2 Marks Questions

Question 1012 Marks

Write a value of $\int\sqrt{\text{x}^2-9}\text{ dx}$

Answer

Let $\text{I}=\int\sqrt{\text{x}^2-9}\text{ dx}$
We know that,
$\int\sqrt{\text{x}^2-\text{a}^2}\text{ dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2-\text{a}^2}-\frac{\text{a}^2}{2}\log\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|+\text{C}$
Here $a = 3, a^2 = 9$
$\therefore\ \text{I}=\frac{\text{x}}{2}\sqrt{\text{x}^2-9}-\frac{9}{2}\log\Big|\text{x}+\sqrt{\text{x}^2+9}\Big|+\text{C}$

View full question & answer→

Question 1022 Marks

Integrate the following integrals:
$\int\cos\text{mx}\cos\text{nx dx m}\neq\text{n}$

Answer

$\int\cos\text{mx}\cos\text{nx dx}$
$=\frac{1}{2}\int2\cos(\text{mx})\cos(\text{nx})\text{dx}$
$=\frac{1}{2}\int[\cos(\text{mx}+\text{nx})+\cos(\text{mx}-\text{nx})]\text{dx}$ $[\therefore2\cos\text{A}\cos\text{B}=\cos(\text{A}+\text{B})+\cos(\text{A}-\text{B})]$
$=\frac{1}{2}\Big[\frac{\sin(\text{m+n})\text{x}}{\text{m}+\text{n}}+\frac{\sin(\text{m}-\text{n})\text{x}}{\text{m}-\text{n}}\Big]+\text{C}$

View full question & answer→

Question 1032 Marks

Evalute the following integrals:
$\int\frac{1}{\sin\text{x}\cos^2\text{x}}\text{dx}$

Answer

Let $\text{I}=\int\frac{1}{\sin\text{x}\cos^2\text{x}}\text{dx},$ then,
$\text{I}=\int\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin\text{x}\cos^2\text{x}}\text{dx}$
$=\int\frac{\sin^2}{\sin\text{x}\cos^2\text{x}}\text{dx}+\int\frac{\cos^2\text{x}}{\sin\text{x}\cos^2\text{x}}\text{dx}$
$=\int\sec\text{x}\tan\text{x }\text{dx}+\int\text{cosec x dx}$
$=\sec\text{x}+\log\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$
$\therefore\text{I}=\sec\text{x}+\log\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$

View full question & answer→

Question 1042 Marks

Evaluate $\int\cos^{-1}(\sin\text{x})\text{dx}$

Answer

$\int\cos^{-1}(\sin\text{x})\text{dx}=\int\cos^{-1}\Big(\cos\Big(\frac{\pi}{2}-\text{x}\Big)\Big)\text{dx}$
$=\int\Big(\frac{\pi}{2}-\text{x}\Big)\text{dx}$
$=\frac{\pi}{2}\text{x}-\frac{1}{2}\text{x}^2+\text{C}$
Hence, $\int\cos^{-1}(\sin\text{x})\text{dx}=\frac{\pi}{2}\text{x}-\frac{1}{2}\text{x}^2+\text{C}$

View full question & answer→

Question 1052 Marks

Evaluate the following integrals:
$\int\big(\text{x}^\text{e}+\text{e}^\text{x}+\text{e}^\text{e}\big)\text{dx}$

Answer

$\int\big(\text{x}^\text{e}+\text{e}^\text{x}+\text{e}^\text{e}\big)\text{dx}$
$=\int\text{x}^\text{e}\text{dx}+\int\text{e}^\text{x}\text{dx}+\text{e}^\text{e}\int1\text{dx}$
$=\frac{\text{x}^{\text{e}+1}}{\text{e}+1}+\text{e}^\text{x}+\text{x}.\text{e}^\text{e}+\text{C}$

View full question & answer→

Question 1062 Marks

Evaluate $\int(1-\text{x})\sqrt{\text{x}}\text{ dx}$

Answer

Let $\sqrt{\text{x}}=\text{t}$
$\text{x}=\text{t}^2$
$1-\text{x}=1-\text{t}^2$
$-\text{dx}=-2\text{tdt}$
$\text{dx}=2\text{tdt}$
$\text{I}=\int(1-\text{x})\sqrt{\text{x}}\text{ dx}$
$=\int(1-\text{t}^2)\text{t}2\text{t dt}$
$=2\int (1-\text{t}^2)\text{t}^2\text{ dt}$
$=2\big(\int\text{t}^2\text{dt}-\int\text{t}^4\text{dt}\big)$
$=2\frac{\text{t}^3}{3}-2\frac{\text{t}^5}{5}+\text{C}$
$=\frac{2}{3}\text{x}^{\frac{3}{2}}-\frac{2}{5}\text{x}^{\frac{5}{2}}+\text{C}$

View full question & answer→

Question 1072 Marks

Evaluate $\int\frac{(1+\log\text{x})^2}{\text{x}}\text{ dx}$

Answer

Let $\text{I}=\int\frac{(1+\log\text{x})^2}{\text{x}}\text{ dx}$
Let $1+\log\text{x}=\text{t}$
$\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{I}=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^3}{3}+\text{C}$
$\text{I}=\frac{(1+\log\text{x})^3}{3}+\text{C}$

View full question & answer→

Question 1082 Marks

$\int\cos^2\text{nx dx}$

Answer

$\int\cos^2\text{nx dx}$
$=\int\Big[\frac{1+\cos2\text{nx}}{2}\Big]\text{dx}$ $\Big[\therefore\cos^2\text{x}=\frac{1+\cos2\text{x}}{2}\Big]$
$=\frac{1}{2}\int(1+\cos2\text{nx})\text{dx}$
$=\frac{1}{2}\Big[\text{x}+\frac{\sin2\text{nx}}{2\text{n}}\Big]+\text{C}$
$=\frac{\text{x}}{2}+\frac{\sin2\text{nx}}{4\text{n}}+\text{C}$

View full question & answer→

Question 1092 Marks

Evaluate the following integrals:$\int\text{xe}^{2\text{x}}\text{dx}$

Answer

Let $\text{I}=\int\text{xe}^{2\text{x}}\text{dx}$
Using integration by parts,
$\text{I}=\text{x}\int\text{e}^{2\text{x}}\text{dx}-\int(1\times\int\text{e}^{2\text{x}}\text{dx})\text{dx+C}$
$=\frac{\text{xe}^{2\text{x}}}{2}-\int\Big(\frac{\text{e}^{2\text{x}}}{2}\Big)\text{dx+C}$
$=\frac{\text{xe}^{2\text{x}}}{2}-\frac{\text{e}^{2\text{x}}}{4}+\text{C}$
$\text{I}=\Big(\frac{\text{x}}{2}-\frac{1}{4}\Big)\text{e}^{2\text{x}}+\text{C}$

View full question & answer→

Question 1102 Marks

Evaluate the following integrals:
$\int\frac{1}{3\sqrt{\text{x}^2}}\text{dx}$

Answer

$\int\frac{\text{dx}}{3\sqrt{\text{x}^2}}$
$=\int\frac{\text{dx}}{\text{x}^\frac{2}{3}}$
$=\int\text{x}^\frac{-2}{3}\text{dx}$
$=\frac{\text{x}^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}+\text{c}$
$=3\text{x}^\frac{1}{3}+\text{c}$

View full question & answer→

Question 1112 Marks

Evaluate the following integrals:
$\int\frac{(\sin^{-1}\text{x})^3}{\sqrt{1-\text{x}^2}}=\text{dx}$

Answer

$\int\frac{(\sin^{-1}\text{x})^3}{\sqrt{1-\text{x}^2}}=\text{dx}$ Let $\sin^{-1}\text{x}=\text{t}$ $\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$ Now, $\int\frac{(\sin^{-1}\text{x})^3}{\sqrt{1-\text{x}^2}}=\text{dx}$$=\int\text{t}^3\text{ dt}$
$=\frac{\text{t}^4}{4}+\text{C}$
$=\frac{(\sin^{-1}\text{x})^4}{4}+\text{C}$

View full question & answer→

Question 1122 Marks

Evaluate the following integrals:
$\int3^{2\log_3\text{x}}\text{dx}$

Answer

$\int3^{2\log_3\text{x}}\text{dx}=\int3^{\log_3\text{x}^2}\text{dx}$
$=\int\text{x}^2\text{dx}$
$=\frac{\text{x}^3}{3}+\text{c}$
$\int\log_\text{x}\text{xdx}=\int1\text{dx}$
$=\text{x}+\text{c}$

View full question & answer→

Question 1132 Marks

Evaluate $\int\frac{\text{x}^2}{1+\text{x}^3}\text{ dx}$

Answer

Let $\text{I}=\int\frac{\text{x}^2}{1+\text{x}^2}\text{ dx}$
Let $1+\text{x}^3=\text{t}$
$3\text{x}^2\text{dx}=\text{dt}$
$\text{x}^2\text{dx}=\frac{1}{3}\text{dt}$
$\therefore\ \frac{1}{3}\int\frac{\text{dt}}{\text{t}}=\frac{1}{3}\log\text{t}+\text{C}$
$\text{I}=\frac{1}{3}\log|1+\text{x}^3|+\text{C}$

View full question & answer→

Question 1142 Marks

Write the primitive or anti-derivative of $\text{f(x)}=\sqrt{\text{x}}=\frac{1}{\sqrt{\text{x}}}$.

Answer

$\text{f(x)}=\sqrt{\text{x}}=\frac{1}{\sqrt{\text{x}}}$
Integrating both sides:
$\int\text{f(x)}\text{dx}=\int\Big(\sqrt{\text{x}}=\frac{1}{\sqrt{\text{x}}}\Big)\text{dx}$
$=\int\Big(\text{x}^{\frac{1}{2}}+\text{x}^{-\frac{1}{2}}\Big)\text{dx}$
$=\bigg[\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg]+\bigg[\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg]+\text{C}$
$=\frac{2}{3}\text{x}^{\frac{3}{2}}+2\text{x}^{\frac{1}{2}}+\text{C}$

View full question & answer→

MATHS 2 Marks Questions