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Question

# Explain why $u=a \sin \theta$ is a useful substitution when an integral contains an expression of the form $\sqrt{a^2-u^2}$.

Solution

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Assume, we need to solve the integral

$\int \sqrt {a^2-u^2}\,du$

By substitution

$u=a\sin \theta$

we have

\begin{align*} du&=a\cos \theta \,d\theta \end{align*}

so, the integral goes to

\begin{align*} \int & \sqrt {a^2-a^2\sin^2 \theta }\cdot a\cos \theta\,d\theta \\[12pt] &=a^2\int \cos^2 \theta \,d\theta \\[12pt] &=a^2\int \frac {1+\cos 2\theta } {2 }\,d\theta \\[12pt] &=\frac {a^2 } {2 }\,\left(\theta + \frac {1} {2}\,\sin 2\theta \right)+C \\[12pt] &=\frac {a^2 } {2 }\,\left(\theta + \sin \theta \cos \theta \right)+C \end{align*}

In terms of $u$, since

\begin{align*} \theta&=\arcsin \left( \frac {u} {a} \right) \\[12pt] \sin \theta &= \frac {u} {a} \quad \Rightarrow \quad \cos \theta = \sqrt {1-\sin^2 \theta }= \sqrt {1-\frac {u^2 } {a^2 } } \end{align*}

we have

$\int \sqrt {a^2-u^2}\,du= \frac {a^2} {2}\, \left[ \arcsin \left( \frac {u} {a} \right) +\frac {u} {a} \, \sqrt { 1 - \left( \frac {u} {a} \right)^2 } \right]+C$

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