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Trigonometric Substitution

Integrals of the form

\int f(\cos\theta, \sin\theta)\,d\theta
\end{displaymath} (1)

can be solved by making the substitution $z = e^{i\theta}$ so that $dz = ie^{i\theta}\,d\theta$ and expressing
\cos\theta = {e^{i\theta}+e^{-i\theta}\over 2} = {z+z^{-1}\over 2}
\end{displaymath} (2)

\sin\theta = {e^{i\theta}-e^{-i\theta}\over 2i} = {z-z^{-1}\over 2i}.
\end{displaymath} (3)

The integral can then be solved by Contour Integration.

Alternatively, making the substitution $t\equiv \tan(\theta/2)$ transforms (1) into

\int f\left({{2t\over 1+t^2}, {1-t^2\over 1+t^2}}\right){2\,dt\over 1+t^2}.
\end{displaymath} (4)

The following table gives trigonometric substitutions which can be used to transform integrals involving square roots.

Form Substitution
$\sqrt{a^2-x^2}$ $x = a \sin\theta$
$\sqrt{a^2+x^2}$ $x = a \tan\theta$
$\sqrt{x^2-a^2}$ $x = a \sec\theta$

See also Hyperbolic Substitution

© 1996-9 Eric W. Weisstein