Trigonometric substitution
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}} In mathematics, a trigonometric substitution replaces a trigonometric function for another expression. In calculus, trigonometric substitutions are a technique for evaluating integrals. In this case, an expression involving a radical function is replaced with a trigonometric one. Trigonometric identities may help simplify the answer.<ref>Template:Cite book</ref><ref>Template:Cite book</ref>
In the case of a definite integral, this method of integration by substitution uses the substitution to change the interval of integration. Alternatively, the antiderivative of the integrand may be applied to the original interval.
Case I: Integrands containing a2 − x2
Let <math>x = a \sin \theta,</math> and use the identity <math>1-\sin^2 \theta = \cos^2 \theta.</math>
Examples of 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.
Case II: Integrands containing a2 + x2
Let <math>x = a \tan \theta,</math> and use the identity <math>1+\tan^2 \theta = \sec^2 \theta.</math>
Examples of Case II
Example 1
In the integral
<math display=block>\int\frac{dx}{a^2+x^2}</math>
we may write
<math display=block>x=a\tan\theta,\quad dx=a\sec^2\theta\, d\theta, \quad \theta=\arctan\frac{x}{a},</math>
so that the integral becomes
<math display=block>\begin{align}
\int\frac{dx}{a^2+x^2} &= \int\frac{a\sec^2\theta\, d\theta}{a^2 + a^2\tan^2\theta} \\[6pt]
&= \int\frac{a\sec^2\theta\, d\theta}{a^2(1+\tan^2\theta)} \\[6pt]
&= \int\frac{a\sec^2\theta\, d\theta}{a^2\sec^2\theta} \\[6pt]
&= \int\frac{d\theta}{a} \\[6pt]
&= \frac{\theta}{a}+C \\[6pt]
&= \frac{1}{a} \arctan \frac{x}{a} + C,
\end{align}</math>
provided <math>a \neq 0.</math>
For a definite integral, the bounds change once the substitution is performed and are determined using the equation <math>\theta = \arctan\frac{x}{a},</math> with values in the range <math>-\frac{\pi}{2} < \theta < \frac{\pi}{2}.</math> Alternatively, apply the boundary terms directly to the formula for the antiderivative.
For example, the definite integral
<math display=block>\int_0^1\frac{4\, dx}{1+x^2}\,</math>
may be evaluated by substituting <math>x = \tan\theta, \,dx = \sec^2\theta\,d\theta,</math> with the bounds determined using <math>\theta = \arctan x.</math>
Since <math>\arctan 0 = 0</math> and <math>\arctan 1 = \pi/4,</math> <math display=block>\begin{align}
\int_0^1\frac{4\,dx}{1+x^2} &= 4\int_0^1\frac{dx}{1 + x^2} \\[6pt]
&= 4\int_0^{\pi/4}\frac{\sec^2\theta\, d\theta}{1+\tan^2\theta} \\[6pt]
&= 4\int_0^{\pi/4}\frac{\sec^2\theta\, d\theta}{\sec^2\theta} \\[6pt]
&= 4\int_0^{\pi/4}d\theta \\[6pt]
&= (4\theta)\Bigg|^{\pi/4}_0 = 4 \left (\frac{\pi}{4} - 0 \right) = \pi.
\end{align}</math>
Meanwhile, direct application of the boundary terms to the formula for the antiderivative yields <math display="block">\begin{align} \int_0^1\frac{4\,dx}{1+x^2}\, &= 4\int_0^1\frac{dx}{1+x^2} \\[6pt] &= 4\left[\frac{1}{1} \arctan \frac{x}{1} \right]^1_0 \\[6pt] &= 4(\arctan x)\Bigg|^1_0 \\[6pt] &= 4(\arctan 1 - \arctan 0) \\[6pt] &= 4 \left (\frac{\pi}{4} - 0 \right) = \pi, \end{align}</math> same as before.
Example 2
The integral
<math display=block>\int\sqrt{a^2+x^2}\,{dx}</math>
may be evaluated by letting <math>x=a\tan\theta,\, dx=a\sec^2\theta\, d\theta, \, \theta=\arctan\frac{x}{a},</math>
where <math>a > 0</math> so that <math>\sqrt{a^2}=a,</math> and <math>-\frac{\pi}{2}<\theta<\frac{\pi}{2}</math> by the range of arctangent, so that <math>\sec \theta > 0</math> and <math>\sqrt{\sec^2 \theta} = \sec \theta.</math>
Then, <math display=block>\begin{align}
\int\sqrt{a^2+x^2}\,dx &= \int\sqrt{a^2 + a^2\tan^2\theta}\,(a \sec^2\theta)\, d\theta \\[6pt]
&= \int\sqrt{a^2 (1+\tan^2\theta)}\,(a \sec^2\theta)\, d\theta \\[6pt]
&= \int\sqrt{a^2 \sec^2\theta}\,(a \sec^2\theta)\, d\theta \\[6pt]
&= \int(a \sec\theta)(a \sec^2\theta)\, d\theta \\[6pt]
&= a^2\int \sec^3\theta\, d\theta. \\[6pt]
\end{align}</math>
The integral of secant cubed may be evaluated using integration by parts. As a result, <math display=block>\begin{align}
\int\sqrt{a^2+x^2}\,dx &= \frac{a^2}{2}(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|)+C \\[6pt]
&= \frac{a^2}{2}\left(\sqrt{1+\frac{x^2}{a^2}}\cdot\frac{x}{a} + \ln\left|\sqrt{1+\frac{x^2}{a^2}}+\frac{x}{a}\right|\right)+C \\[6pt]
&= \frac{1}{2}\left(x\sqrt{a^2+x^2} + a^2\ln\left|\frac{x+\sqrt{a^2+x^2}}{a}\right|\right)+C.
\end{align}</math>
To illustrate an application of this formula, suppose we wish to calculate the arc length of the parabola <math>y=x^2</math> from <math>x=0</math> and <math>x=2</math>. Denoting the arc length as <math>s</math>, we get
<math display=block> s = \int_{0}^{2} \sqrt{1 + (y')^2} dx = \int_{0}^{2} \sqrt{1 + 4x^2} dx = 2 \int_{0}^{2} \sqrt{\frac{1}{4} + x^2} dx. </math>
In this instance, we have <math>a=\frac{1}{2}</math>, which gives
<math display=block> \begin{align} s &= \left(x\sqrt{\frac{1}{4}+x^2} + \frac{1}{4}\ln\left|\frac{x+\sqrt{\frac{1}{4}+x^2}}{\frac{1}{2}}\right|\right)_{x=0}^{x=2} \\
&= 2\sqrt{\frac{17}{4}} + \frac{1}{4}\ln\left( 4 + 2\sqrt{\frac{17}{4}} \right) \\
&= \sqrt{17} + \frac{1}{4}\ln(4 + \sqrt{17}) \\
&\approx 4.64678 \\
\end{align} </math>
Case III: Integrands containing x2 − a2
Let <math>x = a \sec \theta,</math> and use the identity <math>\sec^2 \theta -1 = \tan^2 \theta.</math>
Examples of Case III
Integrals such as
<math display=block>\int\frac{dx}{x^2 - a^2}</math>
can also be evaluated by partial fractions rather than trigonometric substitutions. However, the integral
<math display=block>\int\sqrt{x^2 - a^2}\, dx</math>
cannot. In this case, an appropriate substitution is: <math display=block>x = a \sec\theta,\, dx = a \sec\theta\tan\theta\, d\theta, \, \theta = \arcsec\frac{x}{a},</math>
where <math>a > 0</math> so that <math>\sqrt{a^2}=a,</math> and <math>0 \le \theta < \frac{\pi}{2}</math> by assuming <math>x > 0,</math> so that <math>\tan \theta \ge 0</math> and <math>\sqrt{\tan^2 \theta} = \tan \theta.</math>
Then, <math display=block>\begin{align}
\int\sqrt{x^2 - a^2}\, dx &= \int\sqrt{a^2 \sec^2\theta - a^2} \cdot a \sec\theta\tan\theta\, d\theta \\
&= \int\sqrt{a^2 (\sec^2\theta - 1)} \cdot a \sec\theta\tan\theta\, d\theta \\
&= \int\sqrt{a^2 \tan^2\theta} \cdot a \sec\theta\tan\theta\, d\theta \\
&= \int a^2 \sec\theta\tan^2\theta\, d\theta \\
&= a^2 \int (\sec\theta)(\sec^2\theta - 1)\, d\theta \\
&= a^2 \int (\sec^3\theta - \sec\theta)\, d\theta.
\end{align}</math>
One may evaluate the integral of the secant function by multiplying the numerator and denominator by <math>( \sec \theta + \tan \theta)</math> and the integral of secant cubed by parts.<ref name=":1">Template:Cite book</ref> As a result, <math display=block>\begin{align}
\int\sqrt{x^2-a^2}\,dx &= \frac{a^2}{2}(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|)-a^2\ln|\sec\theta+\tan\theta|+C \\[6pt]
&= \frac{a^2}{2}(\sec\theta \tan\theta - \ln|\sec\theta+\tan\theta|)+C \\[6pt]
&= \frac{a^2}{2}\left(\frac{x}{a}\cdot\sqrt{\frac{x^2}{a^2}-1} - \ln\left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right|\right)+C \\[6pt]
&= \frac{1}{2}\left(x\sqrt{x^2-a^2} - a^2\ln\left|\frac{x+\sqrt{x^2-a^2}}{a}\right|\right)+C.
\end{align}</math>
When <math>\frac{\pi}{2} < \theta \le \pi,</math> which happens when <math>x < 0</math> given the range of arcsecant, <math>\tan \theta \le 0,</math> meaning <math>\sqrt{\tan^2 \theta} = -\tan \theta</math> instead in that case.
Substitutions that eliminate trigonometric functions
Substitution can be used to remove trigonometric functions.
For instance,
<math display="block">\begin{align} \int f(\sin(x), \cos(x))\, dx &=\int\frac1{\pm\sqrt{1-u^2}} f\left(u,\pm\sqrt{1-u^2}\right)\, du && u=\sin (x) \\[6pt] \int f(\sin(x), \cos(x))\, dx &=\int\frac{1}{\mp\sqrt{1-u^2}} f\left(\pm\sqrt{1-u^2},u\right)\, du && u=\cos (x) \\[6pt] \int f(\sin(x), \cos(x))\, dx &=\int\frac2{1+u^2} f \left(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2}\right)\, du && u=\tan\left (\frac{x}{2} \right ) \\[6pt] \end{align}</math>
The last substitution is known as the Weierstrass substitution, which makes use of tangent half-angle formulas.
For example,
<math display=block>\begin{align} \int\frac{4 \cos x}{(1+\cos x)^3}\, dx &= \int\frac2{1+u^2}\frac{4\left(\frac{1-u^2}{1+u^2}\right)}{\left(1+\frac{1-u^2}{1+u^2}\right)^3}\, du = \int (1-u^2)(1+u^2)\, du \\&= \int (1-u^4)\,du = u - \frac{u^5}{5} + C = \tan \frac{x}{2} - \frac{1}{5} \tan^5 \frac{x}{2} + C. \end{align}</math>
Hyperbolic substitution
Substitutions of hyperbolic functions can also be used to simplify integrals.<ref>Template:Cite web</ref>
For example, to integrate <math>1/\sqrt{a^2+x^2}</math>, introduce the substitution <math>x=a\sinh{u}</math> (and hence <math>dx=a\cosh u \,du</math>), then use the identity <math>\cosh^2 (x) - \sinh^2 (x) = 1</math> to find:
<math display="block">\begin{align} \int \frac{dx}{\sqrt{a^2+x^2}} &= \int \frac{a\cosh u \,du}{\sqrt{a^2+a^2\sinh^2 u}} \\[6pt] &=\int \frac{\cosh{u} \,du}{\sqrt{1+\sinh^2{u}}} \\[6pt] &=\int \frac{\cosh{u}}{\cosh u} \,du \\[6pt] &=u+C \\[6pt] &=\sinh^{-1}{\frac{x}{a}} + C. \end{align}</math>
If desired, this result may be further transformed using other identities, such as using the relation <math>\sinh^{-1}{z} = \operatorname{arsinh}{z} = \ln(z + \sqrt{z^2 + 1})</math>: <math display="block">\begin{align} \sinh^{-1}{\frac{x}{a}} + C &=\ln\left(\frac{x}{a} + \sqrt{\frac{x^2}{a^2} + 1}\,\right) + C \\[6pt] &=\ln\left(\frac{x + \sqrt{x^2+a^2}}{a}\,\right) + C. \end{align}</math>
See also
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