"The integrals converge. Evaluate the integrals without using..." solved and explained

Armorikam

Answered question

2020-10-26

The integrals converge. Evaluate the integrals without using tables

\int_{0}^{1} \frac{4 r d r}{\sqrt{1 - r^{4}}}

2abehn

Skilled2020-10-27Added 88 answers

Step 1: Given that
The integrals converge. Evaluate the integrals without using tables

\int_{0}^{1} \frac{4 r d r}{\sqrt{1 - r^{4}}}

Step 2: Solve
We have,

\int_{0}^{1} \frac{4 r d r}{\sqrt{1 - r^{4}}} d r = \int_{0}^{1} \frac{4 r}{\sqrt{1 - {(r^{2})}^{2}}} d r

Substitute,

r^{2} = t

2r.dr=dt

d r = \frac{d t}{2 r}

Plugging all the values into the integral we obtain,

\int_{0}^{1} \frac{4 r}{\sqrt{1 - {(r^{2})}^{2}}} d r \int_{0}^{1} \frac{4 r}{\sqrt{1 - t^{2}}} \times \frac{d t}{2 r}

= 2 \int_{0}^{1} \frac{d t}{\sqrt{1 - t^{2}}}

= 2 {[\sin^{- 1} (t)]}_{0}^{1}

= 2 (\sin^{- 1} (1) - 0)

= 2 \sin^{- 1} \sin (\frac{π}{2})

= 2 \times \frac{π}{2}

= π

Jeffrey Jordon

Expert2021-11-08Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?