Trigonometric substitution

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)


Integrals containing "a"2 − "x"2

In the integral


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


we may write

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

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

so that the integral becomes

:egin{align}& {} quad intfrac{dx}{a^2+x^2} = intfrac{asec^2( heta),d heta}{a^2+a^2 an^2( heta)} = intfrac{asec^2( heta),d heta}{a^2(1+ an^2( heta))} \& {} = int frac{asec^2( heta),d heta}{a^2sec^2( heta)} = int frac{d heta}{a} = frac{ heta}{a}+C = frac{1}{a} arctan left(frac{x}{a} ight)+Cend{align}

(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)

:egin{align}& {} quad intsqrt{x^2 - a^2},dx = intsqrt{a^2 sec^2( heta) - a^2} cdot a sec( heta) an( heta),d heta \& {} = intsqrt{a^2 (sec^2( heta) - 1)} cdot a sec( heta) an( heta),d heta = intsqrt{a^2 an^2( heta)} cdot a sec( heta) an( heta),d heta \& {} = int a^2 sec( heta) an^2( heta),d heta = a^2 int sec( heta) (sec^2( heta) - 1),d heta \& {} = a^2 int (sec^3( heta) - sec( heta)),d heta.end{align}

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

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