Editing Trigonometric substitution (section)

====Example 1====
In the integral

<math display=block>\int\frac{dx}{\sqrt{a^2-x^2}},</math>

we may use

<math display=block>x=a\sin \theta,\quad dx=a\cos\theta\, d\theta, \quad \theta=\arcsin\frac{x}{a}.</math>

Then,
<math display=block>\begin{align}
 \int\frac{dx}{\sqrt{a^2-x^2}} &= \int\frac{a\cos\theta \,d\theta}{\sqrt{a^2-a^2\sin^2 \theta}} \\[6pt]
   &= \int\frac{a\cos\theta\,d\theta}{\sqrt{a^2(1 - \sin^2 \theta )}} \\[6pt]
   &= \int\frac{a\cos\theta\,d\theta}{\sqrt{a^2\cos^2\theta}} \\[6pt]
   &= \int d\theta \\[6pt]
   &= \theta + C \\[6pt]
   &= \arcsin\frac{x}{a}+C.
 \end{align}</math>

The above step requires that <math>a > 0</math> and <math>\cos \theta > 0.</math> We can choose <math>a</math> to be the principal root of <math>a^2,</math> and impose the restriction <math>-\pi /2  < \theta < \pi /2</math> by using the inverse sine function.

For a definite integral, one must figure out how the bounds of integration change.  For example, as <math>x</math> goes from <math>0</math> to <math>a/2,</math> then <math>\sin \theta</math> goes from <math>0</math> to <math>1/2,</math> so <math>\theta</math> goes from <math>0</math> to <math>\pi / 6.</math>  Then,

<math display=block>\int_0^{a/2}\frac{dx}{\sqrt{a^2-x^2}}=\int_0^{\pi/6} d\theta = \frac{\pi}{6}.</math>

Some care is needed when picking the bounds. Because integration above requires that <math>-\pi /2  < \theta < \pi /2</math> , <math>\theta</math> can only go from <math>0</math> to <math>\pi / 6.</math> Neglecting this restriction, one might have picked <math>\theta</math> to go from <math>\pi</math> to <math>5\pi /6,</math> which would have resulted in the negative of the actual value.

Alternatively, fully evaluate the indefinite integrals before applying the boundary conditions. In that case, the antiderivative gives

<math display=block>\int_{0}^{a/2} \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin \left( \frac{x}{a} \right) \Biggl|_{0}^{a/2}
= \arcsin \left ( \frac{1}{2}\right) - \arcsin (0) = \frac{\pi}{6}</math> as before.