# How do we solve an integral using trigonometric substitution?

May 25, 2018

In general trigonometric substitutions are useful to solve the integrals of algebraic functions containing radicals in the form $\sqrt{{x}^{2} \pm {a}^{2}}$ or $\sqrt{{a}^{2} \pm {x}^{2}}$. Consider the different cases:

A. Let $f \left(x\right)$ be a rational function of $x$ and $\sqrt{{x}^{2} + {a}^{2}}$:

$\int f \left(x\right) \mathrm{dx} = \int R \left(x , \sqrt{{x}^{2} + {a}^{2}}\right) \mathrm{dx}$

Substitute: $x = a \tan t$, $\mathrm{dx} = a {\sec}^{2} t \mathrm{dt}$ with $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ and use the trigonometric identity:

$1 + {\tan}^{2} t = {\sec}^{2} t$

Considering that for $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ the secant is positive:

$\sqrt{{x}^{2} + {a}^{2}} = \sqrt{{a}^{2} {\tan}^{2} t + {a}^{2}}$

$\sqrt{{x}^{2} + {a}^{2}} = a \sqrt{{\tan}^{2} t + 1} = a \sec t$

Then:

$\int f \left(x\right) \mathrm{dx} = \int R \left(a \tan t , a \sec t\right) {\sec}^{2} t \mathrm{dt}$

B. Let $f \left(x\right)$ be a rational function of $x$ and $\sqrt{{x}^{2} - {a}^{2}}$:

$\int f \left(x\right) \mathrm{dx} = \int R \left(x , \sqrt{{x}^{2} - {a}^{2}}\right) \mathrm{dx}$

Restrict the function to $x \in \left(a , + \infty\right)$ and substitute: $x = a \sec t$, $\mathrm{dx} = a \sec t \tan t \mathrm{dt}$ with $t \in \left(0 , \frac{\pi}{2}\right)$ and use the trigonometric identity:

${\sec}^{2} t - 1 = {\tan}^{2} t$

Considering that for $t \in \left(0 , \frac{\pi}{2}\right)$ the tangent is positive:

$\sqrt{{x}^{2} - {a}^{2}} = \sqrt{{a}^{2} {\sec}^{2} t - {a}^{2}}$

$\sqrt{{x}^{2} - {a}^{2}} = a \sqrt{{\sec}^{2} t - 1} = a \tan t$

Then:

$\int f \left(x\right) \mathrm{dx} = \int R \left(a \sec t , a \tan t\right) \sec t \tan t \mathrm{dt}$

Normally you can see by differentiation that the solution that is found is valid also for $x \in \left(- \infty , - a\right)$

C. Let $f \left(x\right)$ be a rational function of $x$ and $\sqrt{{a}^{2} - {x}^{2}}$:

$\int f \left(x\right) \mathrm{dx} = \int R \left(x , \sqrt{{a}^{2} - {x}^{2}}\right) \mathrm{dx}$

Substitute: $x = a \sin t$, $\mathrm{dx} = a \cos t$ with $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ and use the trigonometric identity:

$1 - {\sin}^{2} t = {\cos}^{2} t$

Considering that for $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ the cosine is positive:

$\sqrt{{a}^{2} - {x}^{2}} = \sqrt{{a}^{2} - {a}^{2} {\sin}^{2} t}$

$\sqrt{{a}^{2} - {x}^{2}} = a \sqrt{1 - {\sin}^{2} t} = a \cos t$

Then:

$\int f \left(x\right) \mathrm{dx} = \int R \left(a \sin t , a \cos t\right) \cos t \mathrm{dt}$