Trigonometric substitution

Case II: Integrands containing a2 + x2

Let ${\displaystyle x=a\tan \theta ,}$ and use the identity ${\displaystyle 1+{\tan }^2\theta ={\sec }^2\theta .}$

Examples of Case II

Example 1

In the integral

$${\displaystyle \int \frac{dx}{a^2+x^2}}$$

we may write

$${\displaystyle x=a\tan \theta ,dx=a{\sec }^2\theta d\theta ,\theta =\arctan \frac{x}{a},}$$

so that the integral becomes

$${\displaystyle \begin{array}{cc}\int \frac{dx}{a^2+x^2}& =\int \frac{a{\sec }^2\theta d\theta }{a^2+a^2{\tan }^2\theta }\\ & =\int \frac{a{\sec }^2\theta d\theta }{a^2(1+{\tan }^2\theta )}\\ & =\int \frac{a{\sec }^2\theta d\theta }{a^2{\sec }^2\theta }\\ & =\int \frac{d\theta }{a}\\ & =\frac{\theta }{a}+C\\ & =\frac{1}{a}\arctan \frac{x}{a}+C,\end{array}}$$

provided ${\displaystyle a\ne 0.}$

For a definite integral, the bounds change once the substitution is performed and are determined using the equation ${\displaystyle \theta =\arctan \frac{x}{a},}$ with values in the range ${\displaystyle -\frac{\pi }{2}<\theta <\frac{\pi }{2}.}$ Alternatively, apply the boundary terms directly to the formula for the antiderivative.

For example, the definite integral

$${\displaystyle {\int }_0^1\frac{4dx}{1+x^2}}$$

may be evaluated by substituting ${\displaystyle x=\tan \theta ,dx={\sec }^2\theta d\theta ,}$ with the bounds determined using ${\displaystyle \theta =\arctan x.}$

Since ${\displaystyle \arctan 0=0}$ and ${\displaystyle \arctan 1=\pi /4,}$ $${\displaystyle \begin{array}{cc}{\int }_0^1\frac{4dx}{1+x^2}& =4{\int }_0^1\frac{dx}{1+x^2}\\ & =4{\int }_0^{\pi /4}\frac{{\sec }^2\theta d\theta }{1+{\tan }^2\theta }\\ & =4{\int }_0^{\pi /4}\frac{{\sec }^2\theta d\theta }{{\sec }^2\theta }\\ & =4{\int }_0^{\pi /4}d\theta \\ & =(4\theta ){|}_0^{\pi /4}=4\left(\frac{\pi }{4}-0\right)=\pi .\end{array}}$$

Meanwhile, direct application of the boundary terms to the formula for the antiderivative yields $${\displaystyle \begin{array}{cc}{\int }_0^1\frac{4dx}{1+x^2}& =4{\int }_0^1\frac{dx}{1+x^2}\\ & =4{\left[\frac{1}{1}\arctan \frac{x}{1}\right]}_0^1\\ & =4(\arctan x){|}_0^1\\ & =4(\arctan 1-\arctan 0)\\ & =4\left(\frac{\pi }{4}-0\right)=\pi ,\end{array}}$$ same as before.

Example 2

The integral

$${\displaystyle \int \sqrt{a^2+x^2}dx}$$

may be evaluated by letting ${\displaystyle x=a\tan \theta ,dx=a{\sec }^2\theta d\theta ,\theta =\arctan \frac{x}{a},}$

where ${\displaystyle a>0}$ so that ${\displaystyle \sqrt{a^2}=a,}$ and ${\displaystyle -\frac{\pi }{2}<\theta <\frac{\pi }{2}}$ by the range of arctangent, so that ${\displaystyle \sec \theta >0}$ and ${\displaystyle \sqrt{{\sec }^2\theta }=\sec \theta .}$

Then, $${\displaystyle \begin{array}{cc}\int \sqrt{a^2+x^2}dx& =\int \sqrt{a^2+a^2{\tan }^2\theta }(a{\sec }^2\theta )d\theta \\ & =\int \sqrt{a^2(1+{\tan }^2\theta )}(a{\sec }^2\theta )d\theta \\ & =\int \sqrt{a^2{\sec }^2\theta }(a{\sec }^2\theta )d\theta \\ & =\int (a\sec \theta )(a{\sec }^2\theta )d\theta \\ & =a^2\int {\sec }^3\theta d\theta .\end{array}}$$ The integral of secant cubed may be evaluated using integration by parts. As a result, $${\displaystyle \begin{array}{cc}\int \sqrt{a^2+x^2}dx& =\frac{a^2}{2}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)+C\\ & =\frac{a^2}{2}\left(\sqrt{1+\frac{x^2}{a^2}}\cdot \frac{x}{a}+\ln \left|\sqrt{1+\frac{x^2}{a^2}}+\frac{x}{a}\right|\right)+C\\ & =\frac{1}{2}\left(x\sqrt{a^2+x^2}+a^2\ln \left|\frac{x+\sqrt{a^2+x^2}}{a}\right|\right)+C.\end{array}}$$

To illustrate an application of this formula, suppose we wish to calculate the arc length of the parabola ${\displaystyle y=x^2}$ from ${\displaystyle x=0}$ and ${\displaystyle x=2}$. Denoting the arc length as ${\displaystyle s}$, we get

$${\displaystyle s={\int }_0^2\sqrt{1+(y^{\prime }{)}^2}dx={\int }_0^2\sqrt{1+4x^2}dx=2{\int }_0^2\sqrt{\frac{1}{4}+x^2}dx.}$$

In this instance, we have ${\displaystyle a=\frac{1}{2}}$, which gives

$${\displaystyle \begin{array}{cc}s& ={\left(x\sqrt{\frac{1}{4}+x^2}+\frac{1}{4}\ln \left|\frac{x+\sqrt{\frac{1}{4}+x^2}}{\frac{1}{2}}\right|\right)}_{x=0}^{x=2}\\ & =2\sqrt{\frac{17}{4}}+\frac{1}{4}\ln \left(4+2\sqrt{\frac{17}{4}}\right)\\ & =\sqrt{17}+\frac{1}{4}\ln (4+\sqrt{17})\\ & \approx 4.64678\end{array}}$$

Case III: Integrands containing x2 − a2

Let ${\displaystyle x=a\sec \theta ,}$ and use the identity ${\displaystyle {\sec }^2\theta -1={\tan }^2\theta .}$

Examples of Case III

Integrals such as

$${\displaystyle \int \frac{dx}{x^2-a^2}}$$

can also be evaluated by partial fractions rather than trigonometric substitutions. However, the integral

$${\displaystyle \int \sqrt{x^2-a^2}dx}$$

cannot. In this case, an appropriate substitution is: $${\displaystyle x=a\sec \theta ,dx=a\sec \theta \tan \theta d\theta ,\theta =\mathrm{arcsec}\frac{x}{a},}$$

where ${\displaystyle a>0}$ so that ${\displaystyle \sqrt{a^2}=a,}$ and ${\displaystyle 0\le \theta <\frac{\pi }{2}}$ by assuming ${\displaystyle x>0,}$ so that ${\displaystyle \tan \theta \ge 0}$ and ${\displaystyle \sqrt{{\tan }^2\theta }=\tan \theta .}$

Then, $${\displaystyle \begin{array}{cc}\int \sqrt{x^2-a^2}dx& =\int \sqrt{a^2{\sec }^2\theta -a^2}\cdot a\sec \theta \tan \theta d\theta \\ & =\int \sqrt{a^2({\sec }^2\theta -1)}\cdot a\sec \theta \tan \theta d\theta \\ & =\int \sqrt{a^2{\tan }^2\theta }\cdot a\sec \theta \tan \theta d\theta \\ & =\int a^2\sec \theta {\tan }^2\theta d\theta \\ & =a^2\int (\sec \theta )({\sec }^2\theta -1)d\theta \\ & =a^2\int ({\sec }^3\theta -\sec \theta )d\theta .\end{array}}$$

One may evaluate the integral of the secant function by multiplying the numerator and denominator by ${\displaystyle (\sec \theta +\tan \theta )}$ and the integral of secant cubed by parts. As a result, $${\displaystyle \begin{array}{cc}\int \sqrt{x^2-a^2}dx& =\frac{a^2}{2}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)-a^2\ln |\sec \theta +\tan \theta |+C\\ & =\frac{a^2}{2}(\sec \theta \tan \theta -\ln |\sec \theta +\tan \theta |)+C\\ & =\frac{a^2}{2}\left(\frac{x}{a}\cdot \sqrt{\frac{x^2}{a^2}-1}-\ln \left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right|\right)+C\\ & =\frac{1}{2}\left(x\sqrt{x^2-a^2}-a^2\ln \left|\frac{x+\sqrt{x^2-a^2}}{a}\right|\right)+C.\end{array}}$$

When ${\displaystyle \frac{\pi }{2}<\theta \le \pi ,}$ which happens when ${\displaystyle x<0}$ given the range of arcsecant, ${\displaystyle \tan \theta \le 0,}$ meaning ${\displaystyle \sqrt{{\tan }^2\theta }=-\tan \theta }$ instead in that case.

Substitutions that eliminate trigonometric functions

Substitution can be used to remove trigonometric functions.

For instance,

$${\displaystyle \begin{array}{cccc}\int f(\sin (x),\cos (x))dx& =\int \frac{1}{\pm \sqrt{1-u^2}}f\left(u,\pm \sqrt{1-u^2}\right)du& & u=\sin (x)\\ \int f(\sin (x),\cos (x))dx& =\int \frac{1}{\mp \sqrt{1-u^2}}f\left(\pm \sqrt{1-u^2},u\right)du& & u=\cos (x)\\ \int f(\sin (x),\cos (x))dx& =\int \frac{2}{1+u^2}f\left(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2}\right)du& & u=\tan \left(\frac{x}{2}\right)\end{array}}$$

The last substitution is known as the Weierstrass substitution, which makes use of tangent half-angle formulas.

For example,

$${\displaystyle \begin{array}{cc}\int \frac{4\cos x}{(1+\cos x{)}^3}dx& =\int \frac{2}{1+u^2}\frac{4\left(\frac{1-u^2}{1+u^2}\right)}{{\left(1+\frac{1-u^2}{1+u^2}\right)}^3}du=\int (1-u^2)(1+u^2)du\\ & =\int (1-u^4)du=u-\frac{u^5}{5}+C=\tan \frac{x}{2}-\frac{1}{5}{\tan }^5\frac{x}{2}+C.\end{array}}$$

Hyperbolic substitution

Substitutions of hyperbolic functions can also be used to simplify integrals.

For example, to integrate ${\displaystyle 1/\sqrt{a^2+x^2}}$, introduce the substitution ${\displaystyle x=a\sinh u}$ (and hence ${\displaystyle dx=a\cosh udu}$), then use the identity ${\displaystyle {\cosh }^2(x)-{\sinh }^2(x)=1}$ to find:

$${\displaystyle \begin{array}{cc}\int \frac{dx}{\sqrt{a^2+x^2}}& =\int \frac{a\cosh udu}{\sqrt{a^2+a^2{\sinh }^{2}u}}\\ & =\int \frac{\cosh udu}{\sqrt{1+{\sinh }^{2}u}}\\ & =\int \frac{\cosh u}{\cosh u}du\\ & =u+C\\ & ={\sinh }^{-1}\frac{x}{a}+C.\end{array}}$$

If desired, this result may be further transformed using other identities, such as using the relation ${\displaystyle {\sinh }^{-1}z=\mathrm{arsinh}z=\ln (z+\sqrt{z^2+1})}$: $${\displaystyle \begin{array}{cc}{\sinh }^{-1}\frac{x}{a}+C& =\ln \left(\frac{x}{a}+\sqrt{\frac{x^2}{a^2}+1}\right)+C\\ & =\ln \left(\frac{x+\sqrt{x^2+a^2}}{a}\right)+C.\end{array}}$$