Resolved: "Use a change of variables to evaluate the..."

permaneceerc

Answered question

2021-10-22

Use a change of variables to evaluate the following indefinite integrals. Check your work by differentiating.

\int \frac{d x}{\sqrt{36 - 4 x^{2}}}

hesgidiauE

Skilled2021-10-23Added 106 answers

Find integral.

i = \int \frac{d x}{\sqrt{36 - 4 x^{2}}}

x = 3 \sin θ

d x = 3 \cos θ d θ

I = \int \frac{3 \cos θ d θ}{\sqrt{36 - 4 (9 \sin^{2} θ)}}

I = \int \frac{3 \cos θ d θ}{6 \sqrt{1 - \sin^{2} θ}}

I = \frac{1}{2} \int \frac{\cos θ}{\cos θ} d θ = \frac{1}{2} \int d θ

I = \frac{1}{2} θ + C

I = \frac{1}{2} \sin^{- 1} (\frac{x}{3}) + C

Checking by differetiation

\frac{d}{d x} (\frac{1}{2} \sin^{- 1} (\frac{x}{3}) + C) = \frac{1}{2} \frac{d}{d x} (\sin^{- 1} (\frac{x}{3})) + \frac{d}{d x} (c)

= \frac{1}{2} \cdot \frac{1}{3} \frac{1}{\sqrt{1 - \frac{x^{2}}{9}}} + 0

= \frac{1}{2} \cdot \frac{1}{\sqrt{9 - x^{2}}} = \frac{1}{\sqrt{36 - 4 x^{2}}}

Proved

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?