Explained: "Evaluate the integral ∫−22dp4−p2"

generals336

Answered question

2020-12-28

Evaluate the integral

\int_{- 2}^{2} \frac{d p}{\sqrt{4 - p^{2}}}

Dora

Skilled2020-12-29Added 98 answers

Step 1
Consider the provided improper integral,

\int_{- 2}^{2} \frac{d p}{\sqrt{4 - p^{2}}}

Evaluate the following integrals.
Apply the substitution method,

L e t p = 2 \sin u \Rightarrow d p = 2 \cos u d u

p = - 2 \Rightarrow u = \frac{π}{2}

p = 2 \Rightarrow u = - \frac{π}{2}

Step 2
So, the integral becomes.

\int_{- 2}^{2} \frac{d p}{\sqrt{4 - p^{2}}} = \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d u

= {[1 \cdot u]}_{- \frac{π}{2}}^{\frac{π}{2}} 1 d u

= {[u]}_{- \frac{π}{2}}^{\frac{π}{2}} 1 d u

= \frac{π}{2} + \frac{π}{2} = π

Hence.

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