How to calculate the integral ${\int}_{0.5}^{1}\frac{1}{\sqrt{2x-{x}^{2}}}$?

hawatajwizp
2022-06-26
Answered

How to calculate the integral ${\int}_{0.5}^{1}\frac{1}{\sqrt{2x-{x}^{2}}}$?

You can still ask an expert for help

Marlee Guerra

Answered 2022-06-27
Author has **25** answers

Note that

2x-x2=1-(1-2x+x2)=1-(1-x)2

Now you can use a trig substitution.

2x-x2=1-(1-2x+x2)=1-(1-x)2

Now you can use a trig substitution.

minwaardekn

Answered 2022-06-28
Author has **6** answers

Given ${\int}_{0.5}^{1}{\displaystyle \frac{1}{\sqrt{2x-{x}^{2}}}}dx$

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

Apply u substitution u=x-1

$\int \frac{1}{\sqrt{1-{u}^{2}}}du=\mathrm{arcsin}(u)=\mathrm{arcsin}(x-1)$

Now apply the boundaries

$={\textstyle [}\mathrm{arcsin}(x-1){{\textstyle ]}}_{0.5}^{1}=\frac{\pi}{6}$

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

Apply u substitution u=x-1

$\int \frac{1}{\sqrt{1-{u}^{2}}}du=\mathrm{arcsin}(u)=\mathrm{arcsin}(x-1)$

Now apply the boundaries

$={\textstyle [}\mathrm{arcsin}(x-1){{\textstyle ]}}_{0.5}^{1}=\frac{\pi}{6}$

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