Question

$mimcool786\: \:see\:the\:attachment$

$prove \: that \: \: \\ \\ \displaystyle \int_0^1 \frac{dx}{ \sqrt{2x - {x}^{2} } } = \frac{\pi}{2} \:$

solve that with proper explanation.

Answer 1

Step-by-step explanation:

Given:

$\int_0^1 \frac{dx}{ \sqrt{2x - {x}^{2} } } = \frac{\pi}{2} \:$

To find: Prove

Solution:

To prove it,take LHS and integrate it.

$\int_0^1 \frac{dx}{ \sqrt{2x - {x}^{2} } }$

Completing the square

$\int_0^1 \frac{dx}{ \sqrt{2x - {x}^{2} } } = \int_0^1 \frac{dx}{ \sqrt{ - ({x}^{2} - 2x + 1 - 1)} } \\ \\ \implies\int_0^1 \frac{dx}{ \sqrt{ - ({x}^{2} - 2x + 1) + 1} } \\ \\ \implies\int_0^1 \frac{dx}{ \sqrt{ - ({x} -1)^{2} + 1 } } \\ \\ \\ \implies\int_0^1 \frac{dx}{ \sqrt{ 1- ({x} -1)^{2} } }$

Let

$x - 1 = t \\ \\ dx = dt$

Do substitution and integrate

$\int \frac{dt}{ \sqrt{1 - {t}^{2} } } = sin^{ - 1} t$

as it is direct formula.

Undo substitution and place the limits

$= {sin}^{ - 1} (x - 1) \bigg|_0^{1} \\ \\ = {sin}^{ - 1} ( 1 - 1) - {sin}^{ - 1}(0 - 1) \\ \\ = {sin}^{ - 1} ( 0) - {sin}^{ - 1}(- 1) \\ \\ = {sin}^{ - 1} ( sin \: 0) + {sin}^{ - 1}( sin \frac{\pi}{2} ) \\ \\ = 0 + \frac{\pi}{2} \\ \\ = \frac{\pi}{2} \\ \\ = R.H.S.$

Final answer:

It has been proved that

$\int_0^1 \frac{dx}{ \sqrt{2x - {x}^{2} } } = \frac{\pi}{2} \:$

Hope it helps you.

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