Trigonometric Substitution

In mathematics, trigonometric substitution is the substitution of trigonometric functions for other expressions. One may use the trigonometric identities
1-\sin^2\theta\equiv\cos^2\theta
1+\tan^2\theta\equiv\sec^2\theta
\sec^2\theta-1\equiv\tan^2\theta
to simplify certain integrals containing the radical expressions
\sqrt{a^2-x^2}
\sqrt{a^2+x^2}
\sqrt{x^2-a^2}
respectively. In the expression a2 − x2, the substitution of a sin(θ) for x makes it possible to use the identity 1 − sin2θ = cos2θ. In the expression a2 + x2, the substitution of a tan(θ) for x makes it possible to use the identity tan2θ + 1 = sec2θ. Similarly, in x2 − a2, the substitution of sec(θ) for x makes it possible to use the identity sec2 − 1 = tan2.

Examples

In the integral
\int\frac{dx}{\sqrt{a^2-x^2}}
one may use
x=a\sin(\theta)\ \ \mbox{so}\ \mbox{that}\ \sin^{-1}(x/a)=\theta,
dx=a\cos(\theta)\,d\theta,
a^2-x^2=a^2-a^2\sin^2(\theta)=a^2(1-\sin^2(\theta))=a^2\cos^2(\theta),
so that the integral becomes
\int\frac{dx}{\sqrt{a^2-x^2}}=\int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2\cos^2(\theta)}}
=\int d\theta=\theta+C=\sin^{-1}(x/a)+C (provided a > 0; if a < 0 then √a2 would be |a|, which would differ from a). For a definite integral, one must figure out how the bounds of integration change. For example, as x goes from 0 to a/2, then sin(θ) goes from 0 to 1/2, so θ goes from 0 to π/6. Then we have
\int_0^{a/2}\frac{dx}{\sqrt{a^2-x^2}}
=\int_0^{\pi/6}d\theta=\frac{\pi}{6}.
In the integral
\int\frac{1}{a^2+x^2}\,dx
one may write
x=a\tan(\theta),\ \mbox{so}\ \mbox{that}\ \theta=\arctan(x/a),
dx=a\sec^2(\theta)\,d\theta,
a^2+x^2=a^2+a^2\tan^2(\theta)=a^2(1+\tan^2(\theta))
=a^2\sec^2(\theta),
x/a=\tan(\theta),
so that the integral becomes
\int\frac{1}{a^2\sec^2(\theta)}\,a\sec^2(\theta)\,d\theta
=\frac{1}{a}\int\,d\theta=\frac{\theta}{a}+C=\frac{1}{a}\arctan(x/a)+C (provided a > 0).

Substitutions that eliminate trigonometric functions

Substitution can be used to remove trigonometric functions. For instance,
\int f(\sin x,\cos x)\,dx=\int\frac1{\pm\sqrt{1-u^2}}f\left(u,\pm\sqrt{1-u^2}\right)\,du, u=\sin x
\int f(\sin x,\cos x)\,dx=\int\frac{-1}{\pm\sqrt{1-u^2}}f\left(\pm\sqrt{1-u^2},u\right)\,du, u=\cos x
(but be careful with the signs)
\int f(\sin x,\cos x)\,dx=\int\frac2{1+u^2} f\left(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2}\right)\,du, u=\tan\frac x2
Example (see quintic of l'Hospitalhttp://www.mathcurve.com/courbes2d/quintique%20de%20l%27hospital/quintique%20de%20l%27hospital):
\int\frac{\cos x}{(1+\cos x)^3}\,dx
=\int\frac2{1+u^2}\frac{\frac{1-u^2}{1+u^2}}{\left(1+\frac{1-u^2}{1+u^2}\right)^3}\,du =\frac14\int(1-u^4)\,du =\frac14\left(u-\frac15u^5\right)+C =\frac{(1+3\cos x+\cos^2x)\sin x}{5(1+\cos x)^3}+C

 

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