Question

Integration using tables help please.
Integrate sqrt(9-25x^2)dx

Answer 1

So I am using the formula int sqrt(a^2-x^2)dx = x/2*sqrt(a^2-x^2)+a^2/2sin^-1 x/a +c
where a=3 and u=5x

Answer 2

For an integral of the form
\[\int\sqrt{a-bx^2}~dx\]
you'd use a substitution, \(x=\sqrt{\dfrac{a}{b}}\sin u\), then \(dx=\sqrt{\dfrac{a}{b}}\cos u~du\).
\[\int\sqrt{a-b\left(\sqrt{\dfrac{a}{b}}\sin u\right)^2}\left(\sqrt{\dfrac{a}{b}}\cos u~du\right)\]
Simplifying, you have
\[\sqrt{\frac{a}{b}}\int\sqrt{a-b\left(\frac{a}{b}\sin^2 u\right)}\cos u~du\\
\sqrt{\frac{a}{b}}\int\sqrt{a-a\sin^2 u}\cos u~du\\
\sqrt{\frac{a}{b}}\sqrt a\int\sqrt{1-\sin^2 u}\cos u~du\\
\frac{a}{\sqrt b}\int\sqrt{\cos^2 u}\cos u~du\\
\frac{a}{\sqrt b}\int\cos^2 u~du\]
Half-angle identity:
\[\cos^2u=\frac{1}{2}\left(1+\cos2u\right)\]
So, you get
\[\frac{a}{2\sqrt b}\int(1+\cos2u)~du\\
\frac{a}{2\sqrt b}\left(u+\frac{1}{2}\sin2u\right)+C\]
Now you have to back-substitute:
\[x=\sqrt{\frac{a}{b}}\sin u~~\iff~~\sin^{-1}\left(\sqrt{\frac{b}{a}}x\right)=u\]
Let's also rewrite the \(\sin 2u\) using the double angle identity:
\[\sin2u=2\sin u\cos u\]
So, we have
\[\frac{a}{2\sqrt b}\left(u+\sin u\cos u\right)+C\\
\frac{a}{2\sqrt b}\left[\sin^{-1}\left(\sqrt{\frac{b}{a}}x\right)+\sin \left(\sin^{-1}\left(\sqrt{\frac{b}{a}}x\right)\right)\cos \left(\sin^{-1}\left(\sqrt{\frac{b}{a}}x\right)\right)\right]+C\\
\frac{a}{2\sqrt b}\left[\sin^{-1}\left(\sqrt{\frac{b}{a}}x\right)+\sqrt{\frac{b}{a}}x\cos \left(\sin^{-1}\left(\sqrt{\frac{b}{a}}x\right)\right)\right]+C\]
The cosine term is a bit tricky, but you can draw a reference triangle to determine the ratio:

which gives you
\[\cos u=\cos\left(\sin^{-1}\left(\sqrt{\frac{b}{a}}x\right)\right)=\sqrt\frac{a-bx^2}{a}\]
So, you get
\[\frac{a}{2\sqrt b}\left[\sin^{-1}\left(\sqrt{\frac{b}{a}}x\right)+\sqrt{\frac{b}{a}}\sqrt\frac{a-bx^2}{a}~x\right]+C\\
\frac{a}{2\sqrt b}\left[\sin^{-1}\left(\sqrt{\frac{b}{a}}x\right)+\frac{\sqrt b}{a}\sqrt{a-bx^2}~x\right]+C\]
So with \(a=9\) and \(b=25\),
\[\begin{align*}\int\sqrt{9-25x^2}~dx&=\frac{9}{2\sqrt{25}}\left[\sin^{-1}\left(\sqrt{\frac{25}{9}}~x\right)+\frac{\sqrt {25}}{9}\sqrt{9-25x^2}~x\right]+C\\
&=\frac{9}{10}\left[\sin^{-1}\left(\frac{5}{3}x\right)+\frac{5}{9}\sqrt{9-25x^2}~x\right]+C\\
&=\frac{9}{10}\sin^{-1}\left(\frac{5}{3}x\right)+\frac{x}{2}\sqrt{9-25x^2}+C
\end{align*}\]