Mathematics

# Integrate : $\sqrt{\dfrac{2 - x}{x}} \, (0 < x < 2)$

##### SOLUTION
$\int \sqrt{\dfrac{2-x}{x}}dx$

$u=\sqrt{\dfrac{2-x}{x}},v'=1$

$=\sqrt{\dfrac{2-x}{x}}x-\int -\dfrac{1}{x^{\frac{1}{2}}\sqrt{2-x}}dx$

$=x\sqrt{\dfrac{2-x}{x}}+2\sin^{-1}\left ( \dfrac{1}{\sqrt{2}}\sqrt{x} \right )$

$=x\sqrt{\dfrac{2-x}{x}}+2\sin^{-1}\left ( \sqrt{\dfrac{x}{2}} \right )+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Hard
$\int \frac{1}{3x^2+ 13x -10}dx$
• A. $\frac{1}{17}[log(\frac{30}{17}+\frac{6x}{17})-log(\frac{-4}{17}+\frac{6x}{17})]+C$
• B. $\frac{-1}{17}[log(\frac{30}{17}+\frac{6x}{17})+log(\frac{-4}{17}+\frac{6x}{17})]+C$
• C. $\frac{-1}{17}[log(\frac{30}{17}+\frac{6x}{17})-log(\frac{-4}{17}-\frac{6x}{17})]+C$
• D. $\frac{-1}{17}[log(\frac{30}{17}+\frac{6x}{17})-log(\frac{-4}{17}+\frac{6x}{17})]+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate:$\displaystyle\int{\left({e}^{\ln{x}}+\sin{x}\right)\cos{x}dx}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle I_{n}=\int (\ln x)^{n} dx,$ then $I_{n}+nI_{n-1}=$
• A. $\displaystyle \frac{\left ( \ln x \right )^{n}}{x}+C$
• B. $x(\ln x)^{n-1}+C$
• C. none of these
• D. $x(\ln x)^{n}+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle\int e^{\sin x}\cdot\left(\begin{matrix} \dfrac{sin x+1}{sec x}\end{matrix}\right)dx$ is equal to?
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• B. $e^{\sin x}+c$
• C. $e^{\sin x}(\sin x+1)+c$
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simplify $\int {\rm{ }}\left( {{x \over {a + bx}}} \right)dx=$