Integration Techniques: (lesson 4 of 4)
Trigonometric Substitutions
Use the trigonometric substitution to evaluate integrals involving the
radicals,
a2−x2, a2+x2, x2−a2
Case I: a2−x2
substitution: x=asinθ
a2−x2=a2−(asinθ)2=a1−(sinθ)2=acosθ
Example 1:
∫x29−x2dx
Solution:
substitution: x=3sinθ
9−x2=3cosθ
Then,
∫x29−x2dx=∫[3sinθ]2⋅3cosθd(3sinθ)=∫27sin2θcosθ3cosθdθ==31∫sin2θ1dθ=9−cotθ+C==9x−9−x2+C
Example 2:
∫x24−x2dx
Solution:
Let x=2sinθ, dx=2cosθdθ
∫4sin2θ4−4sin2θ2cosθdθ=∫4sin2θdθ=41∫csc2θ dθ=−41cotθ+C
Case II: a2+x2
substitution: x=atanθ
Example 3:
∫(x2+1)3dx
Solution:
substitution: x=tanθ
1+x2=secθ
Then,
Case III: x2−a2
substitution: x=asecθ
a2+x2=a2+(atanθ)2=a1+(sinθ)2=asecθ
Example 4:
∫(x2−4x)3dx
Solution:
x2−4x=x2−4x+4−4=(x−2)2−22↔2x−2=secθ↔x=2+2secθ↔2x2−4x=tanθ↔x2−4=2tanθ
Then,
∫(x2−4x)2dx=∫[2tanθ]3d(2+2secθ)==∫8tan3θ2secθtanθdθ=41∫sin2θsecθdθ==41∫cosθ1⋅sin2θcos2θdθ=41∫sin2θcosθdθ==41∫sin2θ1dsinθ=u=sinΘ41∫u21du==4u−1+C=4sinθ−1+C=4⋅(x−2x2−4x)−1+C==4x2−4x−(x−2)+C