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« Integrals Involving Trigonometric Functions
Integration Techniques: (lesson 4 of 4)

Trigonometric Substitutions

Use the trigonometric substitution to evaluate integrals involving the radicals,

a2x2,  a2+x2,  x2a2 \sqrt{a^2 - x^2} , \ \ \sqrt{a^2 + x^2} , \ \ \sqrt{x^2 - a^2}

Case I: a2x2\sqrt{a^2 - x^2}

substitution: x=asinθ\color{red}{x = a \sin \theta} a2x2=a2(asinθ)2=a1(sinθ)2=acosθ \sqrt{a^2 - x^2} = \sqrt{a^2 - (a \sin \theta)^2} = a \sqrt{1 - (\sin \theta)^2} = a \cos \theta

Example 1:

dxx29x2 \int \frac{dx}{x^2 \sqrt{9 - x^2}}

Solution:

substitution: x=3sinθ\color{red}{x = 3 \sin \theta} 9x2=3cosθ \sqrt{9 - x^2} = 3 \cos \theta

Then,

dxx29x2=d(3sinθ)[3sinθ]23cosθ=3cosθdθ27sin2θcosθ==131sin2θdθ=cotθ9+C==9x29x+C \begin{aligned} \int \frac{dx}{x^2 \sqrt {9 - x^2}} &= \int \frac{d (3 \sin \theta)}{[ 3 \sin \theta]^2 \cdot 3 \cos \theta} = \int \frac{3 \cos \theta d \theta}{27 \sin^2 \theta \cos \theta} = \\ &= \frac{1}{3} \int \frac{1}{sin^2 \theta} d \theta = \frac{- \cot \theta}{9} + C = \\ &= \frac{-\sqrt{9 - x^2}}{9x} + C \end{aligned}

Example 2:

dxx24x2 \int \frac{dx}{x^2 \sqrt{4 - x^2}}

Solution:

Let x=2sinθx = 2 \sin \theta, dx=2cosθdθdx = 2 \cos \theta d \theta

2cosθdθ4sin2θ44sin2θ=dθ4sin2θ=14csc2θ   dθ=14cotθ+C \int \frac{2 \cos \theta d \theta}{4 \sin^2 \theta \sqrt{4 - 4 \sin^2 \theta}} = \int \frac{d \theta}{4 \sin^2 \theta} = \frac{1}{4} \int \csc^2 \theta \ \ \ d \theta = - \frac{1}{4} \cot \theta + C

Case II: a2+x2\sqrt{a^2 + x^2}

substitution: x=atanθ\color{red}{x = a \tan \theta}

Example 3:

dx(x2+1)3 \int \frac{dx}{(\sqrt{x^2 + 1})^3}

Solution:

substitution: x=tanθx = \tan \theta 1+x2=secθ \sqrt{1 + x^2} = \sec \theta

Then,

Case III: x2a2\sqrt{x^2 - a^2}

substitution: x=asecθ\color{red}{x = a \sec \theta} a2+x2=a2+(atanθ)2=a1+(sinθ)2=asecθ \sqrt{a^2 + x^2} = \sqrt{a^2 + (a \tan \theta)^2} = a \sqrt{1 + (\sin \theta)^2} = a \sec \theta

Example 4:

dx(x24x)3 \int \frac{dx}{(\sqrt{x^2 - 4x})^3}

Solution:

x24x=x24x+44=(x2)222x22=secθx=2+2secθx24x2=tanθx24=2tanθ x^2 - 4x = x^2 - 4x + 4 - 4 = (x - 2)^2 - 2^2 \\ \leftrightarrow \frac{x - 2}{2} = \sec \theta \leftrightarrow x = 2 + 2 \sec \theta \\ \leftrightarrow \frac{\sqrt{x^2 - 4x}}{2} = \tan \theta \leftrightarrow \sqrt{x^2 - 4} = 2 \tan \theta

Then,

dx(x24x)2=d(2+2secθ)[2tanθ]3==2secθtanθ8tan3θdθ=14secθsin2θdθ==141cosθcos2θsin2θdθ=14cosθsin2θdθ==141sin2θdsinθ=u=sinΘ141u2du==14u+C=14sinθ+C=14(x24xx2)+C==(x2)4x24x+C \begin{aligned} \int \frac{dx}{(\sqrt{x^2 - 4x})^2} &= \int \frac{d (2 + 2 \sec \theta)}{[2 \tan \theta]^3} = \\ &= \int \frac{2 \sec \theta \tan \theta}{8 \tan^3 \theta}d \theta = \frac{1}{4} \int \frac{\sec \theta}{\sin^2 \theta} d \theta = \\ &= \frac{1}{4} \int \frac{1}{\cos \theta} \cdot \frac{\cos^2 \theta}{\sin^2 \theta} d \theta = \frac{1}{4} \int \frac{\cos \theta}{\sin^2 \theta} d \theta = \\ &= \frac{1}{4} \int \frac{1}{\sin^2 \theta} d \sin \theta \mathop = \limits^{u = \sin \Theta } \frac{1}{4} \int \frac{1}{u^2} du = \\ &= \frac{-1}{4u} + C = \frac{-1}{4 \sin \theta} + C = \frac{-1}{4 \cdot \left( \frac{\sqrt{x^2 - 4x}}{x-2} \right)} + C = \\ &= \frac{- (x - 2)}{4 \sqrt{x^2 - 4x}} + C \end{aligned}