Trigonometric substitution

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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:[1][2]

Substitution 1. If the integrand contains a2 − x2, let

x = a \sin(\theta)

and use the identity

1-\sin^2(\theta) = \cos^2(\theta).

Substitution 2. If the integrand contains a2 + x2, let

x = a \tan(\theta)

and use the identity

1+\tan^2(\theta) = \sec^2(\theta).

Substitution 3. If the integrand contains x2 − a2, let

x = a \sec(\theta)

and use the identity

\sec^2(\theta)-1 = \tan^2(\theta).

Examples[edit]

Integrals containing a2x2[edit]

In the integral

\int\frac{dx}{\sqrt{a^2-x^2}}

we may use

x=a\sin(\theta),\quad dx=a\cos(\theta)\,d\theta, \quad \theta=\arcsin\left(\frac{x}{a}\right)
\begin{align}
\int\frac{dx}{\sqrt{a^2-x^2}} & = \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2-a^2\sin^2(\theta)}} \\
&= \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2(1-\sin^2(\theta))}} \\
&= \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2\cos^2(\theta)}} \\
&= \int d\theta \\
&= \theta+C \\
&= \arcsin \left(\tfrac{x}{a}\right)+C
\end{align}

Note that the above step requires that a > 0 and cos(θ) > 0; we can choose the a to be the positive square root of a2; and we impose the restriction on θ to be −π/2 < θ < π/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(θ) goes from 0 to 1/2, so θ goes from 0 to π/6. Then we have

\int_0^{\frac{a}{2}}\frac{dx}{\sqrt{a^2-x^2}}=\int_0^{\frac{\pi}{6}} d\theta = \tfrac{\pi}{6}.

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

Integrals containing a2 + x2[edit]

In the integral

\int\frac{dx}{{a^2+x^2}}

we may write

x=a\tan(\theta),\quad  dx=a\sec^2(\theta)\,d\theta, \quad \theta=\arctan\left(\tfrac{x}{a}\right)

so that the integral becomes

\begin{align}
\int\frac{dx}{{a^2+x^2}} &= \int\frac{a\sec^2(\theta)\,d\theta}{{a^2+a^2\tan^2(\theta)}} \\
&= \int\frac{a\sec^2(\theta)\,d\theta}{{a^2(1+\tan^2(\theta))}} \\
&= \int \frac{a\sec^2(\theta)\,d\theta}{{a^2\sec^2(\theta)}} \\
&= \int \frac{d\theta}{a} \\
&= \tfrac{\theta}{a}+C \\
&= \tfrac{1}{a} \arctan \left(\tfrac{x}{a}\right)+C
\end{align}

(provided a ≠ 0).

Integrals containing x2a2[edit]

Integrals like

\int\frac{dx}{x^2 - a^2}

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

\int\sqrt{x^2 - a^2}\,dx

can be done by substitution:

x = a \sec(\theta),\quad  dx = a \sec(\theta)\tan(\theta)\,d\theta, \quad \theta = \arcsec\left(\tfrac{x}{a}\right)
\begin{align}
\int\sqrt{x^2 - a^2}\,dx &= \int\sqrt{a^2 \sec^2(\theta) - a^2} \cdot a \sec(\theta)\tan(\theta)\,d\theta \\
&= \int\sqrt{a^2 (\sec^2(\theta) - 1)} \cdot a \sec(\theta)\tan(\theta)\,d\theta \\
&= \int\sqrt{a^2 \tan^2(\theta)} \cdot a \sec(\theta)\tan(\theta)\,d\theta \\
&= \int a^2 \sec(\theta)\tan^2(\theta)\,d\theta \\
&= a^2 \int \sec(\theta)(\sec^2(\theta) - 1)\,d\theta \\
&= a^2 \int (\sec^3(\theta) - \sec(\theta))\,d\theta.
\end{align}

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

Substitutions that eliminate trigonometric functions[edit]

Substitution can be used to remove trigonometric functions. In particular, see Tangent half-angle substitution.

For instance,

\begin{align}
\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}{\mp\sqrt{1-u^2}} f\left(\pm\sqrt{1-u^2},u\right)\,du && u=\cos (x) \\
\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\left (\tfrac{x}{2} \right ) \\
\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 = \int \frac{1-u^2}{1+u^2}\,du
\end{align}

Hyperbolic functions[edit]

Substitutions of hyperbolic functions can also be used to simplify integrals.[3]

In the integral \int \frac{1}{\sqrt{a^2+x^2}}\,dx, make the substitution x=a\sinh{u}, dx=a\cosh{u}\,du.

Then, using the identities \cosh^2 (x) - \sinh^2 (x) = 1 and \sinh^{-1}{x} = \ln(x + \sqrt{x^2 + 1}),

\begin{align}
\int \frac{1}{\sqrt{a^2+x^2}}\,dx &= \int \frac{a\cosh{u}}{\sqrt{a^2+a^2\sinh^2{u}}}\,du\\
&=\int \frac{a\cosh{u}}{a\sqrt{1+\sinh^2{u}}}\,du\\
&=\int \frac{a\cosh{u}}{a\cosh{u}}\,du\\
&=u+C\\
&=\sinh^{-1}{\frac{x}{a}}+C\\
&=\ln\left(\sqrt{\frac{x^2}{a^2} + 1} + \frac{x}{a}\right) + C\\
&=\ln\left(\frac{\sqrt{x^2+a^2} + x}{a}\right) + C
\end{align}

See also[edit]

References[edit]

  1. ^ Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 0-495-01166-5. 
  2. ^ Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th ed.). Addison-Wesley. ISBN 0-321-58876-2. 
  3. ^ Boyadzhiev, Khristo N. "Hyperbolic Substitutions for Integrals". Retrieved 4 March 2013.