M.C.Q (1 Marks)

Question 2011 Mark

Choose the correct answer in Exercise.
$\int\sqrt{1+\text{x}^2}\text{dx}$ is equal to

$\frac{\text{x}}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{log}\Bigg|\Big(\text{x}+\sqrt{1+\text{x}^2}\Big)\Bigg|+\text{C}$
$\frac{2}{3}(1+\text{x}^2)^{\frac{3}{2}}+\text{C}$
$\frac{2}{3}\text{x}(1+\text{x}^2)^{\frac{3}{2}}+\text{C}$
$\frac{\text{x}^2}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{x}^2\text{log}\Bigg|\text{x}+\sqrt{1+\text{x}^2}\Bigg|+\text{C}$

Answer

$\frac{\text{x}}{2}\sqrt{1+\text{x}^2}+\frac{1}{2}\text{log}\Bigg|\Big(\text{x}+\sqrt{1+\text{x}^2}\Big)\Bigg|+\text{C}$

$\int\sqrt{1+\text{x}^2}\text{dx}=\int\sqrt{\text{x}^2+1^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2+1^2}+\frac{1^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+1^2}\Big|+\text{C}$
$=\frac{\text{x}}{2}\sqrt{\text{x}^2+1}+\frac{1}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+1}\Big|+\text{C}$
$\Bigg[\therefore\ \int\sqrt{\text{x}^2+\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2+\text{a}^2}+\frac{\text{a}^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|\Bigg]+\text{c}$

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Question 2021 Mark

Value of $\int\frac{\text{dx}}{\sqrt{2\text{x}-\text{x}^2}}$

$\sin^{-1}(\text{x}-1)+\text{c}$
$\sin^{-1}(1+\text{x})+\text{c}$
$\sin^{-1}(1+\text{x}^2)+\text{c}$
$-\sqrt{2\text{x}-\text{x}^2}+\text{c}$

Answer

$\sin^{-1}(\text{x}-1)+\text{c}$

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Question 2031 Mark

$\int\frac{1}{1+\tan\text{x}}\text{ dx}=$

$\log_\text{e}(\text{x}+\sin\text{x})+\text{C}$
$\log_\text{e}(\sin\text{x}+\cos\text{x})+\text{C}$
$2\sec^2\frac{\text{x}}{2}+\text{C}$
$\frac{1}{2}\big[\text{x}+\log(\sin\text{x}+\cos\text{x})\big]+\text{C}$

Answer

$\frac{1}{2}\big[\text{x}+\log(\sin\text{x}+\cos\text{x})\big]+\text{C}$

Solution:
$\text{I}=\int\frac{1}{1+\tan\text{x}}\text{ dx}$
$=\int\frac{\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
Numerator can be written as,
$\cos\text{x}=\text{A}(\sin\text{x}+\cos\text{x})+\text{B}\frac{\text{d}(\sin\text{x}+\cos\text{x})}{\text{dx}}$
$\cos\text{x}=(\text{A}-\text{B})\sin\text{x}+(\text{A}+\text{B})\cos\text{x}$
$\text{A}-\text{B}=0$ and $\text{A}+\text{B}=1$
$\text{A}=\frac{1}{2}=\text{B}$
$\text{I}=\int\frac{\big[\frac{1}{2}(\sin\text{x}+\cos\text{x})+\frac{1}{2}(\cos\text{x}-\sin\text{x})\big]\text{dx}}{\sin\text{x}+\cos\text{x}}$
$\text{I}=\frac{1}{2}\int\Big(1+\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\Big)\text{dx}$
$\text{I}=\frac{1}{2}\big[1+\int(\sin\text{x}+\cos\text{x})\big]+\text{C}$

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Question 2041 Mark

$\int\sqrt{\frac{\text{x}}{1-\text{x}}}\text{ dx}$ is equal to:

$\sin^{-1}\sqrt{\text{x}}+\text{C}$
$\sin^{-1}\Big\{\sqrt{\text{x}}-\sqrt{\text{x}(1-\text{x})}\Big\}+\text{C}$
$\sin^{-1}\Big\{\sqrt{\text{x}(1-\text{x})}\Big\}+\text{C}$
$\sin^{-1}\sqrt{\text{x}}-\sqrt{\text{x}(1-\text{x})}+\text{C}$

Answer

$\sin^{-1}\sqrt{\text{x}}-\sqrt{\text{x}(1-\text{x})}+\text{C}$

Solution:
$\text{I}=\int\sqrt{\frac{\text{x}}{1-\text{x}}}\text{ dx}$
$\text{I}=\int\sqrt{\frac{\text{x}}{1-\text{x}}\cdot\frac{\text{x}}{\text{x}}}\text{ dx}$
$\text{I}=\int\frac{\text{x dx}}{\sqrt{\text{x}-\text{x}^2}}$
Consider,
$\text{x}=\text{A}\frac{\text{d}(\text{x}-\text{x}^2)}{\text{dx}}+\text{B}$
$\text{x}=\text{A}(1-2\text{x})+\text{B}$
$\text{x}=-2\text{Ax}+\text{A}+\text{B}$
$-2\text{A}=1$
$\text{A}=\frac{-1}{2}$
$\text{I}=\int\frac{\frac{-1}{2}(1-2\text{x})+\frac{1}{2}}{\sqrt{\text{x}-\text{x}^2}}\text{ dx}$
$\text{I}=\int\Big(\frac{-1}{2}\frac{1-2\text{x}}{\sqrt{\text{x}-\text{x}^2}}+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}\Big)\text{dx}$
$\text{I}=\frac{-1}{2}\times2\sqrt{\text{x}-\text{x}^2}+\frac{1}{2}\int\frac{1}{\sqrt{\text{x}-\text{x}^2}}\text{ dx}$
Second term after completing square method you will get as
$\text{I}=-\sqrt{\text{x}-\text{x}^2}+\sin^{-1}\sqrt{\text{x}}+\text{C}$

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Question 2051 Mark

Evaluate: $\int\sqrt{1+\text{y}}^2.\text{2ydy:}$

$\text{I}=\frac{2}{3}(1+\text{y}^2)^\frac{3}{2}+\text{c}$
$\text{I}=\frac{2}{5}(1-\text{y}^2)\frac{3}{2}+\text{c}$
$\text{I}=\frac{2}{3}(1-\text{y}^2)^\frac{3}{2}+\text{c}$
None of these

Answer

$\text{I}=\frac{2}{3}(1+\text{y}^2)^\frac{3}{2}+\text{c}$

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Question 2061 Mark

$\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{ dx}$ is equal to:

$2(\sin\text{x}+\text{x}\cos\theta)+\text{C}$
$2(\sin\text{x}-\text{x}\cos\theta)+\text{C}$
$2(\sin\text{x}+2\text{x}\cos\theta)+\text{C}$
$2(\sin\text{x}-2\text{x}\cos\theta)+\text{C}$

Answer

$2(\sin\text{x}+\text{x}\cos\theta)+\text{C}$

Solution:
$\text{I}=\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{ dx}$
$\text{I}=\int\frac{2\cos^2\text{x}-1-(2\cos^2\theta-1)}{\cos\text{x}-\cos\theta}\text{ dx}$
$\text{I}=\int\frac{2(\cos^2\text{x}-\cos^2\theta)}{\cos\text{x}-\cos\theta}\text{ dx}$
$\text{I}=2\int(\cos\text{x}+\cos\theta)\text{dx}$
$\text{I}=2(\sin\text{x}+\text{x}\cos\theta)+\text{C}$

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Question 2071 Mark

$\int_{0}^{\frac{\pi^2}{4}}\frac{\sin\sqrt{\text{y}}}{\sqrt{\text{y}}}.$

$1$
$2$
$\frac{\pi}{4}$
$\frac{\pi^2}{8}$

Answer

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Question 2081 Mark

$\int\frac{\text{a}}{(1+\text{x}^2)\tan^{-1}\text{x}}\text{dx}=$

$\text{a}\log|\tan^{-1}\text{x}|+\text{c}$
$\frac{1}{2}(\tan^{-1}\text{x})^2+\text{c}$
$\text{a}\log(1+\text{x}^2)+\text{c}$
$\text{None of these}$

Answer

$\text{a}\log|\tan^{-1}\text{x}|+\text{c}$

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Question 2091 Mark

Integrate the following function with respect to x $\int\big(5\text{x}^2+3\text{x}-2\big)\text{dx:}$

$\frac{5\text{x}^3}{3}-\frac{3\text{x}^3}{2}-\text{2x}+\text{c}$
$\frac{5\text{x}^3}{3}+\frac{3\text{x}^2}{2}-\text{2x}+\text{c}$
$\frac{5\text{x}^3}{3}-\frac{3\text{x}^2}{2}+\text{2x}-\text{c}$
$\frac{5\text{x}^3}{3}+\frac{3\text{x}^3}{2}+\text{2x}+\text{c}$

Answer

$\frac{5\text{x}^3}{3}+\frac{3\text{x}^2}{2}-\text{2x}+\text{c}$

Solution:
$\int\big(5\text{x}^2+3\text{x}-2\big)\text{dx}=5\int{\text{x}^2}\text{dx}+3\int\text{xdx}-2\int\text{dx}$
$=\frac{5\text{x}^3}{3}+\frac{3\text{x}^2}{2}-\text{2x}+\text{c}$

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Question 2101 Mark

$\int\frac{\sin^2\text{x}}{\cos^4\text{x}}\text{ dx}=$

$\frac{1}{3}\tan^2\text{x}+\text{C}$
$\frac{1}{2}\tan^2\text{x}+\text{C}$
$\frac{1}{3}\tan^3\text{x}+\text{C}$
none of these.

Answer

$\frac{1}{3}\tan^3\text{x}+\text{C}$

Solution:
$\text{I}=\int\frac{\sin^2\text{x}}{\cos^4\text{x}}\text{ dx}$
$\text{I}=\int\tan^{2}\text{x}\sec^2\text{x dx}$
Put $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\text{t}^2\text{dt}$
$\text{I}=\frac{\text{t}^3}{3}+\text{C}$
$\text{I}=\frac{\tan^3\text{x}}{3}+\text{C}$

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MATHS M.C.Q (1 Marks)