Question 511 Mark
Find an anti derivative (or integral) of the following function by the method of inspection.
$\text{e}^{2\text{x}}$
Answer$\because \ \ \ \ \ \ $$\frac{\text{d}}{\text{dx}}\text{e}^{2\text{x}}=\text{e}^ {2\text{x}}\text{ }\frac{\text{d}}{\text{dx}}(2\text{x})=2\text{e}^{2\text{x}}$ $\ \ \ \ \ \ \ \Rightarrow \ \ \ \ \frac{\text{1}}{\text{2}}\frac{\text{d}}{\text{dx}}\text{e}^{2\text{x}}=\text{e}^{2\text{x}}$ $ \ \ \ \ \ \ \Rightarrow \ \ \ \ \frac{\text{d}}{\text{dx}}\bigg(\text{ }\frac{\text{1}}{\text{2}}\text{e}^{2\text{x}} \bigg)=\text{e}^{2\text{x}}$$\because \ \ \ \ \ \ $An anti-derivative of $\text{e}^{2\text{x}} \text { is } \frac{1}{2}\text{e}^{2\text{x}}. $
View full question & answer→Question 521 Mark
Integrate the functions in Exercises:
$\frac{1}{1+\cot\text{x}}$
Answer$\text{Let I}=\int\frac{1}{1+\cot\text{x}}\text{ dx} =\int\frac{1}{1+\frac{\cos\text{x}}{\sin\text{x}}}\text{ dx} $
$=\int\frac{1}{\bigg(\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}}\bigg)}\text{ dx} =\int\frac{\sin\text{x}}{\sin\text{x}+\cot\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{2\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}=\frac{1}{2}\int\frac{\sin\text{x}+\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx} $
Adding and subtracting $\cos\text{x}$ in the numerator,
$=\frac{1}{2}\int\frac{\sin\text{x}+\cos\text{x}-\cos\text{x}+\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{(\sin\text{x}+\cos\text{x})-(\cos\text{x}-\sin\text{x})}{\sin\text{x}+\cos\text{x}}\text{ dx} $
$=\frac{1}{2}\int\bigg(\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}+\cos\text{x}}-\frac{\cos\text{x}-\sin\text{x}}{\sin \text{x}+\cos\text{x}}\bigg)\text{ dx} $
$=\frac{1}{2}\int\bigg(1-\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\bigg)\text{ dx} $
$=\frac{1}{2}\bigg[\int1\text{ dx}-\int\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}\bigg]=\frac{1} {2}[\text{x}-\text{I}_{1}]$$\ \ \ \ \ \text{where }\text{I}_{1}=\int\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx} \ \ \ ...\text{(i)}$
Putting $\sin\text{ x + cos x = t}\ \ \ \Rightarrow\ \ \ \cos\text{x}-\sin\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow\ \ \ \ (\cos\text{x}-\sin\text{x})\text{ dx = dt} $
$ \therefore\ \ \ \ \ $$\text{I}_1=\int\frac{\text{dt}}{\text{t}}=\log\begin{vmatrix}\text{t}\end{vmatrix}=\log\begin{vmatrix}\sin\text{x}+\cos\text{x}\end{vmatrix} $
Putting this value in eq. (i), we get required integral,
$=\frac{1}{2}[\text{x}-\log\begin{vmatrix}\sin\text{x + cosx}\end{vmatrix}]+\text{c} $
View full question & answer→Question 531 Mark
Find the following integrals in Exercises:
$\int(\text{ax}^2+\text{bx}+\text{c})\text { dx}$
Answer$\int(\text{ax}^2+\text{bx}+\text{c)}\text{ dx}$
$=\int\text{ax}^2 \text{ dx}+\int\text{bx}\text{ dx}+\int\text{c}\text{ dx}$
$=\text{a}\int\text{x}^2\text{ dx}+\text{b}\int\text{x}+\text{dx}+c\int1\text{ dx}$
$=\text{a}\frac{\text{x}^3}{3}+\text{b}\frac{\text{x}^2}{2}+\text{cx}+\text{c}_{1}$
where $\text{c}_{1}$ is the constant of integration.
View full question & answer→Question 541 Mark
Integrate the functions in Exercises:
$\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}$
Answer$\text{Let I}=\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{ dx}=\int\frac{\sec^2\text{x}}{(1 - \tan\text{x})^2}\text{ dx}$
$=-\int\frac{-\sec^2\text{x}}{(1 - \tan\text{x})^2}\text{ dx}\ \ \ \ \ \ \ \ \ \ ...\text{(i)} $
Putting $1-\tan\text{x = t}\ \ \Rightarrow\ \ \ \ -\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow\ \ \ \ -\sec^2\text{x}\text{ dx}=\text{dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=-\int\frac{\text{dt}}{\text{t}^2}=-\int\text{t}^{-2}\text{ dt}=\frac{-\text{t}^{-1}}{-1}+\text{c}=\frac{1}{\text{t}}+\text{c}$
$=\frac{1}{1 - \tan \text{x}}+\text{c} $
View full question & answer→Question 551 Mark
Find an anti derivative (or integral) of the following function by the method of inspection.
$\sin2\text{x}-4\text{ e}^{3\text{x}}$
Answer$\because \ \ \ \ \ \ \ \text{ }\frac{\text{d}}{\text{dx}}(\cos2\text{x})=-2\sin2\text{x}$ $\Rightarrow\frac{1}{-2}\text{ }\frac{\text{d}} {\text{dx}}(\cos2\text{x})=\sin2\text{x}$ $\Rightarrow\text{ }\frac{\text{d}}{\text{dx}}\bigg(\frac{-1}{2}\cos2\text{x} \bigg)=\sin2\text{x} \ \ \ ....\text{(i)}$Again$ \ \ \ \ \ \ \ \ \ \text{ }\frac{\text{d}}{\text{dx}}\text{e}^{3\text{x}}=3\text{e}^{3\text{x}}$
$\ \ \ \ \Rightarrow\ \ \ \ \ \text{ }\frac{\text{d}}{\text{dx}}\bigg (\text{ }\frac{\text{1}}{\text{3}}\text{e}^{3\text{x}}\bigg)=\text{e}^{3\text{x}}$
$\frac{\text{d}}{\text{dx}}\bigg( \text{ }\frac{\text{-4}}{\text{3}}\text{e}^{3\text{x}}\bigg)=-4\text{e}^{3\text{x } } \ \ \ \ \ \ \ \text{[Multiplying both sides by -4 ]} \ \ \ \ \ \ \ ...\text{(ii)}$ Adding eq. (i) and (ii), we get
$\Rightarrow \ \ \ \ \text{ }\frac{\text{d}}{\text{dx}}\bigg(\text{ }\frac{\text{-1}}{\text{2}}\cos2\text{x}\bigg)+\text{ }\frac {\text{d}}{\text{dx}}\bigg(\text{ }\frac{\text{-4}}{\text{3}}\text{e}^{3\text{x}}\bigg)= \sin2\text{x}+\bigg(-4\text{e}^ {3\text{x}}\bigg)$
$\Rightarrow\text{ }\frac{\text{d}}{\text{dx}}\bigg(\text{ }\frac{\text{-1}}{\text{2}}\cos2\text{x}-\text{ }\text{ }\frac {\text{4}}{\text{3}}\text{e}^{3\text{x}}\bigg)= \sin2\text{x}-4\text{e}^{3\text{x}}$ $\therefore\ \ \ \ \ \ \ \ $An anti-derivative of $\sin2\text{x}-4\text{e}^{3\text{x}} \text{ is }\frac{-1}{2} \cos2\text{x}-\frac{4}{3} \text{e}^{3\text{x}}.$
View full question & answer→Question 561 Mark
Integrate the functions in Exercises:
$\text{x}\sqrt{1+2\text{x}^2}$
Answer $\text{Let}\text{ I}=\int\text{x}\sqrt{1+2\text{x}^2}\text{ dx} =\frac{1}{4}\int\sqrt{1+2\text{x}^2}(4\text{x dx}) \ \ \ \ \ \ ....\text{(i)} $
Putting $ 1+2\text{x}^2=\text{t}\ \ \ \ \Rightarrow \ \ \ \ \ 4\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \ \ \Rightarrow \ \ \ \ 4\text{x}\text{ dx}=\text{dt} $
$\therefore\ \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{4}\int\sqrt{\text{t}}\text{ dt}=\frac{1}{4}\int\text{t}^{\frac{1}{2}}\text{ dt}$
$=\frac{1}{4}\frac{\text{t}^{^3/_2}}{{^3/_2}}+\text{c}=\frac{1}{4}\dot\ \frac{2}{3}\text{t}^{^3/_2 } +\text{c} $
$=\frac{1}{6}\big(1 + 2\text{x}^2\big)^{^3/_2}+\text{c} $
View full question & answer→Question 571 Mark
Evaluate $\int\sec^2(7-4\text{x})\text{dx}$
Answer$\int\sec^2(7-4\text{x})\text{dx}$
$=\frac{\tan(7-4\text{x})}{-4}+\text{C}$ $(\because\sec^2\text{x}=\tan\text{x}+\text{C})$
View full question & answer→Question 581 Mark
Find an anti derivative (or integral) of the following function by the method of inspection.
$\sin2\text{x}$
Answer$\because\text{ }\frac{\text{d}}{\text{dx}}(\cos2\text{x})=-2\sin2\text{x} $$\Rightarrow \frac{1}{-2}\frac{\text{d}}{\text{dx}}(\cos2\text{x})=\sin2{\text{x}} $
$\Rightarrow \text{ }\frac{\text{d}}{\text{dx}}\bigg( \frac{-1}{2}\cos2\text{x}\bigg)=\sin2\text{x}$ $\therefore \ \ \ \ \ \ \ $An anti-derivative of $\sin2\text{x is}\frac{-1}{2}\cos2\text{x}.$
View full question & answer→Question 591 Mark
Find the following integrals in Exercises: $\int\sec\text{x}(\sec\text{x}+\tan\text{x})\text{ dx}$
Answer$\int\sec\text{x}(\sec\text{x}+\tan\text{x})\text{ dx}$
$=\int(\sec^2\text{x}+\sec\text{x }\tan\text{x})\text{ dx} $
$=\int\sec^2\text{x}\text{ dx}+\int\sec\text{x}\tan\text{x}\text{ dx} $
$=\tan\text{x}+\sec\text{x}+\text{c}$
View full question & answer→Question 601 Mark
Integrate the functions in Exercises:
$\frac{1}{\text{x}(\log\text{x})^\text{m}},\text{x}>0$
Answer$\text{Let I}=\int\frac{1}{\text {x}(\log \text{x})^\text{m}} \text{ dx}=\int\frac{\frac{1}{\text{x}}\text{dx}}{(\log \text{x)}^\text{m}} \ \ \ \ ....\text{(i)} $
Putting $\log\text{x}=\text{t}\ \ \ \Rightarrow \ \ \ \ \frac {1}{\text{x}}=\frac{\text{dt}}{\text {dx}}\ \ \ \ \Rightarrow \ \ \ \ \frac{\text{dx}}{\text{x}}=\text{dt}$
$\therefore \ \ \ \ \ $ From eq. (i), $\text{I}=\int\frac{\text{dt}}{\text{t}^{\text{m}}}=\int\text{t}^\text{-m}\text{ dt}=\frac{\text{t}^{-\text{m}+1}}{-\text {m + 1}}+\text{c}$
$=\frac{(\log\text{x})^{1-\text {m}}}{1-\text{m}}+\text{c} $
View full question & answer→Question 611 Mark
Integrate the functions in Exercises:
$\frac{1}{\text{x}+\text{x}\log\text{x}}$
Answer Putting $1 + \log \text{x}=\text{t}\ \ \ \ \ \Rightarrow\ \ \ \ \ \ \frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}\ \ \ \ \ \Rightarrow\ \ \ \ \ \frac{\text{dx}}{\text{x}}=\text{dt} $
$\therefore\ \ \ \ \ \int \frac{1}{\text{x + x log x }}\text{ dx}$
$\int \frac{1}{\text{1 + log x }}\frac{\text{dx}}{\text{x}}==\int\frac{1}{\text{t}}\text{ dt}=\log\mid\text{t}\mid+\text{c}$
$=\log\mid\text{1+ log x}\mid+\text{ c}$
View full question & answer→Question 621 Mark
Integrate the functions in Exercises:
$\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}$
Answer$\text{Let I}=\int\frac{\sqrt{\tan\text{x}}}{\sin\text{x}\cos\text{x}}\text{ dx}=\int\frac{\sqrt{\tan\text{x}}}{\frac{\sin\text{x}}{\cos\text{x}}\cos\text{x}.\cos\text{x}}\text{ dx} $ $=\int\frac{\sqrt{\tan\text{x}}}{{\tan\text{x}\cos^2\text{x}}}\text{ dx} =\int\frac{\sec^2\text{x}}{\sqrt{\tan\text{x}}}\text{ dx} \ \ \ \ ...\text{(i)}$ Putting $\tan\text{x}=\text{t}\ \ \ \ \Rightarrow \ \ \ \ \sec^2\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow \ \ \ \ \sec^2\text{x}\text{ dx = dt}$ $\therefore \ \ \ \ \ $From eq. (i), $\text{I}=\int\frac{\text{dt}}{\sqrt{\text{t}}}=\int\text{t}^{\frac{-1}{2}}\text{dt}=\frac{\text{t}^{^1/_2}}{^1/_2}+\text{c}$$=2\sqrt{\text{t}}+\text{c}=2\sqrt{\tan\text{x}}+\text{c}$
View full question & answer→Question 631 Mark
Integrate the functions in Exercises:
$\frac{\text{x}^3\sin\big(\tan^{-1}\text{x}^4\big)}{1+\text{x}^8}$
Answer$\text{Let I}=\int\frac{\text{x}^3\sin{(\tan^{-1}\text{x}^4)}}{1 +\text{x}^8}\text{ dx} $
$=\frac{1}{4}\int\sin(\tan^{-1}\text{x}^4)\dot\ \frac{4\text{x}^3}{1 + \text{x}^8}\text{ dx} \ \ \ \ ...\text{(i)} $
Putting $\tan^{-1} \text{x}^4=\text{t}\ \ \ $$\Rightarrow\ \ \ \frac{1}{1+(\text{x}^4)^2}\frac{\text{d}}{\text{dx}}\text{x}^4=\frac{\text{dt}}{\text{dx}}\ \ \ $$\Rightarrow \ \ \ \frac{4\text{x}^3}{1+\text{x}^8}\text{ dx = dt} $
$\therefore \ \ \ $From eq. (i), $\text{ I}=\frac{1}{4}\int\sin\text{t}\text{ dt}=\frac{-1}{4}\text{cos t}+\text{c}$
$=\frac{-1}{4}\cos(\tan^{-1}\text{x}^4)+\text{c}$
View full question & answer→Question 641 Mark
Find the following integrals in Exercises: $\int(1-\text{x})\sqrt{\text{x}}\text{ dx}$
Answer$\int(1-\text{x})\sqrt{\text{x }}\text{ dx} =\int\big(\sqrt{\text{x}}-\text{x}\sqrt{\text{x}}\big)\text{ dx}$
$=\int\bigg(\text{x}^\frac{1}{2}-\text{x}^\frac{3}{2}\bigg)\text{ dx}=\int\text{x}^\frac{1}{2}\text{ dx}-\int\text{x}^\frac{3}{2}\text{ dx} $
$=\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{\text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\text{c} =\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}+ \frac{\text{x}^{\frac{5}{2}}}{\frac{5}{2}}$
$=\frac{2}{3}\text{x}^\frac{3}{2}-\frac{2}{5}\text{x}^{\frac{5}{2}}+\text{c}$
View full question & answer→Question 651 Mark
Integrate the functions in Exercises:
$\sin(\text{ax+b})\cos(\text{ax+b})$
Answer$\int\sin(\text{ax+b})\cos(\text{ax+b})\text{ dx}$
$=\frac{1}{2}\int2\sin(\text{ax+b})\cos(\text{ax+b})\text { dx} $
$=\frac{1}{2}\int\sin2(\text{ax+b})\text{ dx} $
$=\frac{1}{2}\int\sin(2\text{ax+2b})\text{ dx} $
$=\frac{1}{2}\frac{[-\cos(2\text{ax}+2\text{b})]}{2\text{a}\rightarrow\text{Coeff. of x}}+\text{c} $
$=\frac{-1}{4\text{a}}\cos2(\text{ax+b})+\text{c} $
View full question & answer→Question 661 Mark
Find the following integrals in Exercises. $\int\frac{\text{x}^3+3\text{x}+4}{\sqrt{\text{x}}}\text{ dx}$
Answer$\int\frac{\text{x}^3+3\text{x}+4}{\sqrt{\text{x}}}\text{ dx}$
$=\int\Bigg(\frac{\text{x}^3}{\text{x}^{\frac{1}{2}}}+\frac{3\text{x}}{\text{x}^{\frac{1}{2}}}+\frac{4}{\text{x}^{\frac{1}{2}}}\Bigg)\text{ dx} $
$=\int\bigg(\text{x}^{3-\frac{1}{2}}+3\text{x}^{1-\frac {1}{2}}+4\text{x}^{\frac{-1}{2}}\bigg)\text{ dx}$
$=\int\bigg(\text{x}^{\frac{5}{2}}+3\text{x}^{\frac{1}{2}}+4\text{x}^{\frac{-1}{2}}\bigg)\text{ dx}$
$=\int\text{x}^\frac{5}{2}\text{ dx}+3\int\text{x}^\frac{1}{2}\text{ dx}+4\int\text{x}^\frac{-1}{2}\text{ dx} $
$=\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}+3\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+4 \frac{\text{x}^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+\text{c} $
$=\frac{\text{x}^{\frac{7}{2}}}{\frac{7}{2}}+3 \frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}+4 \frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\text{c} $
$=\frac{2}{7}\text{x}^{\frac{7}{2}}+2\text{x}^{\frac{3}{2}}+8\text{x}^{\frac{1}{2}}+\text{c}$
View full question & answer→Question 671 Mark
Find the following integrals in Exercises. $\int\bigg(\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}\bigg)^2$
Answer$\int\bigg(\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}\bigg)^2\text{ dx}$
$=\int \bigg\{\big(\sqrt{\text{x } } \big)^2+\bigg(\frac{1}{\sqrt{\text{x}}}\bigg)^2-2\sqrt{\text{x}}\frac{1}{\sqrt{\text{x}}}\bigg\}\text{ dx}$
$ =\int\bigg(\text{x}+\frac{1}{\text{x}}-2\bigg)\text{ dx}$
$=\int\text{x}\text{ dx}+\int\frac{1}{\text{x}}\text{ dx}-\int 2\text{ dx}=\frac{\text{x}^2}{2}+\log\mid\text{x}\mid-2\text{x}+\text{c}$
View full question & answer→Question 681 Mark
Integrate the functions in Exercises:
$\frac{\text{x}}{\text{e}^{\text{x}^2}}$
Answer$\text{Let I}=\int\frac{\text{x}}{\text{e}^{\text{x}^{2}}} \text{ dx}=\frac{1}{2}\int\frac{\text{2x}}{\text{e}^{\text{x}^{2}}}\text{ dx}\ \ \ \ ....\text{(i)} $
$\text{Putting}\text{ x}^2=\text{t}\ \ \ \ \Rightarrow \ \ \ \ 2\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow \ \ \ \ \ \text{2x}{\text{ dx}}=\text{ dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{e}^{\text{t}}}=\frac{1}{2}\int\text{e}^{\text{-t}}\text{ dt}$
$=\frac{1}{2}\cdot\frac{\text{e}^{-\text{t}}}{-1\rightarrow \text{Coeff. of t}}+\text{c}$
$=\frac{-1}{2(\text{e}^{\text{t}})}+\text{c}=\frac{-1}{2(\text{e}^{\text{x}^{2}})}+\text{c} $
View full question & answer→Question 691 Mark
Integrate the functions in Exercises:
$\sqrt{\sin2\text{x}}\cos2\text{x}$
Answer$\text{Let I}=\int\sqrt{\sin2\text{x }}\cos 2\text{x dx}=\frac{1}{2}\int\sqrt{\sin2\text{x }}(2\cos\text{2x dx}) \ \ \ \ \ \ \ ...\text{(i)} $
Putting $\sin2\text{x}=\text{t}\ \ \ \Rightarrow\ \ \ \ \cos2 \text{x}\frac{\text{d}}{\text{dx}}(2\text{x})=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow\ \ \ 2\cos 2\text{x dx = dt} $
$\therefore\ \ \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{2}\int\sqrt{\text{t}}\text{ dt}=\frac{1}{2}\int\text{t}^{\frac{1}{2}}\text{ dt}=\frac{1}{2}\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{c}$
$=\frac{1}{2}\dot\ \frac{\text{t}^{^3/_2}}{^3/_2}+\text{c}=\frac{1}{3}(\sin 2\text{x})^{\frac{3}{2}}+\text{c} $
View full question & answer→Question 701 Mark
Find the following integrals in Exercises:
$\int\text{x}^2 \bigg(1-\frac{1}{\text{x}^2}\bigg)\text{dx}$
Answer$\int\text{x}^2 \bigg(1-\frac{1}{\text{x}^2}\bigg)\text{dx}=\int\bigg(\text{x}^2-\frac{\text{x}^2}{\text{x}^2}\bigg)\text {dx}$
$=\int(\text{x}^2-1)\text{dx}=\int\text{x}^2\text{ dx}-\int1\text{ dx}$
$\frac{\text{x}^3}{3}-\text{x+c}\ \ \ \ \ \ \ \ \ \ \ \bigg[\because\int\text{x}^\text{n} \text{ dx}=\frac{\text {x}^{\text{n}+1}}{\text{n+1}}\text{ if}\text{ n}\neq-1\bigg]$
View full question & answer→Question 711 Mark
Integrate the functions in Exercises:
$\frac{\text{e}^{\tan^{-1}\text {x}}}{1+\text{x}^2}$
Answer$\text{Let I}=\int\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx} \ \ \ \ \ \ \ ...\text{(i)} $
$\text{Putting }\tan^{-1}\text{x}=\text{t}\ \ \ \Rightarrow \ \ \ \frac{1}{1+\text{x}^2}=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow \ \ \ \ \frac{\text{dx}}{1+\text{x}^2}=\text{ dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=\int\text{e}^{\text{t}}\text{ dt}=\text{e}^{\text{t}}+\text{c}=\text{e}^{{\tan}^{-1}\text{x}}+\text{c}$
View full question & answer→Question 721 Mark
Integrate the functions in Exercises:
$\frac{(1+\log\text{x})^2}{\text{x}}$
Answer$\text{Let I}=\int\frac{(1+\log\text{x})^2}{\text{x}}\text{ dx} \ \ \ \ ... \text{(i)}$
Putting $1+\log\text{x}=\text{t} \ \ \ \ \Rightarrow \ \ \ \frac{1}{\text{x}}=\frac{\text{dt}{}}{\text{dx}}\ \ \Rightarrow \ \ \ \frac{\text{dx}{}}{\text{x}}=\text{dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=\int\text{t}^2\text{ dt}=\frac{\text{t}^3}{3}+\text{c}= \frac{1}{3}(1+\log\text{x})^3+\text{c}$
View full question & answer→Question 731 Mark
By Using properties of definite integral, evaluate the following integral in Exercise:
$\int^{2\pi}_{0}\cos^{5}\text{x}\ \text{dx}$
Answer$\int^{2\pi}\limits_{0}\cos^{5}\ \text{x}\ \text{dx}=2\int^{\pi}\limits_{0}\cos^{5}\text{x}\ \bigg[\because\int^{2\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx}=2\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx},\text{if}\ \text{f}(2\text{a}-\text{x})=\text{f}\text{(x)}\bigg]$
$\text{Here}\ \text{f}\text{(x)}=\cos^{5}\text{x}\ \therefore\ \text{f}(2\pi-\text{x}=\cos^{5}(2\pi-\text{x})=\cos^{5}\text{x}$
$\Rightarrow\ \ \text{f}\ \text{(x)}=2(0)=0$
View full question & answer→Question 741 Mark
Integrate the functions in Exercises:
$\frac{\sin\text{x}}{(1+\cos\text{x})^2}$
Answer$ \text{Let I}=\int\frac{\sin\text{x}}{(1 +\cos\text{x})^2}\text{ dx}=-\int\frac{-\sin\text{x}}{(1+\cos\text{x})^2}\text{ dx} \ \ \ \ \ \ \ \ ...\text{(i)} $
Putting $1+\cos\text{x}=\text{t}\ \ \ \Rightarrow\ \ \ -\sin\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow\ \ \ -\sin\text{x dx}=\text{dt} $
$\therefore \ \ \ \ $From eq. (i), $\text{I}=-\int\frac{\text{dt}}{\text{t}^2}=-\int\text{t}^{-2}\text{ dt}=\frac{-\text{t}^{-1}}{-1}+\text{c}$
$=\frac{1}{1+\cos\text{x}}+\text{c}$
View full question & answer→Question 751 Mark
Find the following integrals in Exercises:
$\int(4\text{ e}^{3\text{x}} + 1)\text{ dx}$
Answer$\int\bigg(4\text{e}^{3\text{x}}+1\bigg)\text{dx}$
$=\int4\text{e}^{3\text{x}}\text{ dx}+\int1 \text{ dx}=4\int{e}^{3\text {x}}\text{ dx}+\text{x}$
$=4 \frac{\text{e}^{3\text{x}}}{3}+\text{x}+\text{c} \ \ \ \ \ \ \ \ \ \ \bigg[\because \text{e}^ \text{ax}\text{ dx}=\frac{\text{e}^\text{ax}}{\text{x}}\text{and}\int1\text{ dx}=\text{x}\bigg]$
View full question & answer→Question 761 Mark
Integrate the functions in Exercises:
$(\text{x}^3-1)^{\frac{1}{3}}\text{ x}^5$
Answer$\text{Let I}=\int(\text{x}^3 - 1)^{\frac{1}{3}}\text{x}^5\text{ dx}$ $=\int(\text{x}^3 - 1\big)^{\frac{1}{3}}\text{x}^3\text{x}^2\text{ dx}=\frac{1}{3}\int(\text{x}^3 - 1)^{\frac{1}{3}}\text{ x}^3(3\text{x}^2\text{ dx})\ \ \ \ \ \ .....\text{(i)}$ Putting $\text{ x}^3-1=\text{t}\ \ \ \ \Rightarrow \ \ \ \ \ \text{ x}^3=\text{t}+1\ \ \ \Rightarrow \ \ \ \ \text{3x}^2=\frac{\text{dt}}{\text{dx}}\ \ \ \ \Rightarrow \ \ \ \ 3\text{x}^2\text{ dx}=\text{dt} $ From eq. (i), $\text{I}=\frac{1}{3}\int\text{t}^{\frac{1}{3}}(\text{t}+1)\text{ dt}$ $=\frac{1}{3}\int\bigg(\text{t}^{\frac{4}{3}}+\text{t}^{\frac{1}{3}}\bigg)\text{ dt}=\frac{1}{3}\bigg(\int\text{t}^{\frac{4} {3}}\text{ dt}+\int\text{t}^{\frac{1}{3}}\text{ dt}\bigg) $$=\frac{1}{3}\Bigg(\frac{\text{t}^{\frac{7}{3}}}{\frac{7}{3}}+\frac{\text{t}^{\frac{4}{3}}}{\frac{4}{3}}\Bigg)+\text{ c} =\frac{1}{3}\bigg(\frac{3}{7}\text{ t}^{\frac{7}{3}}+\frac{3}{4}\text{ t}^{\frac{4}{3}}\bigg)+\text{c}$
$=\frac{1}{7}\text{t}^{\frac{7}{3}}+\frac{1}{4}\text{t}^{\frac{4}{3}}+ \text{c} $ $=\frac{1}{7}(\text{x}^3 - 1)^{\frac{7}{3}}+\frac{1}{4}(\text{x}^3 - 1)^{\frac{4}{3}}+\text{c} $
View full question & answer→Question 771 Mark
Integrate the functions in Exercises:
$\frac{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}$
Answer$\text{Let I}=\int\frac{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}} \text{ dx}=\frac{1}{2}\int\frac{\text 2\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)}{\text {e}^{\text{2x}}+\text{e}^{-\text{2x}}}\text{ dx} \ \ \ \ \ \ ...\text{(i)} $
Putting$\text{ e}^{\text{2x}}+\text{e}^{-2\text{x}}=t\ \ \ \Rightarrow \ \ \ \ \text{ e}^{\text{2x}}\frac{\text{d}}{\text{dx }} 2\text{x} +\text{e}^{-2\text{x}}\frac{\text{d}}{\text{dx}}(-2\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{2\text{x}}.2-2\text{e}^{-2\text{x}}=\frac{\text{dt}}{\text{dx}} \ \ \ \ \Rightarrow\ \ \ \ \ 2\big(\text{e}^{\text{2x}}-\text{e}^{-\text{2x}}\big)\text{ dx}=\text{dt} $
$\therefore \ \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}} $
$=\frac{1}{2}\log\begin{vmatrix}t\end{vmatrix}+\text{c}=\frac{1}{2}\log\begin{vmatrix}\text{e}^\text {2x}+\text{e}^{-\text{2x}}\end{vmatrix}+\text{c} $
$=\frac{1}{2}\log\big(\text{e}^\text{2x}+\text{e}^{-\text{2x}}\big)+\text{c} $
View full question & answer→Question 781 Mark
Integrate the functions in Exercises:
$\sec^2(7-4\text{x})$
Answer$\int\sec^2(7 -4\text{x})\text{ dx}$
$=\frac{\tan(7 - 4\text{x})}{-4\rightarrow\text{Coeff. of x}}+\text{c} $
$=\frac{-1}{4}\tan(7 - 4\text{x})+\text{c} $
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Evaluate $\int\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{ dx}$
$=\int\sec^2\text{x dx}-\int\sec\text{x}\tan\text{x dx}$
$\text{I}=\tan\text{x}-\sec\text{x}+\text{C}$
View full question & answer→Question 801 Mark
Integrate the functions in Exercises:
$\frac{1}{1-\tan\text{x}}$
Answer$\text{Let I}=\int\frac{1}{1-\tan\text{x}}\text{ dx} =\int\frac{1}{1-\frac{\sin\text{x}}{\cos\text{x}}}\text{ dx} $
$=\int\frac{1}{\bigg(\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}}\bigg)}\text{ dx} =\int\frac{\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{2\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx} =\frac{1}{2}\int\frac{\cos\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx} $
Adding and subtracting $\sin\text{x}$ in the numerator,
$=\frac{1}{2}\int\frac{\cos\text{x}-\sin\text{x}+\sin\text{x}+\cos\text{x}} {\cos\text{x}-\sin\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{(\cos\text{x}-\sin\text{x})+(\sin\text{x}+\cos\text{x})}{\cos\text{x}-\sin\text{x}}\text{ dx} $
$=\frac{1}{2}\int\bigg(\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}-\sin\text{x}}-\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\bigg)\text{ dx} $
$=\frac{1}{2}\int\bigg(1-\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\bigg)\text{ dx} $
$\frac{1}{2}\bigg[\int1\text{ dx}-\int\frac{-\sin\text{x}-\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx}\bigg]$
$=\frac{1}{2}\big[\text{x}-\log\begin{vmatrix}\cos\text{x} - \sin \text{x} \end{vmatrix}\big]+\text{c} $
View full question & answer→Question 811 Mark
Integrate the functions in Exercises:
$\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}$
Answer$\text{Let I }=\int\frac{2\cos\text{x}-3\sin\text{x}}{6\cos\text{x}+4\sin\text{x}}\text{ dx}=\int\frac{2\cos\text{x}-3\sin\text{x}}{2(2\sin\text{x}+3\cos\text{x})}\text{ dx} $
$=\frac{1}{2}\int\frac{2\cos\text{x}-3\sin\text{x}}{2\sin\text{x}+3\cos\text{x}}\text{ dx} \ \ \ \ \ \ ...\text{(i)} $
Putting $2\sin\text{x}+3\cos\text{x}=\text{t}\ \ \ \ \Rightarrow \ \ \ \ 2\cos\text{x}-3\sin\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow \ \ \ \ (2\cos\text{x}-3\sin\text{x})\text{ dx}=\text{dt} $
$ \therefore \ \ \ \ $From eq. (i), $\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}} =\frac{1}{2}\log\begin{vmatrix}t\end{vmatrix}+\text{c}$
$=\frac{1}{2}\log\begin{vmatrix}2\sin\text{x}+3\cos \text{x}\end{vmatrix}+\text{c}$
View full question & answer→Question 821 Mark
Find the following integrals in Exercises: $\int\sqrt{\text{x}}(3\text{x}^2+2\text{x+3})\text{ dx}$
Answer$\int\sqrt{\text{x}}(3\text{x}^2+2\text{x+3})\text{ dx}$
$=\int\text{x}^\frac{1}{2}(3\text{x}^2+2\text{x+3})\text{ dx}$
$=\int\bigg(3\text{x}^\frac{5}{2}+2\text{x}^{\frac{3}{2}}+3\text{x}^\frac{1}{2}\bigg)\text{ dx}$
$=\int3\text{x}^{\frac{5}{2}}\text{ dx }+\int2\text{x}^{\frac{3}{2}}\text{ dx}+\int3\text{x}^{\frac{1}{2}}\text{ dx}$
$=3\frac{\text{x}^{\frac{5}{2}+1}}{\frac{5}{2}+1}+2 \frac { \text{x}^{\frac{3}{2}+1}}{\frac{3}{2}+1}+3 \frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{c} $
$=3\frac{\text{x}^{\frac{7}{2}}}{\frac{7}{2}}+2 \frac { \text{x}^{\frac{5}{2}}}{\frac{5}{2}}+3 \frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}+\text{c} $
$\frac{6}{7}\text{x}^\frac{7}{2 }+ \frac{4}{5}\text{x}^\frac{5}{2 }+ 2\text{x}^\frac{3}{2}+\text{c} $
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Evaluate the following integrals:
$\int\log_\text{x}\text{xdx}$
Answer$\int\log_\text{x}\text{xdx}$
$=\int1.\text{dx}$
$=\text{x}+\text{c}$
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Evaluate $\int2^{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int2^{\text{x}}\text{ dx}$
$\text{I}=\frac{2^{\text{x}}}{\log_\text{e}2}+\text{C}$
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