Editing Trigonometric substitution (section)

==Case I: Integrands containing ''a''<sup>2</sup> − ''x''<sup>2</sup>==
Let <math>x = a \sin \theta,</math> and use the [[list of trigonometric identities|identity]] <math>1-\sin^2 \theta = \cos^2 \theta.</math>

===Examples of Case I===
[[File:Trig Sub Triangle 1.png|thumb|Geometric construction for Case I]]

====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.

====Example 2====
The integral

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

may be evaluated by letting <math display="inline">x=a\sin \theta,\, dx=a\cos\theta\, d\theta,\, \theta=\arcsin\dfrac{x}{a},</math> where <math>a > 0</math> so that <math display="inline">\sqrt{a^2}=a,</math> and <math display="inline">-\pi/2 \le \theta \le \pi/2</math> by the range of arcsine, so that <math>\cos \theta \ge 0</math> and <math display="inline">\sqrt{\cos^2 \theta} = \cos \theta.</math>

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

For a definite integral, the bounds change once the substitution is performed and are determined using the equation <math display="inline">\theta = \arcsin\dfrac{x}{a},</math> with values in the range <math display="inline">-\pi/2 \le \theta \le \pi/2.</math> Alternatively, apply the boundary terms directly to the formula for the antiderivative.

For example, the definite integral

<math display=block>\int_{-1}^1\sqrt{4-x^2}\,dx,</math>

may be evaluated by substituting <math>x = 2\sin\theta, \,dx = 2\cos\theta\,d\theta,</math> with the bounds determined using <math display="inline">\theta = \arcsin\dfrac{x}{2}.</math>

Because <math>\arcsin(1/{2}) = \pi/6</math> and <math>\arcsin(-1/2) = -\pi/6,</math>
<math display=block>\begin{align}
 \int_{-1}^1\sqrt{4-x^2}\,dx &= \int_{-\pi/6}^{\pi/6}\sqrt{4-4\sin^2\theta}\,(2\cos\theta) \,d\theta \\[6pt]
   &= \int_{-\pi/6}^{\pi/6}\sqrt{4(1-\sin^2\theta)}\,(2\cos\theta) \,d\theta \\[6pt]
   &= \int_{-\pi/6}^{\pi/6}\sqrt{4(\cos^2\theta)}\,(2\cos\theta) \,d\theta \\[6pt]
   &= \int_{-\pi/6}^{\pi/6}(2\cos\theta)(2\cos\theta) \,d\theta \\[6pt]
   &= 4\int_{-\pi/6}^{\pi/6}\cos^2\theta\,d\theta \\[6pt]
   &= 4\int_{-\pi/6}^{\pi/6}\left(\frac{1+\cos 2\theta}{2}\right)\,d\theta \\[6pt]
   &= 2 \left[\theta+\frac{1}{2} \sin 2\theta \right]^{\pi/6}_{-\pi/6}
 = [2\theta+\sin 2\theta] \Biggl |^{\pi/6}_{-\pi/6} \\[6pt]
 &= \left(\frac{\pi}{3}+\sin\frac{\pi}{3}\right)-\left(-\frac{\pi}{3}+\sin\left(-\frac{\pi}{3}\right)\right)
 = \frac{2\pi}{3}+\sqrt{3}.
 \end{align}</math>

On the other hand, direct application of the boundary terms to the previously obtained formula for the antiderivative yields
<math display=block>\begin{align}
\int_{-1}^1\sqrt{4-x^2}\,dx &= \left[ \frac{2^2}{2}\arcsin\frac{x}{2}+\frac{x}{2}\sqrt{2^2-x^2} \right]_{-1}^{1}\\[6pt]
&= \left( 2 \arcsin \frac{1}{2} + \frac{1}{2}\sqrt{4-1}\right) - \left( 2 \arcsin \left(-\frac{1}{2}\right) + \frac{-1}{2}\sqrt{4-1}\right)\\[6pt]
&= \left( 2 \cdot \frac{\pi}{6} + \frac{\sqrt{3}}{2}\right) - \left( 2\cdot \left(-\frac{\pi}{6}\right) - \frac{\sqrt 3}{2}\right)\\[6pt]
&= \frac{2\pi}{3} + \sqrt{3}
\end{align} </math>
as before.