Question

Can someone help me find the to this problem: Integrate the square root of 1-36x^2

Answer 1

\[\int\sqrt{1-36x^2}~dx\]
Make a substitution: \(x=\dfrac{1}{6}\sin u\), so that \(dx=\dfrac{1}{6}\cos u~du\):
\[\int\sqrt{1-36\left(\frac{1}{6}\sin u\right)^2}~\left(\frac{1}{6}\cos u~du\right)\\
\frac{1}{6}\int\sqrt{1-\frac{36}{36}\sin^2 u}~\cos u~du\\
\frac{1}{6}\int\sqrt{\cos^2 u}~\cos u~du\\
\frac{1}{6}\int\cos u~\cos u~du\\
\frac{1}{6}\int\cos^2 u~du\]

Answer 2

From here, use an identity: \(\cos^2u=\dfrac{1+\cos2u}{2}\)