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Concept Version 9
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Trigonometric Substitution

Trigonometric functions can be substituted for other expressions to change the form of integrands and simplify the integration.

Learning Objective

  • Use trigonometric substitution to solve an integral


Key Points

    • If the integrand contains $a^2 − x^2$, let x=asin(θ)x = a \sin(\theta)x=asin(θ).
    • If the integrand contains a2+x2a^2 + x^2a​2​​+x​2​​, let x=atan(θ)x = a \tan(\theta)x=atan(θ).
    • If the integrand contains x2−a2x^2 - a^2x​2​​−a​2​​, let x=asec(θ)x = a \sec(\theta)x=asec(θ).

Term

  • trigonometric

    relating to the functions used in trigonometry: sin\sinsin, cos\coscos, tan\tantan, csc\csccsc, cot\cotcot, sec\secsec


Full Text

Trigonometric functions can be substituted for other expressions to change the form of integrands. One may use the trigonometric identities to simplify certain integrals containing radical expressions (or expressions containing nnnth roots). The following are general methods of trigonometric substitution, depending on the form of the function to be integrated.

Substitution Rule #1

If the integral contains a2−x2a^2-x^2a​2​​−x​2​​, let x=asin(θ)x = a \sin(\theta)x=asin(θ) and use the identity: 

1−sin2(θ)=cos2(θ)1-\sin^2(\theta) = \cos^2(\theta)1−sin​2​​(θ)=cos​2​​(θ)

Substitution Rule #2

If the integrand contains a2+x2a^2+x^2a​2​​+x​2​​, let x=atan(θ)x = a \tan(\theta)x=atan(θ) and use the identity:

1+tan2(θ)=sec2(θ)1+\tan^2(\theta) = \sec^2(\theta)1+tan​2​​(θ)=sec​2​​(θ)

Substitution Rule #3

If the integrand contains x2−a2x^2-a^2x​2​​−a​2​​, let x=asec(θ)x = a \sec(\theta)x=asec(θ) and use the identity:

sec2(θ)−1=tan2(θ)\sec^2(\theta)-1 = \tan^2(\theta)sec​2​​(θ)−1=tan​2​​(θ)

Note that, for a definite integral, one must figure out how the bounds of integration change due to the substitution.

Examples

In order to better understand these substitutions, let's go over the derivation of some of them.

Example 1: Integrals where the integrand contains $a^2 − x^2$ (where aaa is positive)

In the integral 

∫dxa2−x2\displaystyle{\int\frac{dx}{\sqrt{a^2-x^2}}}∫​√​a​2​​−x​2​​​​​​​dx​​

we may use:

 $\displaystyle{x=a\sin(\theta)\\ dx=a\cos(\theta)\,d\theta\\ \theta=\arcsin\left(\frac{x}{a}\right)}$

With the substitution, we get:

∫dxa2−x2=∫acos(θ)dθa2−a2sin2(θ)=∫acos(θ)dθa2(1−sin2(θ))=∫acos(θ)dθa2cos2(θ)=∫dθ=θ+C=arcsin(xa)+C\begin{aligned} \int\frac{dx}{\sqrt{a^2-x^2}} & = \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2-a^2\sin^2(\theta)}} \\ &= \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2(1-\sin^2(\theta))}} \\ &= \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2\cos^2(\theta)}} \\ &= \int d\theta \\ &= \theta+C \\ &= \displaystyle{\arcsin \left(\frac{x}{a}\right)}+C \end{aligned}​∫​√​a​2​​−x​2​​​​​​​dx​​​​​​​​​​=∫​√​a​2​​−a​2​​sin​2​​(θ)​​​​​acos(θ)dθ​​​=∫​√​a​2​​(1−sin​2​​(θ))​​​​​acos(θ)dθ​​​=∫​√​a​2​​cos​2​​(θ)​​​​​acos(θ)dθ​​​=∫dθ​=θ+C​=arcsin(​a​​x​​)+C​​

Example 2: Integrals where the integrand contains $a^2 − x^2$ (where aaa is not zero)

In the integral 

∫dxa2+x2\displaystyle{\int\frac{dx}{{a^2+x^2}}}∫​a​2​​+x​2​​​​dx​​

we may use:

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

With the substitution, we get:

∫dxa2+x2=∫asec2(θ)dθa2+a2tan2(θ)=∫asec2(θ)dθa2(1+tan2(θ))=∫asec2(θ)dθa2sec2(θ)=∫dθa=θa+C=1aarctan(xa)+C\begin{aligned}\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 \left(\frac{x}{a}\right)+C\end{aligned}​∫​a​2​​+x​2​​​​dx​​​​​​​​​​=∫​a​2​​+a​2​​tan​2​​(θ)​​asec​2​​(θ)dθ​​​=∫​a​2​​(1+tan​2​​(θ))​​asec​2​​(θ)dθ​​​=∫​a​2​​sec​2​​(θ)​​asec​2​​(θ)dθ​​​=∫​a​​dθ​​​=​a​​θ​​+C​=​a​​1​​arctan(​a​​x​​)+C​​

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