Question 1513 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 1523 Marks
Evalute the following integrals:
$\int\frac{1}{\cos3\text{x}-\cos\text{x}}\text{dx}$
Answer$\int\frac{1}{\cos3\text{x}-\cos\text{x}}\text{dx}$
$=\int\frac{1}{4\cos^2\text{x}-4\cos\text{x}}\text{dx}\big[\because\cos3\text{x}=\text{4}\cos^3\text{x}\cos\text{x}\big]$
$=\int\frac{1}{4\cos\text{x}(\cos^2\text{x}-1)}\text{dx}$
$=\frac{-1}{4}\int\frac{1}{\cos\text{x}\sin^2\text{x}}\text{dx}$
$=\frac{-1}{4}\int\Big(\frac{\sin^2\text{x}+\cos^2\text{x}}{\cos\text{x}\sin^2\text{x}}\text{dx}\Big)$
$=\frac{-1}{\text{4}}\Big[\int\sec\text{ x dx}+\int\cot\text{x cosec x dx}$
$=\frac{-1}{4}\big(\text{ln }|\sec\text{x}+\tan\text{x}|-\text{cosec x}\big)+\text{C}$
$=\frac{1}{4}\big(\text{cosec x}-\text{ln}|\sec\text{x}+\tan\text{x}|\big)+\text{C}$
View full question & answer→Question 1533 Marks
Evaluate the following integrals:
$\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\Rightarrow\text{I}=\int\sqrt{4-\text{t}^2}\text{dt}$
$=\int\sqrt{2^2-\text{t}^2}\text{dt}$
$=\frac{\text{t}^2}{2}\sqrt{2^2-\text{t}^2}+\frac{4}{2}\sin^{-1}\frac{\text{t}}{2}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\sin\text{x}\sqrt{4-\sin^2\text{x}}+2\sin^{-1}\Big(\frac{\sin\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 1543 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
Answer$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{-(\text{x}^2-\text{x})}\text{dx}$
$=\int\sqrt{-\bigg\{\text{x}^2-\text{x}+\Big(\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2\bigg\}}\text{dx}$
$=\int\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$
$=\Big(\frac{\text{x}-\frac{1}{2}}{2}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}\bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\bigg)+\text{C}$
$\Big[\because\ \int\sqrt{\text{a}^2-\text{x}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{a}^2-\text{x}^2}+\frac{1}{2}\text{a}^2\sin^{-1}\frac{\text{x}}{\text{a}}=\text{C}\Big]$
$=\Big(\frac{2\text{x}-1}{4}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}(2\text{x}-1)+\text{C}$
View full question & answer→Question 1553 Marks
Evaluate the following integrals:
$\int\limits^\text{b}_{\text{a}}\frac{\text{f(x)}}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\text{ dx}$
AnswerLet $\text{I}=\int\limits^\text{b}_{\text{a}}\frac{\text{f(x)}}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\text{ dx}\ ....(\text{i})$
$=\int\limits^\text{b}_{\text{a}}\frac{\text{f}(\text{a}+\text{b}-\text{x})}{\text{f}(\text{a}+\text{b}-\text{x})+\text{f}(\text{a}+\text{b}-\text{a}-\text{b}+\text{x})}\text{ dx}$
$=\int\limits^\text{b}_{\text{a}}\frac{\text{f}(\text{a}+\text{b}-\text{x})}{\text{f}(\text{a}+\text{b}-\text{x})+\text{f(x)}}\text{ dx}$
$\therefore\ \text{I}=\int\limits^\text{b}_{\text{a}}\frac{\text{f}(\text{a}+\text{b}-\text{x})}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^\text{b}_{\text{a}}\bigg[\frac{\text{f(x)}}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}+\frac{\text{f}(\text{a}+\text{b}-\text{x})}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\bigg]\text{dx}$
$=\int\limits^\text{b}_{\text{a}}\frac{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}{\text{f(x)}+\text{f}(\text{a}+\text{b}-\text{x})}\text{ dx}$
$=\big[\text{x}\big]^{\text{b}}_\text{a}$
$=\text{b}-\text{a}$
Hence, $\text{I}=\frac{\text{b}-\text{a}}{2}$
View full question & answer→Question 1563 Marks
Integrate the function in exercise.
$\text{x}^2 \ \text{e}^\text{x}$
AnswerLet $\text{I}=\int\text{x}\sin3\text{x dx}$
Taking $x^2$ as first function and ex as second function and integrating by parts, we obtain.
$\text{I}=\text{x}\int\sin3\text{x dx}-\int\Bigg\{\Big(\frac{\text{d}}{\text{dx}}\text{x}\Big)\int\sin3\text{x dx}\Bigg\}$
$=\text{x}\Big(\frac{-\cos3\text{x}}{3}\Big)-\int1.\Big(\frac{-\cos3\text{x}}{3}\Big)\text{dx}$
$=\frac{-\text{x}\cos3\text{x}}{3}+\frac{1}{3}\int\cos3\text{x} \ \text{dx}$
$=\frac{-\text{x}\cos3\text{x}}{3}+\frac{1}{9}\sin3\text{x} \ \text{dx}+\text{C}$
View full question & answer→Question 1573 Marks
Write a value of $\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$ $=\int\text{e}^{\text{x}}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}$ Let $\text{e}^{\text{x}}\sec\text{x}=\text{t}$ $(\text{e}^{\text{x}}\sec\text{x}+\text{e}^{\text{x}}\sec\text{x}\tan\text{x})\text{dx}=\text{dt}$ $\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}=\text{dt}$ $\therefore\text{I}=\int\text{dt}$ $=\text{t}+\text{C}$$=\text{e}^{\text{x}}\sec\text{x}+\text{C}$ $(\because\text{t}=\text{e}^{\text{x}}\sec\text{x})$
View full question & answer→Question 1583 Marks
Evalute the following integrals:
$\int\frac{1+\tan\text{x}}{1-\tan\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1+\tan\text{x}}{1-\tan\text{x}}\text{dx}$
$=\int\bigg(\frac{1+\frac{\sin\text{x}}{\cos\text{x}}}{1-\frac{\sin\text{x}}{\cos\text{x}}}\bigg)\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}-\sin\text{x}}\Big)\text{dx}$
Putting $\cos\text{x}-\sin\text{x}=\text{t}$
$\Rightarrow(-\sin\text{x}-\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow(\sin\text{x}+\cos\text{x})\text{dx}=-\text{dt}$
$\therefore\text{I}=-\int\frac{1}{\text{t}}\text{dt}$
$=-\text{ln}|\text{t}|+\text{C}$
$=-\text{ln}|\cos\text{x}-\sin\text{x}|+\text{C}\ \big[\because\text{t}\cos\text{x}-\sin\text{x}\big]$
View full question & answer→Question 1593 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 1603 Marks
Evaluate the following integrals:
$\int^\limits4_1\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}4\text{x}+3,&\text{if }\ 1\leq\text{x}\leq2\\3\text{x}+5,&\text{if }\ 2\leq\text{x}\leq4\end{cases}$
AnswerWe have,
$\int^\limits4_1\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}4\text{x}+3,&\text{if }\ 1\leq\text{x}\leq2\\3\text{x}+5,&\text{if }\ 2\leq\text{x}\leq4\end{cases}$
$\text{I}=\int^\limits4_1\text{f(x)}\text{dx}$
$\Rightarrow\text{I}=\int^\limits2_1\text{f(x)}\text{dx}+\int^\limits4_2\text{f(x)}\text{dx}$ [Additive property]
$\Rightarrow\text{I}=\int^\limits2_1(4\text{x}+3)\text{dx}+\int^\limits4_2(3\text{x}+5)\text{dx}$
$\Rightarrow\text{I}=\Big[2\text{x}^2+3\text{x}\Big]^2_1+\Big[\frac{3\text{x}^2}{2}+5\text{x}\Big]^4_2$
$\Rightarrow\text{I}=8+6-2-3+24+20-6-10$
$\Rightarrow\text{I}=37$
View full question & answer→Question 1613 Marks
Evalute the following integrals:
$\int\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{10^\text{x}+\text{x}^{10}}\text{dx}$
AnswerLet $\text{I}=\int\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{10^\text{x}+\text{x}^{10}}\text{dx}\ .....(\text{i})$
Let $10^\text{x}+\text{x}^{10}=\text{t}$ then,
$\text{d}(10^\text{x}+\text{x}^{10})=\text{dt}$
$\Rightarrow(10^\text{x}\log_\text{e}10+10\text{x}^9)\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{(10\text{x}^9+10^\text{x}\log_\text{e}10)}$
Putting $10^x + x^{10} = t$ and $\text{dx}=\frac{\text{dt}}{10\text{x}^9+10^\text{x}\log_\text{e}10}$ in equation $(i),$ we get,
$\text{I}=\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{\text{t}}\times\frac{\text{dt}}{10\text{x}^9+10^\text{x}\log_\text{e}10}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|10^\text{x}+\text{x}^{10}|+\text{C}$
$\therefore\text{I}=\log|10^\text{x}+\text{x}^{10}|+\text{C}$
View full question & answer→Question 1623 Marks
$\int\frac{1}{1+\cos3\text{x}}\text{dx}$
Answer$\int\frac{\text{dx}}{1+\cos3\text{x}}$
$=\int\frac{(1-\cos3\text{x})}{(1+\cos3\text{x})(1-\cos3\text{x})}\text{dx}$
$=\int\Big(\frac{1-\cos3\text{x}}{1-\cos^23\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1-\cos3\text{x}}{\sin^23\text{x}}\Big)\text{dx}$
$=\int\text{cosec}^23\text{x}\text{ dx}-\int\text{cosec}3\text{x}\cot3\text{x dx}$
$=-\frac{\cot3\text{x}}{3}+\frac{\text{cosec}3\text{x}}{3}+\text{c}$
$=\frac{1}{3}[\text{cosec}3\text{x}-\cot3\text{x}]+\text{c}$
$=\frac{1}{3}\bigg[\frac{1}{\sin3\text{x}}-\frac{\cos3\text{x}}{\sin3\text{x}}\bigg]+\text{c}$
$=\frac{1}{3}\bigg[\frac{1-\cos3\text{x}}{\sin3\text{x}}\bigg]+\text{c}$
View full question & answer→Question 1633 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^{\frac{3}{2}}\text{x}\ \text{dx}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^{\frac{3}{2}}\text{x}\ }{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}\text{dx}$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^{\frac{3}{2}}\bigg(\frac{\pi}{2}-\text{x}\bigg)}{\sin^{\frac{3}{2}}\bigg(\frac{\pi}{2}-\text{x}\bigg)+\cos^{\frac{3}{2}}\bigg(\frac{\pi}{2}-\text{x}\bigg)}\text{dx}\ \ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\ \text{dx}=\int\limits_{0}^{\text{a}}\text{f}(\text{a}-\text{x})\text{dx}=\bigg]$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\cos^{\frac{3}{2}}\text{x}}{\cos^{\frac{3}{2}}\text{x}+\sin^{\frac{3}{2}}\text{x}}\text{dx}$Adding eq. (i) and (ii),
$21=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\frac{\sin^{\frac{3}{2}}\text{x}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}+\frac{\cos^{\frac{3}{2}}\text{x}}{\cos^{\frac{3}{2}}\text{x}+\sin^{\frac{3}{2}}\text{x}}\bigg]\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\frac{\sin^{\frac{3}{2}}+\cos^{\frac{3}{2}}\text{x}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}\bigg]\text{dx}$
$\Rightarrow\ \ 21=\int\limits_{0}^{\frac{\pi}{2}}1\ \text{dx}=\bigg(\text{x}^{\frac{\pi}{2}}_{0}\bigg)\ \Rightarrow\ \ 21=\frac{\pi}{2}\ \Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1643 Marks
Evaluate the integral in Exercise:
$\int^{2}_{1}\bigg(\frac{1}{\text{x}}-\frac{1}{2\text{x}^{2}}\bigg)\text{e}^{2\text{x}}\text{dx}$
Answer$\text{Let}$
$\text{I}=\int_{1}^{2}\bigg(\frac{1}{\text{x}}-\frac{1}{2\text{x}^{2}}\bigg)\text{e}^{2\text{x}}\text{dx}$
$=\int^{2}_{1}\frac{1}{\text{x}}\text{e}^{2\text{x}}\text{dx}+\int\frac{-1}{2\text{x}^{2}}\text{e}^{2\text{x}}\text{dx}$
$=\bigg[\frac{1}{\text{x}}\frac{\text{e}^{2\text{x}}}{2}\bigg]^{2}_{1}-\int^{2}_{1}\frac{-1}{\text{x}^{2}}\frac{\text{e}^{2\text{x}}}{2}\text{dx}+\int^{2}_{1}\frac{-1}{2\text{x}^{2}}\text{e}^{2\text{x}}\text{dx}$ [integrating by parts]
$=\frac{1}{2}\bigg[\frac{\text{e}^{2\text{x}}}{\text{x}}\bigg]^{2}_{1}-\int^{2}_{1}\frac{-1}{2\text{x}^{2}}{\text{e}^{2\text{x}}}{}\text{dx}+\int^{2}_{1}\frac{-1}{2\text{x}^{2}}\text{e}^{2\text{x}}\text{dx}$
$=\frac{1}{2}\bigg[\frac{\text{e}^{4}}{2}-\frac{\text{e}^{2}}{1}\bigg]=\frac{1}{4}\big(\text{e}^{4}-2\text{e}^{2}\big)=\frac{1}{4}\big(\text{e}^{2}-2\big)$
View full question & answer→Question 1653 Marks
$\int(5\text{x}+3)\sqrt{2\text{x}-1}\text{ dx}$
Answer$\text{Let I}=\int(5\text{x}+3)\sqrt{2\text{x}-1}\text{ dx}$
$\text{Putting}\ \ 2\text{x}-1=\text{t}$
$\Rightarrow2\text{x}=\text{t}+1$
$\Rightarrow\text{x}=\frac{\text{t}+1}{2}$
$\&\ 2\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\int\Big[5\Big(\frac{\text{t}+1}{2}\Big)+3\Big]\times\sqrt{\text{t}}\times\frac{\text{dt}}{2}$
$=\int\Big(\frac{5\text{t}}{5}+\frac{5}{2}+3\Big)\times\frac{\sqrt{\text{t}}\text{ dt}}{2}$
$=\frac{1}{4}\int(5\text{t}+11)\text{t}^\frac{1}{2}\text{ dt}$
$=\frac{1}{4}\int(5\text{t}^\frac{3}{2}+11\text{t}^\frac{1}{2})\text{ dt}$
View full question & answer→Question 1663 Marks
Integrate the function in Exercise:
$\tan^{-1}\text{x}$
AnswerLet $\text{I}=\int1.\tan^{-1}\text{x dx}$
Taking $\tan^{-1 }x$ as first function and $1$ as second function and integrating by parts, we obtain.
$\text{I}=\tan^{-1}\text{x}\int1.\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\tan^{-1}\text{x}\Bigg)\int1. \ \text{dx}\Bigg\}\text{dx}$
$=\tan^{-1}\text{x}.\text{x}-\int\frac{1}{1+\text{x}^2}.\text{x dx}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\int\frac{2\text{x}}{1+\text{x}^2}\text{dx}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\text{log}|1+\text{x}^2|+\text{C}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\text{log}(1+\text{x}^2)+\text{C}$
View full question & answer→Question 1673 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}^2}-\frac{2}{\text{x}^3}\Big)\text{dx}$
Answer$\text{I}=\int\text{e}^{\text{x}}\big(\text{x}^{-2}-2\text{x}^{-3}\big)\text{dx}$
$=\int\text{e}^{\text{x}}\text{x}^{-2}\text{dx}-2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}$
Integration by parts
$=\text{e}^{\text{x}}\text{x}^{-2}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\big(\text{x}^{-2}\big)\Big)\text{dx}-2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}$
$=\text{e}^{\text{x}}\text{x}^{-2}+2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}-2\int\text{e}^{\text{x}}\text{x}^{-3}\text{dx}$
$=\frac{\text{e}^{\text{x}}}{\text{x}^2}+\text{C}$
$\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}^2}-\frac{2}{\text{x}^3}\Big)\text{dx}=\frac{\text{e}^\text{x}}{\text{x}^2}+\text{C}$
View full question & answer→Question 1683 Marks
Evalute the following integrals:
$\int\frac{\cos\text{x}}{\cos(\text{x}-\text{a})}\text{dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{\cos(\text{x}-\text{a})}\text{dx}$
Putting x - a = t
⇒ x = a + t
⇒ dx = dt
$\therefore\text{I}=\int\frac{\cos(\text{a}+\text{t})\text{dt}}{\cos\text{t}}$
$=\frac{\cos\text{a}\cos\text{t}}{\cos\text{t}}-\frac{\sin\text{a}\sin\text{t}}{\cos\text{t}}\text{dt}$
$=\int\big(\cos\text{a}-\sin\text{a}\tan\text{t}\big)\text{dt}$
$=\text{t}\cos\text{a}-\sin\text{a ln}|\sec\text{t}|+\text{C}$
$=(\text{x}-\text{a})\cos\text{a}-\sin\text{a ln}|\sec(\text{a}-\text{a})|+\text{C}\ \big[\text{t}=\text{x}-\text{a}\big]$
View full question & answer→Question 1693 Marks
Write a value of $\int\frac{\cos\text{x}}{3+2\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{3+2\sin\text{x}}\text{ dx}$
Let $3+2\sin\text{x}=\text{t}$
$2\cos\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{2}\log\text{t}+\text{C}$
$=\frac{1}{2}\log(3+2\sin\text{x})+\text{C}$
$\therefore\ \int\frac{\cos\text{x}}{3+2\sin\text{x}}\text{ dx}=\frac{1}{2}\log(3+2\sin\text{x})+\text{C}$
View full question & answer→Question 1703 Marks
Evaluate the following definite integrals:
$\int_{1}^\limits{2}\text{e}^{2\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{2\text{x}^2}\big)\text{dx}$
AnswerLet $\text{I}=\int_{1}^\limits{2}\text{e}^{2\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{2\text{x}^2}\big)\text{dx}$ Then,
$\text{I}=\int_{1}^\limits{2}\text{e}^{2\text{x}} \frac{1}{\text{x}}-\int_{1}^\limits{2}\text{e}^{2\text{x}} \frac{1}{2\text{x}^2}\text{ dx}$
Integrating first term by parts,
$\Rightarrow\text{I}=\bigg\{\Big[\frac{\text{e}^{2\text{x}}}{2\text{x}}\Big]^2_1-\int_{1}^\limits{2}-\text{e}^{2\text{x}} \frac{1}{2\text{x}^2}\bigg\}-\int_{1}^\limits{2}\text{e}^{2\text{x}} \frac{1}{2\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{e}^{2\text{x}}}{2\text{x}}\Big]^2_1$
$\Rightarrow\text{I}=\frac{\text{e}^4}{4}-\frac{\text{e}^2}{2}$
$\Rightarrow\text{I}=\frac{\text{e}^4-2\text{e}^2}{4}$
View full question & answer→Question 1713 Marks
Write a value of $\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{1+\sin2\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}+\cos\text{x}}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sqrt{(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
$=\int\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
Let $\sin\text{x}+\cos\text{x}=\text{t}$
$(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}}$
$=-\log\text{t}+\text{C}$
$\text{I}=\log|\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 1723 Marks
Integrate the function in Exercise:
$\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}$
Answer$\text{Ler I}=\int\frac{\sec^2\text{x}}{\sqrt{\tan^2\text{x}+4}}\text{ dx}\ \ \ \ ...\text{(i)}$
$\text{Putting }\tan\text{x}=\text{t}\ \ \Rightarrow\ \ \sec^2\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \Rightarrow\ \ \sec^2\text{x}\text{ dx}=\text{dt}$
$\therefore\ \ \text{From eq. (i), }\text{ I}=\int\frac{\text{dt}}{\sqrt{\text{t}^2+4}}=\int\frac{1}{\sqrt{\text{t}^2+(2)^2}}\text{ dt}$
$=\log\bigg|\text{t}+\sqrt{\text{t}^2+(2)^2}\bigg|+\text{c}$
$=\log\Big|\tan\text{x}+\sqrt{\tan^2\text{x}+4}\Big|+\text{c}$
View full question & answer→Question 1733 Marks
Evaluate the following integrals:
$\int\log(\text{x}+1)\text{dx}$
AnswerLet $\text{I}=\int\log(\text{x}+1)\text{dx}$
$=\int1\times\log(\text{x}+1)\text{dx}$
Using integration by parts,
$\text{I}=\log(\text{x}+1)\int1\text{dx}-\int\Big(\frac{1}{\text{x}+1}\times\int1\text{dx}\Big)\text{dx+C}$
$=\text{x}\log(\text{x}+1)-\int\Big(\frac{\text{x}}{\text{x}+1}\Big)\text{dx+C}$
$=\text{x}\log(\text{x}+1)-\int\Big(1-\frac{1}{\text{x}+1}\Big)\text{dx+C}$
$\text{I}=\text{x}\log(\text{x}+1)-\text{x}+\log(\text{x}+1)+\text{C}$
View full question & answer→Question 1743 Marks
Evaluate the following integrals:
$\int\big(2-3\text{x}\big)\big(3+2\text{x}\big)\big(1-2\text{x}\big)\text{dx}$
Answer$\int\big(2-3\text{x}\big)\big(3+2\text{x}\big)\big(1-2\text{x}\big)\text{dx}$
$=\int\big(6+4\text{x}-9\text{x}-6\text{x}^2\big)\big(1-2\text{x}\big)\text{dx}$
$=\int\big(-6\text{x}^2-5\text{x}+6\big)\big(1-2\text{x}\big)\text{dx}$
$=\int\big(-6\text{x}^2+12\text{x}^3-5\text{x}+10\text{x}^2+6-12\text{x}\big)\text{dx}$
$=\int\big(4\text{x}^2+12\text{x}^3-17\text{x}+6\big)\text{dx}$
$=\int\big(12\text{x}^3+4\text{x}^2-17\text{x}+6\big)\text{dx}$
$=\frac{12}{4}\text{x}^4+\frac{4}{3}\text{x}^3-\frac{17}{2}\text{x}^2+6\text{x}+\text{C}$
$=3\text{x}^4+\frac{4}{3}\text{x}^3-\frac{17}{2}\text{x}^2+6\text{x}+\text{C}$
View full question & answer→Question 1753 Marks
Find the integrals of the functions in Exercises:
$\sin^3\text{x}\cos^3\text{x}$
Answer$\text{Let}\text{ I}=\int\sin^3\text{x}\cos^3\text{x}\text{ dx}$
$=\int\cos^3\text{x }\sin^2\text{x }\sin\text{x}\text{ dx}$
$=\int\cos^3\text{x}\big(1-\cos^2\text{x}\big)\sin\text{x}\text{ dx}$
$\text{Let}\cos\text{x}=\text{t}$
$\Rightarrow -\sin\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\text{I}=-\int\text{t}^3\big(1-\text{t}^2\big)\text{ dt}$
$=-\int\big(\text{t}^3-\text{t}^5\big)\text{dt}$
$=-\Bigg\{\frac{\text{t}^4}{4}-\frac{\text{t}^6}{6}\Bigg\}+\text{C}$
$=-\Bigg\{\frac{\cos^4\text{x}}{4}-\frac{\cos^6\text{x}}{6}\Bigg\}+\text{C}$
$=\frac{\cos^6\text{x}}{6}-\frac{\cos^4\text{x}}{4}+\text{C}$
View full question & answer→Question 1763 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{2\text{x}}{1+\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{2\text{x}}{1+\text{x}^4}\text{ dx}$
Let $\text{x}^2=\text{t}$ Then, $2\text{xdx}=\text{dt}$
When $\text{x}=10,\text{t}=0$ and $\text{x}=1,\text{t}=1$
$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{2\text{x}}{1+\text{x}^4}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{1}\frac{1}{1+\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=\big[\tan^{-1}\text{t}\big]^1_0$
$\Rightarrow\text{I}=\tan^{-1}1-\tan^{-1}0$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1773 Marks
Evaluate the following integrals:
$\int\cot^{\text{n}}\text{cosec}^2\text{x}\text{ dx},\text{ n}\neq-1$
Answer$\int\cot^{\text{n}}\text{cosec}^2\text{x}\text{ dx},\text{ n}\neq-1$
Let $\cot\text{x}=\text{t}$
$-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$
$\text{cosec}^2\text{x}\text{ dx}=-\text{dt}$
Now, $\int\cot^{\text{n}}\text{x }\text{cosec}^2\text{x}\text{ dx}$
$=-\int\text{t}^{\text{n}}\text{dt}$
$=\frac{-\text{t}^{\text{n}+1}}{\text{n}+1}+\text{C}$
$=-\frac{\cot^{\text{n}+1}}{\text{n}+1}+\text{C}$
View full question & answer→Question 1783 Marks
Evaluate the following integrals:
$\int^\limits{2}_{1}|\text{x}-3|\text{dx}$
Answer$\int^\limits{2}_{1}|\text{x}-3|\text{dx}$
We know that,
$|\text{x}+1|=\begin{cases}-(\text{x}+1),&1\leq\text{x}\leq3\$\text{x}+1),&\text{x}>3\end{cases}$
$\therefore\ \text{I}=\int^\limits{2}_{1}|\text{x}-3|\text{dx}$
$\Rightarrow\text{I}=\int^\limits{2}_{1}-(\text{x}-3)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{-\text{x}^2}{2}-3\text{x}\Big]^2_1$
$\Rightarrow\text{I}=-2-6+\frac{1}{2}+3$
$\Rightarrow\text{I}=\frac{3}{2}$
View full question & answer→Question 1793 Marks
Evalute the following integrals:
$\int\frac{\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}}{\text{e}^\text{x}+\text{x}^\text{e}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}}{\text{e}^\text{x}+\text{x}^\text{e}}\text{dx}$
Putting $e^x + x^e = t$
$\Rightarrow\text{e}^\text{x}+\text{ex}^{\text{e}-1}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}\big(\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}\big)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\big(\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}\big)\text{dx}=\frac{\text{dt}}{\text{e}}$
$\therefore\text{I}=\frac{1}{\text{e}}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{\text{e}}\text{ ln}|\text{t}|+\text{C}$
$=\frac{1}{\text{e}}\text{ ln}\Big|\text{e}^\text{x}+\text{x}^\text{e}\Big|+\text{C}\big[\because\text{t}=\text{e}^\text{x}+\text{x}^\text{e}\big]$
View full question & answer→Question 1803 Marks
Write a value of $\int\tan^6\text{x}\sec^2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\tan^6\text{x}\sec^2\text{x}\text{ dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\text{t}^6\text{dt}$
$=\frac{\text{t}^7}{7}+\text{C}$
Thus, $\text{I}=\frac{\tan^7\text{x}}{7}+\text{C}$
View full question & answer→Question 1813 Marks
Evaluate the definite integral in Exercise:
$\int^{1}_{0}\frac{\text{dx}}{\sqrt{1+\text{x}}-\sqrt{\text{x}}}$
Answer$\text{Let I}=\int^{1}\limits_{0}\frac{\text{dx}}{\sqrt{1+\text{x}}-\sqrt{\text{x}}}$
$\text{I}=\int^{1}\limits_{0}\frac{1}{\Big(\sqrt{1+\text{x}}-\sqrt{\text{x}}\Big)}\times\frac{\Big(\sqrt{1+\text{x}}+\sqrt{\text{x}}\Big)}{\Big(\sqrt{1+\text{x}}+\sqrt{\text{x}}\Big)}\text{dx}$
$=\int^{1}_{0}\frac{\sqrt{1+\text{x}}+\sqrt{\text{x}}}{1+\text{x}-\text{x}}\text{dx}$
$=\int^{1}_{0}{\sqrt{1+\text{x}}\text{dx}+\int^{1}\limits_{0}\sqrt{\text{x}}}\ \text{dx}$
$=\bigg[\frac{2}{3}\big(1+\text{x}\big)^{\frac{3}{2}}\bigg]^{1}_{0}+\bigg[\frac{2}{3}(\text{x)}^{\frac{3}{2}}\bigg]^{1}_{0}$
$=\frac{2}{3}\bigg[\big(2)^{\frac{3}{2}}-1\bigg]+\frac{2}{3}[1]$
$=\frac{2}{3}(2)^{\frac{3}{2}}$
$=\frac{2.2\sqrt{2}}{3}$
$=\frac{4\sqrt{2}}{3}$
View full question & answer→Question 1823 Marks
Evaluate the following integrals:
$\int\limits^{\text{a}}_0\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\text{a}}_0\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\text{ dx}\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\text{a}}_0\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{a}-\text{x}}+\sqrt{\text{x}}}\text{ dx}$
$\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$\text{I}=\int\limits^{\text{a}}_0\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\text{a}}_0\frac{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\text{ dx}$
$=\int\limits^{\text{a}}_0\text{dx}=\big[\text{x}\big]^{\text{a}}_0=\text{a}$
Hence, $\text{I}=\frac{\text{a}}{2}$
View full question & answer→Question 1833 Marks
Evaluate the definite integral in Exercise:
$\int\limits^{\frac{\pi}{4}}_0\frac{\sin\text{x}\cos\text{x}}{\cos^{4}\text{x}+\sin^{4}\text{x}}\text{dx}$
Answer$\text{Let I}=\int^{\frac{\pi}{4}}\limits_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^{4}\text{x}+\sin^{4}\text{x}}\text{dx}$
$\Rightarrow\text{ I}=\int^{\frac{\pi}{4}}\limits_{0}\frac{\frac{(\sin\text{x}\cos\text{x)}}{\cos^{4}\text{x}}}{\frac{\cos^{4}\text{x}+\sin^{4}\text{x}}{\cos^{4}\text{x}}}\text{dx}$
$\Rightarrow\text{ I}=\int^{\frac{\pi}{4}}\limits_{0}\frac{\tan\text{x}\sec^{2}\text{x}}{1+\tan^{4}\text{x}}\text{dx}$
$\text{Let}\ \tan^{2}\text{x}=\text{t}\ \Rightarrow2\tan\text{x}\text{dx}=\text{dt}$
when $\text{x}=0,\text{t}=0$ and when $\text{x}=\frac{\pi}{4},\text{t}=1$
$\therefore\ \text{I}=\frac{1}{2}\int^{1}\limits_{0}\frac{\text{dt}}{1+\text{t}^{2}}$
$=\frac{1}{2}\Big[\tan^{-1}\text{t}\Big]^{1}_{0}$
$=\frac{1}{2}\Big[\tan^{-1}-\tan^{-1}0\Big]$
$=\frac{1}{2}\Big[\frac{\pi}{4}\Big]$
$=\frac{\pi}{8}$
View full question & answer→Question 1843 Marks
Evaluate the following integrals:
$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(2\sin|\text{x}|+\cos|\text{x}|\big)\text{dx}$
Answer$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(2\sin|\text{x}|+\cos|\text{x}|\big)\text{dx}$
$=\int^{0}_{-\frac{\pi}{4}}\big(-2\sin\text{x}+\cos\text{x}\big)\text{dx}+\int_{0}^{\frac{\pi}{2}}\big(2\sin\text{x}+\cos\text{x}\big)\text{dx}$
$=\big[2\cos\text{x}+\sin\text{x}\big]^0_{-\frac{\pi}{4}}+\big[-2\cos\text{x}+\sin\text{x}\big]_0^{\frac{\pi}{2}}$
$=2+0-0+1+0+1+2-0$
$=6$
View full question & answer→Question 1853 Marks
Evaluate the following integrals:
$\int^\limits6_{-6}\big|\text{x}+2\big|\text{dx}$
Answer$\int^\limits6_{-6}\big|\text{x}+2\big|\text{dx}$
We know that,
$|\text{x}+2|=\begin{cases}-(\text{x}+2),&-6\leq\text{x}\leq-2\\\text{x}+2,&-2<\text{x}\leq6\end{cases}$
$\therefore\ \text{I}=\int^\limits6_{-6}\big|\text{x}+2\big|\text{dx}$
$\Rightarrow\text{I}=\int^\limits{-2}_{-6}\big(\text{x}+2\big)\text{dx}+\int^\limits6_{-2}\big(\text{x}+2\big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{-\text{x}^2}{2}-2\text{x}\Big]^{-2}_{-6}+\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^6_{-2}$
$\Rightarrow\text{I}=-2+4+18-12+18+12-2+4$
$\Rightarrow\text{I}=40$
View full question & answer→Question 1863 Marks
Evaluate the following intregals:
$\int\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
Answer Let $\text{I}=\int\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
$=\int\sqrt{\frac{(1-\text{x})(1-\text{x})}{(1+\text{x})(1-\text{x})}}\text{dx}$
$=\int\Big(\frac{1-\text{x}}{\sqrt{1-\text{x}^2}}\Big)\text{dx}$
$=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}-\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}$
Putting $1-\text{x}^2=\text{t}$
$\Rightarrow-2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\text{x dx}=-\frac{\text{dt}}{2}$
Then
$\text{I}=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}+\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\sin^{-1}(\text{x})+\frac{1}{2}\times2\sqrt{\text{t}}+\text{C}$
$=\sin^{-1}(\text{x})+\sqrt{1-\text{x}^2}+\text{C}$
View full question & answer→Question 1873 Marks
Integrate the function in Exercise:$\sqrt{\frac{1-\sqrt{\text{x}}}{1+\sqrt{\text{x}}}}$
Answer$\text{I}=\sqrt{\frac{1-\sqrt{\text{x}}}{1+\sqrt{\text{x}}}}\text{dx}$
$\text{Let }\ \text{x}=\cos^{2}\theta\Rightarrow\text{dx}=-2\sin\theta\cos\theta\ \text{d}\theta$
$\text{I}=\int\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}(-2\sin\theta\cos\theta)\text{d}\theta$
$=-\int\sqrt{\frac{2\sin^{2}\frac{\theta}{2}}{2\cos^{2}\frac{\theta}{2}}}\sin2\theta\ \text{d}\theta$
$=-\int\tan\frac{\theta}{2}.\sin\theta\cos\theta\ \text{d}\theta$
$=-2\int\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\bigg(2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\bigg)\cos\theta\ \text{d}\theta$
$=-4\int\sin^{2}\frac{\theta}{2}\cos\theta\ \text{d}\theta$
$=-4\int\sin^{2}\frac{\theta}{2}.\bigg(2\cos^{2}\frac{\theta}{2}-1\bigg) \text{d}\theta$
$=-4\int\bigg(2\sin^{2}\frac{\theta}{2}\cos^{2}\frac{\theta}{2}-\sin^{2}\frac{\theta}{2}\bigg) \text{d}\theta$
$=-8\int\sin^{2}\frac{\theta}{2}.\cos^{2}\frac{\theta}{2}\text{d}\theta+4\int\sin^{2}\frac{\theta}{2}\text{d}\theta$
$=-2\int\sin^{2}\theta\text{d}\theta+4\int\sin^{2}\frac{\theta}{2}\text{d}\theta$
$=-2\int\bigg(\frac{1-\cos2\theta}{2}\bigg)\text{d}\theta+4\int\frac{1-\cos\theta}{2}\text{d}\theta$
$=-2\bigg[\frac{\theta}{2}-\frac{\sin2\theta}{4}\bigg]+4\bigg[\frac{\theta}{2}-\frac{\sin\theta}{2}\bigg]+\text{C}$
$=-\theta+\frac{\sin2\theta}{2}+2\theta-2\sin\theta+\text{C}$
$=\theta+\frac{\sin2\theta}{2}-2\sin\theta+\text{C}$
$=\theta+\frac{2\sin\theta\cos\theta}{2}-2\sin\theta+\text{C}$
$=\theta+\sqrt{1-\cos^{2}\theta}.\cos\theta-2\sqrt{1-\cos^{2}\theta}+\text{C}$
$=\cos^{-1}\sqrt{\text{x}}+\sqrt{1-\text{x}}.\sqrt{\text{x}}-2\sqrt{1-\text{x}}+\text{C}$
$=-2\sqrt{1-\text{x}}+\cos^{-1}\sqrt{\text{x}}+\sqrt{\text{x}(1-\text{x)}}+\text{C}$
$=-2\sqrt{1-\text{x}}+\cos^{-1}\sqrt{\text{x}}+\sqrt{\text{x}-\text{x}^{2}}+\text{C}$
View full question & answer→Question 1883 Marks
Write a value of $\int\frac{1+\log\text{x}}{3+\text{x}\log\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1+\log\text{x}}{3+\text{x}\log\text{x}}\text{ dx}$
Let $3+\text{x}\log\text{x}=\text{t}$
$\Big(\log\text{x}+\text{x}\cdot\frac{1}{\text{x}}\Big)\text{dx}=\text{at}$
$(1+\log\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log\text{t}+\text{C}$
$\text{I}=\log(3+\text{x}\log\text{x})+\text{C}$
View full question & answer→Question 1893 Marks
$\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{5}{2}}}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{5}{2}}}$
$\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^2\sqrt{1-\cos\text{x}}}\text{dx}$
$\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\frac{1}{(1-\cos\text{x})^2}\text{dx}=\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\frac{1}{(\sin^2\text{x})}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\cos\text{ec}^2\text{x dx}=\big[-\cot\text{x}\big]^{\frac{\pi}{2}}_{\frac{\pi}{3}}$
$=-\Big[\cot\frac{\pi}{2}-\cot\frac{\pi}{2}\Big]$ $=-\Big[0-\frac{1}{\sqrt{3}}\Big]=+\frac{1}{\sqrt{3}}$
View full question & answer→Question 1903 Marks
Evaluate the following integrals:
$\int\text{x}^2\cos\text{x dx}$
Answer$\int\text{x}^2\cos\text{x dx}$
Taking $x^2$ as the first function and $\cos x$ as the second function.
$=\text{x}^2\int\cos\text{x dx}-\int\big(\frac{\text{d}}{\text{dx}}\text{x}^2\int\cos\text{x dx}\big)\text{dx}$
$=\text{x}^2\sin\text{x}-\int2\text{x}\sin\text{x dx}$
$=\text{x}^2\sin\text{x}-2\big[\text{x}\int\sin\text{x}-\int\big\{\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{x dx}\big\}\text{dx}\big]$
$=\text{x}^2\sin\text{x}-2[-\text{x}\cos\text{x}+\int\cos\text{x dx}]$
$=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x+C}$
View full question & answer→Question 1913 Marks
Evaluate the following integrals:
$\int^\limits9_0\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}\sin\text{x},&0\leq\text{x}\leq\frac{\pi}{2}\\1,&\frac{\pi}{2}\leq\text{x}\leq3\\\text{e}^{\text{x}-3},&3\leq\text{x}\leq9\end{cases}$
AnswerWe have,
$\int^\limits9_0\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}\sin\text{x},&0\leq\text{x}\leq\frac{\pi}{2}\\1,&\frac{\pi}{2}\leq\text{x}\leq3\\\text{e}^{\text{x}-3},&3\leq\text{x}\leq9\end{cases}$
$\text{I}=\int^\limits9_0\text{f(x)}\text{dx}$
$\Rightarrow\text{I}=\int^\limits{\frac{\pi}{2}}_1\text{f(x)}\text{dx}+\int^\limits3_\frac{\pi}{2}\text{f(x)}\text{dx}+\int^\limits9_3\text{e}^{\text{x}-3}\text{ dx}$ [Additive property]
$\Rightarrow\text{I}=\int^\limits{\frac{\pi}{2}}_1\sin\text{x dx}+\int^\limits3_\frac{\pi}{2}\text{1 }\text{dx}+\int^\limits9_3\text{e}^{\text{x}-3}\text{ dx}$
$\Rightarrow\text{I}=\big[-\cos\text{x}\big]^{\frac{\pi}{2}}_0+\big[\text{x}\big]^3_\frac{\pi}{2}+\text{e}^6-\text{e}^0$
$\Rightarrow\text{I}=0+1+3-\frac{\pi}{2}+\text{e}^6-\text{e}^0$
$\Rightarrow\text{I}=3-\frac{\pi}{2}+\text{e}^6$
View full question & answer→Question 1923 Marks
$\int \text{(2x} - 3)^{5} + \sqrt{3\text{x + 2}}\text{ dx}$
Answer$\int\big[(2\text{x}-3)^5+\sqrt{3\text{x}+2}\big]\text{dx}$
$=\int(2\text{x}-3)^5\text{dx}+\int{(3\text{x}+2)^{\frac{1}{2}}}\text{dx}$
$=\frac{(2\text{x}-3)^{5+1}}{2(5+1)}+\frac{(3\text{x}+2)^{\frac{1}{2}{+1}}}{3\Big(\frac{1}{2}+1\Big)}+\text{c}$
$=\frac{(2\text{x}-3)^6}{12}+\frac{2}{9}(3\text{x}+2)^{\frac{3}{2}}+\text{c}$
View full question & answer→Question 1933 Marks
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\sin5\text{x}\sin3\text{x}}\text{dx}$
Answer$\int\frac{\sin2\text{x}}{\sin5\text{x}\sin3\text{x}}\text{dx}$
$=\int\frac{\sin(5\text{x}-3\text{x})}{\sin5\text{x}\sin3\text{x}}\text{dx}$
$=\int\frac{5\text{x}\cos3\text{x}-\cos5\text{x}\sin3\text{x}}{\sin5\text{x}\sin3\text{x}}$
$=\int\frac{\sin5\text{x}\cos3\text{x}}{\sin5\text{x}\sin3\text{x}}-\frac{\cos5\text{x}\sin3\text{x}}{\sin5\text{x}\sin3\text{x}}\text{dx}$
$=\int\big[\cot3\text{x}-\cot5\text{x}\big]\text{dx}$
$=\int\cot3\text{x dx}-\int\cot5\text{x dx}$
$=\frac{1}{3}\text{ln}|\sin3\text{x}|-\frac{1}{5}\text{ln}|\sin5\text{x}|+\text{C}$
View full question & answer→Question 1943 Marks
Evalute the following integrals:
$\int\sqrt{\frac{1-\sin2\text{x}}{1+\sin2\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\frac{1-\sin2\text{x}}{1+\sin2\text{x}}}\text{dx}$ then,
$=\int\sqrt{\frac{1-\cos\Big(\frac{\pi}{2}-2\text{x}\Big)}{1+\cos\Big(\frac{\pi}{2}-2\text{x}\Big)}}\text{dx}$
$=\int\sqrt{\frac{2\sin^2\Big(\frac{\pi}{4}-\text{x}\Big)}{2\cos^2\Big(\frac{\pi}{4}-\text{x}\Big)}}\text{dx}$
$=\int\sqrt{\tan^2\Big(\frac{\pi}{4}-\text{x}\Big)}\text{dx}$
$=\int\tan\Big(\frac{\pi}{4}-\text{x}\Big)\text{dx}$
$=\log\Big|\cos\Big(\frac{\pi}{4}-\text{x}\Big)\Big|+\text{C}$
View full question & answer→Question 1953 Marks
Evaluate the following integrals:
$\int\text{x}^3\log\text{x dx}$
AnswerLet $\text{I}=\int\text{x}^3\log\text{x dx}$
Using integration by parts,
$\text{I}=\log\text{x}\int\text{x}^3\text{dx}-\int\Big(\frac{1}{\text{x}}\times\int\text{x}^3\text{dx}\Big)\text{dx+C}$
$=\frac{\text{x}^4}{4}\log\text{x}-\int\frac{\text{x}^4}{4\text{x}}\text{dx+C}$
$=\frac{\text{x}^4}{4}\log\text{x}-\frac{1}{4}\int\text{x}^3\text{dx+C}$
$=\frac{\text{x}^4}{4}\log\text{x}-\frac{1}{4}\int\frac{\text{x}^4}{4}\text{dx+C}$
$\text{I}=\frac{\text{x}^4}{4}\log\text{x}-\frac{1}{16}\text{x}^4+\text{C}$
View full question & answer→Question 1963 Marks
Write a value of $\int\text{e}^{3\log\text{x}}\text{x}^{4}\text{ dx}.$
AnswerLet $\text{I}=\int\text{e}^{3\log\text{x}}\text{x}^{4}\text{ dx}$
$=\int\text{e}^{\log\text{x}^3}\cdot\text{x}^{4}\text{ dx}$
$=\int\text{x}^3\cdot\text{x}^4\text{ dx}$ $\big[\because\text{e}^{\log\text{x}}=\text{x}\big]$
$=\int\text{x}^{7}\text{ dx}$
$\therefore\ \text{I}=\frac{\text{x}^{8}}{8}+\text{C}$
View full question & answer→Question 1973 Marks
Evaluate the following integrals:
$\int\tan^{-1}\Big(\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big)\text{dx}$
Answer$\int\tan^{-1}\Big[\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big]\text{dx}$
$=\int\tan^{-1}\Big[\frac{2\sin\text{x}\cos\text{x}}{2\cos^2\text{x}}\Big]\text{dx}$
$=\int\tan^{-1}\Big[\frac{\sin\text{x}}{\cos\text{x}}\Big]\text{dx}$
$=\int\tan^{-1}(\tan\text{x})\text{dx}$
$=\int\text{x dx}$
$=\frac{\text{x}^2}{2}+\text{C}$
$\therefore\ \int\tan^{-1}\Big[\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big]\text{dx}=\frac{\text{x}^2}{2}+\text{C}$
View full question & answer→Question 1983 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{4}}_{\frac{-\pi}{4}}|\sin\text{x}|\text{dx}$
Answer$\int^\limits{\frac{\pi}{4}}_{\frac{-\pi}{4}}|\sin\text{x}|\text{dx}$
We know that,
$|\sin\text{x}|=\begin{cases}-\sin\text{x},&-\frac{\pi}{4}\leq\text{x}\leq0\\\sin\text{x},&0<\text{x}\leq\frac{\pi}{4}\end{cases}$
$\therefore\ \text{I}=\int^\limits{\frac{\pi}{4}}_{\frac{-\pi}{4}}|\sin\text{x}|\text{dx}$
$\Rightarrow\text{I}=\int^\limits0_{-\frac{\pi}{4}}-\sin\text{x dx}+\int\limits^{\frac{\pi}{4}}_0\sin\text{x dx}$
$\Rightarrow\text{I}=\big[\cos\text{x}\big]^0_{\frac{-\pi}{4}}-\big[\cos\text{x}\big]^{\frac{-\pi}{4}}_0$
$\Rightarrow\text{I}=1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+1$
$\Rightarrow\text{I}=2-\frac{2}{\sqrt{2}}$
$\Rightarrow\text{I}=2-\sqrt{2}$
View full question & answer→Question 1993 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})\text{dx}$
Answer$\int\limits^{\frac{\pi}{2}}_0\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})\text{dx}$
$=-\int\limits^{\frac{\pi}{2}}_0\text{e}^{\text{x}}\Big[\cos\text{x}+(-\sin\text{x})\Big]\text{dx}$
$=-\big[\text{e}^{\text{x}}\cos\text{x}\big]^{\frac{\pi}{2}}_0$ $\Big\{\int\text{e}^{\text{x}}\big[\text{f(x)}+\text{f}'(\text{x})\big]\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}\Big\}$
$=-\Big(\text{e}^{\frac{\pi}{2}}\cos\frac{\pi}{2}-\text{e}^0\cos0\Big)$
$=-\Big(\text{e}^{\frac{\pi}{2}}\times0-1\times1\Big)$
$=-(0-1)$
$=1$
View full question & answer→Question 2003 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^{\frac{-1}{3}}+\sqrt{\text{x}}+2}{\sqrt[3]{\text{x}}}\text{dx}$
Answer$\int\Bigg(\int\frac{\text{x}^{-\frac{1}{3}}+\sqrt{\text{x}}+2}{\text{x}^{\frac{1}{3}}}\Bigg)\text{dx}$
$=\int\Bigg(\frac{\text{x}^{-\frac{1}{3}}}{\text{x}^{\frac{1}{3}}}+\frac{\text{x}^{\frac{1}{2}}}{\text{x}^{\frac{1}{3}}}+\frac{2}{\text{x}^{\frac{1}{3}}}\Bigg)\text{dx}$
$=\int\Big(\text{x}^{-\frac{2}{3}}+\text{x}^{\frac{1}{6}}+2\text{x}^{-\frac{1}{3}}\Big)\text{dx}$
$=\Bigg[\frac{\text{x}^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}+\frac{\text{x}^{\frac{1}{6}+1}}{\frac{1}{6}+1}+2\frac{\text{x}^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}\Bigg]$
$=\Bigg[\frac{\text{x}^{\frac{1}{3}}}{\frac{1}{3}}+\frac{\text{x}^{\frac{7}{6}}}{\frac{7}{6}}+3\text{x}^{\frac{2}{3}}\Bigg]+\text{C}$
$=3\text{x}^{\frac{1}{3}}+\frac{6}{7}\text{x}^{\frac{7}{6}}+3\text{x}^{\frac{2}{3}}+\text{C}$
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