# Trigonometric substitution

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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:

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

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

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

Examples

Integrals containing "a"2 − "x"2

In the integral

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

we may use

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

so that the integral becomes

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

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

:$int_0^\left\{a/2\right\}frac\left\{dx\right\}\left\{sqrt\left\{a^2-x^2=int_0^\left\{pi/6\right\}d heta=frac\left\{pi\right\}\left\{6\right\}.$

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

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

we may write

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

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

so that the integral becomes

:

(provided "a" > 0).

Integrals containing "x"2 − "a"2

Integrals like

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

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

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

can be done by substitution:

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

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

:

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,

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

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

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

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

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

ee also

* Tangent half-angle formula

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