# Integration by Trigonometric Substitution

Integrals by Trig Substitution Examples

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by turksvids

## Key Questions

• 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:

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

• Useful Trigonometric Identities

${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

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

$\sin 2 \theta = 2 \sin \theta \cos \theta$

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

${\cos}^{2} \theta = \frac{1}{2} \left(1 + \cos 2 \theta\right)$

${\sin}^{2} \theta = \frac{1}{2} \left(1 - \cos 2 \theta\right)$

I hope that this was helpful.

• By completing the square,
${t}^{2} - 6 t + 13 = \left({t}^{2} - 6 t + 9\right) + 4 = {\left(t - 3\right)}^{2} + {2}^{2}$

So, we can rewrite the integral as
$\int \frac{\mathrm{dt}}{\sqrt{{\left(t - 3\right)}^{2} + {2}^{2}}}$

Let $t - 3 = 2 \tan \theta$.
Rightarrow {dt}/{d theta}=2sec^2 theta Rightarrow dt=2sec^2 theta d theta

by the above substitution,
=int{2sec^2theta d theta}/{sqrt{(2tan theta)^2+2^2)} =int{sec^2theta}/{sqrt{tan^2theta+1}}d theta
by the identity ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$,
$= \int \frac{{\sec}^{2} \theta}{\sqrt{{\sec}^{2} \theta}} d \theta = \int \sec \theta d \theta = \ln | \sec \theta + \tan \theta | + {C}_{1}$
by using
$\tan \theta = \frac{t - 3}{2}$ and $\sec \theta = \sqrt{{\tan}^{2} \theta + 1} = \frac{\sqrt{{t}^{2} - 6 + 13}}{2}$,
we have
$= \ln | \frac{\sqrt{{t}^{2} - 6 + 13}}{2} + \frac{t - 3}{2} | + {C}_{1}$
$= \ln | \frac{\sqrt{{t}^{2} - 6 t + 13} + t - 3}{2} | + {C}_{1}$
by log property,
$= \ln | \sqrt{{t}^{2} - 6 t + 13} + t - 3 | - \ln 2 + {C}_{1}$
by setting $C = \ln 2 + {C}_{1}$,
$= \ln | \sqrt{{t}^{2} - 6 t + 13} + t - 3 | + C$

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