2.2: Trigonometric Substitution
( \newcommand{\kernel}{\mathrm{null}\,}\)
When we have integrals that involve the square root term
√a2+x2
we may be able to trigonometric substitution to solve the integral.
Solve
∫√1−x2dx
by substituting x=sinθ and dx=cosθdθ.
The integrand then becomes
√1−x2=√1−sin2θ=√cos2θ=cosθ
We have
∫√1−x2dx=∫cosθcosθdθ=∫cos2θdθ=∫(12+12cos2θ)dθ=12θ+14sin2θ+C=12arcsinx+12sinθcosθ+C=12arcsinx+12x√1−x2+C
- ∫√1−x2x4dx
- ∫1√4−9x2dx
Two Key Formulas
From Trigonometry, we have the following two key formulas:
sec2x=1+tan2x
so
secx=√1+tan2x
and
tan2x=sec2x−1
so
tanx=√sec2x−1.
When we have integrals that involve any of the above square roots, we can use the appropriate substitution.
∫x3√1+x2dxx=tanθ,dx=sec2θdθ√1+x2=√1+tan2θ=√sec2θ=secθ=∫tan3θsecθsec2θdθ=∫tan3θsecθdθ=∫tan2θtanθsecθdθ=∫(sec2θ−1)secθtanθdθu=secθ,du=secθtanθdθ=∫(u2−1)du=u33−u+C=sec3θ3−secθ+C=(1+x2)323−√1+x2+C
- ∫x3√x2−1dx
- ∫x2√9+4x2dx
- ∫1√x2+2xdx
Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.