Question

find the following integrals​

Answer 1

To integrate :

$\star \: \frac{dx}{ \sqrt{2x - {x}^{2} } }$

Here , Add and subtract 1 in denominator :

$\implies \: \frac{dx}{ \sqrt{2x - {x}^{2} + 1 - 1 } } \\ \\ \implies \: \frac{dx}{ \sqrt{1 - {x}^{2} + 2x - 1} } \\ \\ \implies \: \frac{dx}{ \sqrt{ 1 - ( {x}^{2} - 2x + 1)} } \\ \\ \implies \: \frac{dx}{ \sqrt{ {1}^{2} - ( {x - 1)}^{2} } } \\ \\ \implies \: { \sin}^{ - 1} \frac{x - 1}{1} + c \\ \\ \implies \: { \sin}^{ - 1} (x - 1) + c$

Formula used:

$\star \boxed{ \red{\bold{ \int \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = { \sin}^{ - 1} \frac{x}{a} + c}}}$

Answer 2

Given ,

The function is

$\tt \frac{1}{ \sqrt{2x - {x}^{2} } }$

Here , the quadratic equation is

• 2x - x²

It can be written as

2x - x² - 1 + 1

(1)² - (x - 1)²

Thus ,

$\tt \implies \int{ \tt \frac{1}{ \sqrt{2x - {x}^{2} } } \: dx }$

$\tt \implies\int{ \tt \frac{1}{ \sqrt{ {(1)}^{2} - {(x - 1)}^{2} } } \: dx }$

Let , x - 1 = t => dx = dt

Thus ,

$\tt \implies \int{ \tt \frac{1}{ \sqrt{ {(1)}^{2} - {(t)}^{2} } } \: dt }$

$\tt \implies {sin}^{ - 1} \frac{t}{1} + c$

$\tt \implies {sin}^{ - 1}t + c$

Since , t = x - t

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

Remmember :

$\tt \implies \int{ \tt \frac{1}{ \sqrt{ {(a)}^{2} - {(x)}^{2} } } \: dx} = {sin}^{ - 1} \frac{x}{a} + c$

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