Integration by substitution

Changing the variable can turn an unfamiliar integral into a standard one. Put x = g(t), so dx = g′(t) dt, and ∫f(x) dx becomes ∫f(g(t)) g′(t) dt.

In practice, look for a function inside the integrand whose derivative also appears as a factor. Call the inner function t; its derivative and dx together become dt.

Linear inside: ∫sin mx dx = −(1/m) cos mx + C. With t = mx, dx = dt/m.

Worked example: in ∫2x sin(x² + 1) dx the factor 2x is the derivative of x² + 1. With t = x² + 1 the integral is ∫sin t dt, giving −cos(x² + 1) + C.

Worked example: in ∫sin(tan⁻¹x)/(1 + x²) dx put t = tan⁻¹x, dt = dx/(1 + x²). The answer is −cos(tan⁻¹x) + C.

A useful pattern: when the numerator is the derivative of the denominator, ∫f′(x)/f(x) dx = log|f(x)| + C.

Four results that follow: ∫tan x dx = log|sec x| + C, ∫cot x dx = log|sin x| + C, ∫sec x dx = log|sec x + tan x| + C, ∫cosec x dx = log|cosec x − cot x| + C.

Odd power of sine or cosine: peel off one factor. ∫sin³x cos²x dx = ∫(1 − cos²x) cos²x sin x dx; with t = cos x this is −cos³x/3 + cos⁵x/5 + C.

Worked example: 1/(1 + tan x) = cos x/(cos x + sin x). Writing the numerator as ½[(cos x + sin x) + (cos x − sin x)] gives x/2 + ½ log|cos x + sin x| + C.

Always return to the original variable at the end of an indefinite integral.