Trigonometric Integrals and Trigonometric Substitution
Handle powers of sine and cosine with identities, and clear expressions such as √(a² - x²) by substituting a trigonometric function for x.
What a learner can do afterwards
- Reduce an odd power of sine by splitting off one factor and substituting
- Choose the substitution that matches a sum or difference of squares
- Convert the answer back to the original variable using a right triangle
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Powers of sine and cosine fall to one identity: sin^2 x + cos^2 x = 1. When a power is odd, split off a single factor to serve as du, convert the rest with the identity, and substitute. When the powers are even, half-angle formulas lower the degree instead. The formulas are sin^2 x = (1 - cos 2x) / 2 and cos^2 x = (1 + cos 2x) / 2.
Take the integral of sin^3 x dx. Split it as sin^2 x sin x and convert: (1 - cos^2 x) sin x. Set u = cos x, so du = -sin x dx. The integral becomes an easy polynomial in u. For cos^5 x you mirror it: split off one cos x and set u = sin x.
Roots of squares invite a trig substitution, and the root shape picks it. Use x = a sin theta for sqrt(a^2 - x^2), x = a tan theta for sqrt(a^2 + x^2), and x = a sec theta for sqrt(x^2 - a^2). Convert dx too, such as dx = a cos theta d theta, integrate in theta, then draw a right triangle to return to x.
Matching the root shape is the whole decision: sqrt(16 - x^2) wants x = 4 sin theta. If theta stands alone at the end, write it as sin^-1(x/a). Never skip the trip back to x.
Split odd powers, halve even ones, and match each root to its trig substitution.
More to read
3.2 Trigonometric Integrals
Powers of sine and cosine fall to one workhorse identity: sine squared plus cosine squared equals 1. With an odd power, split off a single factor to serve as du, convert the remaining even power with the identity, and substitute for the other trig function. With even powers, the half angle formulas lower the degree instead. Tangent and secant integrals follow the same spirit using the secant tangent identity. The pattern to memorise is split, convert, substitute.
3.3 Trigonometric Substitution
Roots of sums and differences of squares invite a trig substitution that the Pythagoras theorem then cleans up. Use x equals a sin theta for roots like sqrt of a squared minus x squared, x equals a tan theta for a squared plus x squared, and x equals a sec theta for x squared minus a squared. Convert dx as well, integrate in theta, and finally draw a right triangle to translate back to x.
Key Concepts
These techniques share one theme: rewrite the integrand into a form whose antiderivative you know. Odd trig powers give up one factor for du, even powers yield to half angle identities, and rooted quadratics yield to trig substitution plus a reference triangle. Keep the two Pythagoras identities, sine squared plus cosine squared and one plus tan squared equals sec squared, on instant recall. Every solution ends with a check: differentiate, or test the triangle translation at a friendly angle.
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