# Evaluate the integral int_-2^2(dp)/(sqrt(4-p^2))

generals336 2020-12-28 Answered
Evaluate the integral
${\int }_{-2}^{2}\frac{dp}{\sqrt{4-{p}^{2}}}$
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## Expert Answer

Dora
Answered 2020-12-29 Author has 98 answers
Step 1
Consider the provided improper integral,
${\int }_{-2}^{2}\frac{dp}{\sqrt{4-{p}^{2}}}$
Evaluate the following integrals.
Apply the substitution method,
$Letp=2\mathrm{sin}u⇒dp=2\mathrm{cos}udu$.
$p=-2⇒u=\frac{\pi }{2}$
$p=2⇒u=-\frac{\pi }{2}$
Step 2
So, the integral becomes.
${\int }_{-2}^{2}\frac{dp}{\sqrt{4-{p}^{2}}}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}1du$
$={\left[1\cdot u\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}1du$
$={\left[u\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}1du$
$=\frac{\pi }{2}+\frac{\pi }{2}=\pi$
Hence.
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