2 Marks Questions

Question 1012 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,
$\text{I}=\int\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin\text{x}\cos^2\text{x}}\text{dx}$
$=\int\frac{\sin^2}{\sin\text{x}\cos^2\text{x}}\text{dx}+\int\frac{\cos^2\text{x}}{\sin\text{x}\cos^2\text{x}}\text{dx}$
$=\int\sec\text{x}\tan\text{x }\text{dx}+\int\text{cosec x dx}$
$=\sec\text{x}+\log\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$
$\therefore\text{I}=\sec\text{x}+\log\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$

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Question 1022 Marks

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

Answer

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

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Question 1032 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 $1+\log\text{x}=\text{t}$
$\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{I}=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^3}{3}+\text{C}$
$\text{I}=\frac{(1+\log\text{x})^3}{3}+\text{C}$

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Question 1042 Marks

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

Answer

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

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Question 1052 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,
$\text{I}=\text{x}\int\text{e}^{2\text{x}}\text{dx}-\int(1\times\int\text{e}^{2\text{x}}\text{dx})\text{dx+C}$
$=\frac{\text{xe}^{2\text{x}}}{2}-\int\Big(\frac{\text{e}^{2\text{x}}}{2}\Big)\text{dx+C}$
$=\frac{\text{xe}^{2\text{x}}}{2}-\frac{\text{e}^{2\text{x}}}{4}+\text{C}$
$\text{I}=\Big(\frac{\text{x}}{2}-\frac{1}{4}\Big)\text{e}^{2\text{x}}+\text{C}$

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Question 1062 Marks

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

Answer

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

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Question 1072 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}$
$=\frac{\text{t}^4}{4}+\text{C}$
$=\frac{(\sin^{-1}\text{x})^4}{4}+\text{C}$

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Question 1082 Marks

Write tha value of $\int\sec\text{x}(\sec\text{x}+\tan\text{x})\text{dx}$

Answer

$\int\sec\text{x}(\sec\text{x}+\tan\text{x})\text{dx}$
$=\int(\sec^2\text{x}+\sec\text{x}\tan\text{x})\text{dx}$
$=\tan\text{x}+\sec\text{x}+\text{C}$

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Question 1092 Marks

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

Answer

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

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Question 1102 Marks

$\int\sin^2\text{bx dx}$

Answer

$\int\sin^2\text{bx dx}$
$=\int\Big[\frac{1-\cos2\text{bx}}{2}\Big]\text{dx}$ $\Big[\therefore\sin^2\text{x}=\frac{1-\cos2\text{x}}{2}\Big]$
$=\frac{1}{2}\int(1-\cos2\text{bx})\text{dx}$
$=\frac{1}{2}\Big[\text{x}-\frac{\sin2\text{bx}}{2\text{b}}\Big]+\text{C}$
$=\frac{\text{x}}{2}-\frac{\sin2\text{bx}}{4\text{b}}+\text{C}$

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Question 1112 Marks

Evaluate the following integral:
$\int\frac{1}{\sqrt{\text{a}^2+\text{b}^2\text{x}^2}}\text{ dx}$

Answer

$\int\frac{1}{\sqrt{\text{a}^2+\text{b}^2\text{x}^2}}\text{ dx}$
$=\int\frac{\text{dx}}{\sqrt{\text{b}^2\Big({\frac{\text{a}^2}{\text{b}^2}+\text{x}^2\Big)}}}$
$=\frac{1}{\text{b}}\int\frac{\text{dx}}{\sqrt{\text{x}^2+\big(\frac{\text{a}}{\text{b}}\big)^2}}$
$=\frac{1}{\text{b}}\log\Big|\text{x}+\sqrt{\text{x}^2+\frac{\text{a}^2}{\text{b}^2}}\Big|+\text{C}$
$=\frac{1}{\text{b}}\Big[\log\Big|\text{x}+\frac{\sqrt{\text{b}^2\text{x}^2+\text{a}^2}}{\text{b}}\Big|\Big]+\text{C}$
$=\frac{1}{\text{b}}\Big[\log\Big|\frac{\text{bx}+\sqrt{\text{b}^2\text{x}^2+\text{a}^2}}{\text{b}}\Big|\Big]+\text{C}$
$=\frac{1}{\text{b}}\Big[\log\big|\text{bx}+\sqrt{\text{b}^2\text{x}^2+\text{a}^2}\big|-\log\text{b}\Big]+\text{C}$
$=\frac{1}{\text{b}}\log\big|\text{bx}+\sqrt{\text{b}^2\text{x}^2+\text{a}^2}\big|-\frac{\log\text{b}}{\text{b}}+\text{C}$
Let $\text{C}-\frac{\log\text{b}}{\text{b}}+\text{C}'$
$=\frac{1}{\text{b}}\log\big|\text{bx}+\sqrt{\text{b}^2\text{x}^2+\text{a}^2}\big|+\text{C}'$

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Question 1122 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, $\int\cos^{-1}(\sin\text{x})\text{dx}=\frac{\pi}{2}\text{x}-\frac{1}{2}\text{x}^2+\text{C}$

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Question 1132 Marks

Evaluate $\int\frac{1}{\text{x}^2+16}\text{ dx}$

Answer

Since, $\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}$
Thus, $\int\frac{1}{\text{x}^2+16}\text{ dx}=\frac{1}{4}\tan^{-1}\Big(\frac{\text{x}}{4}\Big)+\text{C}$

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Question 1142 Marks

Evalute the following integrals:
$\int\frac{1}{\text{x}\log\text{x}}\text{dx}$

Answer

Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.
Let $\text{I}=\int\frac{1}{\text{x}\log\text{x}}\text{dx}$
Putting $\log\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\log\text{x}|+\text{C}$

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MATHS 2 Marks Questions