Trigonometric substitution

In mathematics, trigonometric substitution is the substitution of trigonometric functions for other expressions. One may use the trigonometric identities to simplify certain integrals containing radical expressions:

:$ext\{For\; \}\; sqrt\{a^2-x^2\}\; ext\{\; let\; \}\; x=a\; sin(\; heta)\; ext\{\; and\; use\; \}\; 1-sin^2(\; heta)\; =\; cos^2(\; heta)$

:$ext\{For\; \}\; sqrt\{a^2+x^2\}\; ext\{\; let\; \}\; x=a\; an(\; heta)\; ext\{\; and\; use\; \}\; 1+\; an^2(\; heta)\; =\; sec^2(\; heta)$

:$ext\{For\; \}\; sqrt\{x^2-a^2\}\; ext\{\; let\; \}\; x=a\; sec(\; heta)\; ext\{\; and\; use\; \}\; sec^2(\; heta)-1\; =\; an^2(\; heta)$

Examples

Integrals containing "a"² − "x"²

In the integral

:$intfrac\{dx\}\{sqrt\{a^2-x^2$

we may use

:$x=asin(\; heta),\; dx=acos(\; heta),d\; heta$:$heta=arcsinleft(frac\{x\}\{a\}\; ight)$

so that the integral becomes

:$intfrac\{dx\}\{sqrt\{a^2-x^2\; =\; intfrac\{acos(\; heta),d\; heta\}\{sqrt\{a^2-a^2sin^2(\; heta)\; =\; intfrac\{acos(\; heta),d\; heta\}\{sqrt\{a^2(1-sin^2(\; heta))$:$\{\}\; =\; intfrac\{acos(\; heta),d\; heta\}\{sqrt\{a^2cos^2(\; heta)\; =\; int\; d\; heta=\; heta+C=arcsinleft(frac\{x\}\{a\}\; ight)+C$

Note that the above step requires that "a" > 0 and cos("&theta;") > 0; we can choose the "a" to be the positive square root of "a"²; and we impose the restriction on "&theta;" to be −&pi;/2 < "&theta;" < &pi;/2 by using the arcsin function.

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(&theta;) goes from 0 to 1/2, so &theta; goes from 0 to &pi;/6. Then we have

:$int\_0^\{a/2\}frac\{dx\}\{sqrt\{a^2-x^2=int\_0^\{pi/6\}d\; heta=frac\{pi\}\{6\}.$

Some care is needed when picking the bounds. The integration above requires that −&pi;/2 < "&theta;" < &pi;/2, so "&theta;" going from 0 to &pi;/6 is the only choice. If we had missed this restriction, we might have picked "&theta;" to go from &pi; to 5&pi;/6, which would result in the negative of the result.

Integrals containing "a"² + "x"²

In the integral

:$intfrac\{dx\}\{a^2+x^2\}$

we may write

:$x=a\; an(\; heta),\; dx=asec^2(\; heta),d\; heta$

:$heta=arctanleft(frac\{x\}\{a\}\; ight)$

so that the integral becomes

:

(provided "a" > 0).

Integrals containing "x"² − "a"²

Integrals like

:$intfrac\{dx\}\{x^2\; -\; a^2\}$

should be done by partial fractions rather than trigonometric substitutions. However, the integral

:$intsqrt\{x^2\; -\; a^2\},dx$

can be done by substitution:

:$x\; =\; a\; sec(\; heta),\; dx\; =\; a\; sec(\; heta)\; an(\; heta),d\; heta$

:$heta\; =\; arcsecleft(frac\{x\}\{a\}\; ight)$

:

We can then solve this using the formula for the integral of secant cubed.

ubstitutions that eliminate trigonometric functions

Substitution can be used to remove trigonometric functions. For instance,

:$int\; f(sin\; x,cos\; x),dx=intfrac1\{pmsqrt\{1-u^2fleft(u,pmsqrt\{1-u^2\}\; ight),du,\; qquad\; qquad\; u=sin\; x$

:$int\; f(sin\; x,cos\; x),dx=intfrac\{-1\}\{pmsqrt\{1-u^2fleft(pmsqrt\{1-u^2\},u\; ight),du\; qquad\; qquad\; u=cos\; x$(but be careful with the signs)

:$int\; f(sin\; x,cos\; x),dx=intfrac2\{1+u^2\}\; fleft(frac\{2u\}\{1+u^2\},frac\{1-u^2\}\{1+u^2\}\; ight),du\; qquad\; qquad\; u=\; anfrac\; x2$

:$intfrac\{cos\; x\}\{(1+cos\; x)^3\},dx\; =\; intfrac2\{1+u^2\}frac\{frac\{1-u^2\}\{1+u^2\{left(1+frac\{1-u^2\}\{1+u^2\}\; ight)^3\},du\; =$

:$frac\{1\}\{4\}int(1-u^4),du\; =\; frac\{1\}\{4\}left(u-frac15u^5\; ight)\; +\; C\; =\; frac\{(1+3cos\; x+cos^2x)sin\; x\}\{5(1+cos\; x)^3\}\; +\; C$

ee also

* Tangent half-angle formula