int\sec^2\text{x dx}=\tan\text{x}\big]$
$=\tan\frac{\text{x}}{2}+\text{C}$
View full question & answer→Question 943 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 953 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int^{\frac{\pi}{2}}_{0}\frac{\cos^{5}\text{x}\ \text{dx}}{\sin^{5}\text{x}+\cos^{5}\text{x}}$
Answer$\text{Let}\text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\cos^{5}\text{x}\ }{\sin^{5}\text{x}+\cos^{5}\text{x}}\text{dx}$ $\Rightarrow\ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\cos^{5}\bigg(\frac{\pi}{2}-\text{x}\bigg)}{\sin^{5}\bigg(\frac{\pi}{2}-\text{x}\bigg)+\cos^{5}\bigg(\frac{\pi}{2}-\text{x}\bigg)}\text{dx}\ \ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\ \text{dx}=\int\limits_{0}^{\text{a}}\text{f}(\text{(a}-\text{x)}\text{dx}=\bigg]$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^{5}\text{x}}{\cos^{5}\text{x}+\sin^{5}\text{x}}\text{dx}$Adding eq. (i) and (ii),
$21=\int\limits_{0}^{\frac{\pi}{2}}\bigg(\frac{\cos^{5}\text{x}}{\sin^{5}\text{x}+\cos^{5}\text{x}}+\frac{\sin^{5}\text{x}}{\cos^{5}\text{x}+\sin^{5}\text{x}}\bigg)\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\bigg(\frac{\cos^{5}\text{x}+\sin^{5}\text{x}}{\sin^{5}\text{x}+\cos{5}\text{x}}\bigg)\text{dx}$ $\Rightarrow\ \ 21=\int\limits_{0}^{\frac{\pi}{2}}1\ \text{dx}=\bigg(\text{x}^{\frac{\pi}{2}}_{0}\bigg)\ \Rightarrow21=\frac{\pi}{2}\ \Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 963 Marks
Evaluate the following integrals:
$\int\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{1+\cos4\text{x}}}\text{dx}$
Answer$\int\Big(\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{1+\cos4\text{x}}}\Big)\text{x}$
$=\int\frac{\cos(2\text{x})}{\sqrt{2\cos^2(2\text{x})}}\text{dx}$ $\Big[\therefore1+\cos\text{A}=2\cos^2\Big(\frac{\text{A}}{2}\Big)\text{ and }\cos^2\text{A}-\sin^2\text{A}=\cos\text{2A}\Big]$
$=\frac{1}{\sqrt2}\int\Big(\frac{\cos2\text{x}}{\cos\text{2x}}\Big)\text{dx}$
$=\frac{1}{\sqrt{2}}[\text{x}]+\text{C}$
$=\frac{\text{x}}{\sqrt{2}}+\text{C}$
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Evaluate the following integrals:
$\int_{4}^\limits{12}\text{x}(\text{x}-4)^{\frac{1}{3}}\text{dx}$
AnswerLet $\text{I}=\int_{4}^\limits{12}\text{x}(\text{x}-4)^{\frac{1}{3}}\text{dx}$
Let $\text{x}-4=\text{t}$ Then, $\text{dx}=\text{dt}$
When $\text{x}=4,\text{t}=0$ and $\text{x}=12,\text{t}=8$
$\therefore\ \text{I}=\int\limits^8_0(\text{t}+4)\text{t}^{\frac{1}{3}}\text{dt}$
$\Rightarrow\text{I}=\int\limits^8_0\Big(\text{t}^{\frac{4}{3}}+4\text{t}^{\frac{1}{3}}\Big)\text{dt}$
$\Rightarrow\text{I}=\Big[\frac{3}{7}\text{t}^{\frac{7}{3}}+\frac{3}{1}\text{t}^{\frac{4}{3}}\Big]^8_0$
$\Rightarrow\text{I}=\frac{384}{7}+48$
$\Rightarrow\text{I}=\frac{720}{7}$
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Evaluate the following integrals:
$\int\limits^2_0\text{x}\sqrt{2-\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^2_0\text{x}\sqrt{2-\text{x}}\text{ dx}$
$=\int\limits^2_0(2-\text{x})\sqrt{2-2+\text{x}}\text{ dx}$
$=\int\limits^2_0(2-\text{x})\sqrt{\text{x}}\text{ dx}$
$=\int\limits^2_0\big(2\sqrt{\text{x}}-\text{x}\sqrt{\text{x}}\big)\text{dx}$
$=\int\limits^2_0\Big(2\text{x}^{\frac{1}{2}}-\text{x}^{\frac{3}{2}}\Big)\text{dx}$
$=\Bigg[2\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{\text{x}^{\frac{5}{2}}}{\frac{5}{2}}\Bigg]$
$=\bigg[\frac{4}{3}\text{x}^{\frac{3}{2}}-\frac{2}{5}\text{x}^{\frac{5}{2}}\bigg]^2_0$
$=\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5}$
$=\frac{16\sqrt{2}}{15}$
View full question & answer→Question 993 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 1003 Marks
Evaluate the following integrals:
$\int\frac{\sin2\text{x}}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
Answer$\int\frac{\sin2\text{x}}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
$\text{Let a}+\text{b}\cos2\text{x}=\text{t}$
$\Rightarrow-\text{b}\sin(2\text{x})\text{dx}\times2=\text{dt}$
$\Rightarrow\sin(2\text{x})\text{dx}=\frac{-\text{dt}}{2\text{b}}$
$\text{Now,}\int\frac{\sin(2\text{x})}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
$=-\frac{1}{2\text{b}}\int\frac{\text{dt}}{\text{t}^2}$
$=\frac{-1}{2\text{b}}\int\text{t}^{-2}\text{dt}$
$=\frac{-1}{2\text{b}}\Big[\frac{\text{t}^{-2+1}}{-2+1}\Big]+\text{C}$
$=\frac{1}{2\text{b}}\times\frac{1}{\text{t}}+\text{C}$
$=\frac{1}{2\text{b}(\text{a}+\text{b}\cos2\text{x})}+\text{C}$
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