integral 1/2a (a^2- 4x^2)^(1/2) - 1/4 (a^2-4x^2) dx
- Thread starter spdrmncoo
- Start date
Hello, spdrmncoo!
Trig Substitution
Let \(\displaystyle 2x\,=\,a\sin\theta\;\;\Rightarrow\;\;dx\,=\,\frac{a}{2}\cos\theta\,d\theta\)
then: \(\displaystyle \,\sqrt{a^2\,-\,(2x)^2}\:=\:\sqrt{a^2\,-\,a^2\sin^2\theta}\:=\:\sqrt{a^2(1\,-\,\sin^2\theta)}\:=\:\sqrt{a^2\cos^2\theta} \:=\:a\cos\theta\)
The integral becomes: \(\displaystyle \L\:\frac{a}{2}\int\left(a\cos\theta\right)\,\left(\frac{a}{2}\cos\theta\,d\theta\right) \;=\;\frac{a^3}{4}\int\cos^2\theta\,d\theta\)
Can you finish it now?
Trig Substitution
Let \(\displaystyle 2x\,=\,a\sin\theta\;\;\Rightarrow\;\;dx\,=\,\frac{a}{2}\cos\theta\,d\theta\)
then: \(\displaystyle \,\sqrt{a^2\,-\,(2x)^2}\:=\:\sqrt{a^2\,-\,a^2\sin^2\theta}\:=\:\sqrt{a^2(1\,-\,\sin^2\theta)}\:=\:\sqrt{a^2\cos^2\theta} \:=\:a\cos\theta\)
The integral becomes: \(\displaystyle \L\:\frac{a}{2}\int\left(a\cos\theta\right)\,\left(\frac{a}{2}\cos\theta\,d\theta\right) \;=\;\frac{a^3}{4}\int\cos^2\theta\,d\theta\)
Can you finish it now?