Mathematics

# Solve : $\displaystyle \int \, \dfrac{dx}{\sqrt{2x - x^2}}$

##### SOLUTION
$I=\displaystyle\int \dfrac{dx}{\sqrt{2x-x^2}}=\displaystyle\int \dfrac{dx}{\sqrt{1-1+2x-x^2}}\cdot \displaystyle\int \dfrac{dx}{\sqrt{1-(x^2-2x+1)}}$

$I=\displaystyle\int\dfrac{dx}{\sqrt{1^2-(x-1)^2}}$

Put $(x-1)=t$

$dx=dt$.
$I=\displaystyle\int\dfrac{dt}{\sqrt{1-t^2}}$

Put $t=\sin\theta$

$dt=\cos\theta d\theta$

$I=\displaystyle\int \dfrac{\cos \theta d\theta}{\sqrt{1-\sin^2\theta}}=\displaystyle\int \dfrac{\cos\theta d\theta}{\cos\theta}=\displaystyle\int d\theta$
$I=\theta +c=\sin^{-1}(t)+c=\sin^{-1}(x-1)+c$
$\therefore \displaystyle\int \dfrac{dx}{\sqrt{2x-x^2}}=\sin^{-1}(x-1)+c$.

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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 Medium

$\displaystyle \int_{\pi^{2}/16}^{\pi^{2}/4}\frac{\sin\sqrt{x}}{\sqrt{x}}dx=$
• A. $1/\sqrt{2}$
• B. 2 $\sqrt{2}$
• C. ${\pi}/2$
• D. $\sqrt{2}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Integrate the rational function   $\cfrac {2}{(1-x)(1+x^2)}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle I = \int_{0}^{\pi/4} \frac{ \sin 2\theta }{ \sin^{2} \theta +\cos ^{4} \theta } d \theta$
then $I$ is equal to?
• A. $\dfrac{\pi}{\sqrt{2}}$
• B. $\dfrac{\pi}{\sqrt{3}}$
• C. $\dfrac{\pi}{2\sqrt{3}}$
• D. $\dfrac{\pi}{3\sqrt{3}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
The value of $\displaystyle \lim_{n \rightarrow \infty} \Sigma_1^n \cos\left(\dfrac{\pi}{2} +\dfrac{\pi i}{2n} \right) \dfrac{\pi}{2n}=?$
• A. $\displaystyle \int_{\frac{\pi}{3}}^{\pi} \cos x$
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• C. $\displaystyle \int_{\frac{\pi}{2}}^{5\pi} \cos x$
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