Trigonometric substitution explained

In mathematics, a trigonometric substitution replaces a trigonometric function for another expression. In calculus, trigonometric substitutions are a technique for evaluating integrals. In this case, an expression involving a radical function is replaced with a trigonometric one. Trigonometric identities may help simplify the answer.^[1] ^[2] Like other methods of integration by substitution, when evaluating a definite integral, it may be simpler to completely deduce the antiderivative before applying the boundaries of integration.

Case I: Integrands containing a² − x²

Let

x=a\sin\theta,

and use the identity

1-\sin²\theta=\cos²\theta.

Examples of Case I

Example 1

In the integral

$\int\frac,$

we may use

$x=a\sin \theta,\quad dx=a\cos\theta\, d\theta, \quad \theta=\arcsin\frac.$

Then, $\begin \int\frac &= \int\frac \\[6pt] &= \int\frac \\[6pt] &= \int\frac \\[6pt] &= \int d\theta \\[6pt] &= \theta + C \\[6pt] &= \arcsin\frac+C. \end$

The above step requires that

a>0

and

\cos\theta>0.

We can choose

to be the principal root of

a^2,

and impose the restriction

-\pi/2<\theta<\pi/2

by using the inverse sine function.

For a definite integral, one must figure out how the bounds of integration change. For example, as

goes from

a/2,

then

\sin\theta

goes from

1/2,

\theta

goes from

\pi/6.

Then,

$\int_0^\frac=\int_0^ d\theta = \frac.$

Some care is needed when picking the bounds. Because integration above requires that

-\pi/2<\theta<\pi/2

\theta

can only go from

\pi/6.

Neglecting this restriction, one might have picked

\theta

to go from

\pi

5\pi/6,

which would have resulted in the negative of the actual value.

Alternatively, fully evaluate the indefinite integrals before applying the boundary conditions. In that case, the antiderivative gives

$\int_^ \frac = \arcsin \left(\frac \right) \Biggl|_^= \arcsin \left (\frac\right) - \arcsin (0) = \frac$ as before.

Example 2

The integral

$\int\sqrt\,dx,$

may be evaluated by letting $x=a\sin \theta,\, dx=a\cos\theta\, d\theta,\, \theta=\arcsin\dfrac,$ where

a>0

so that

\sqrt=a,

and

-\pi/2 \le \theta \le \pi/2

by the range of arcsine, so that

\cos\theta\ge0

and

\sqrt = \cos \theta.

Then, $\begin \int\sqrt\,dx &= \int\sqrt\,(a\cos\theta) \,d\theta \\[6pt] &= \int\sqrt\,(a\cos\theta) \,d\theta \\[6pt] &= \int\sqrt\,(a\cos\theta) \,d\theta \\[6pt] &= \int(a\cos\theta)(a\cos\theta) \,d\theta \\[6pt] &= a^2\int\cos^2\theta\,d\theta \\[6pt] &= a^2\int\left(\frac\right)\,d\theta \\[6pt] &= \frac \left(\theta+\frac\sin 2\theta \right) + C \\[6pt] &= \frac(\theta+\sin\theta\cos\theta) + C \\[6pt] &= \frac\left(\arcsin\frac+\frac\sqrt\right) + C \\[6pt] &= \frac\arcsin\frac+\frac\sqrt+C. \end$

For a definite integral, the bounds change once the substitution is performed and are determined using the equation $\theta = \arcsin\dfrac,$ with values in the range $-\pi/2 \le \theta \le \pi/2.$ Alternatively, apply the boundary terms directly to the formula for the antiderivative.

For example, the definite integral

$\int_^1\sqrt\,dx,$

may be evaluated by substituting

x=2\sin\theta,dx=2\cos\thetad\theta,

with the bounds determined using

\theta = \arcsin\dfrac.

Because

\arcsin(1/{2})=\pi/6

and

\arcsin(-1/2)=-\pi/6,

\begin \int_^1\sqrt\,dx &= \int_^\sqrt\,(2\cos\theta) \,d\theta \\[6pt] &= \int_^\sqrt\,(2\cos\theta) \,d\theta \\[6pt] &= \int_^\sqrt\,(2\cos\theta) \,d\theta \\[6pt] &= \int_^(2\cos\theta)(2\cos\theta) \,d\theta \\[6pt] &= 4\int_^\cos^2\theta\,d\theta \\[6pt] &= 4\int_^\left(\frac\right)\,d\theta \\[6pt] &= 2 \left[\theta+\frac{1}{2} \sin 2\theta \right]^_ = [2\theta+\sin 2\theta] \Biggl |^_ \\[6pt] &= \left(\frac+\sin\frac\right)-\left(-\frac+\sin\left(-\frac\right)\right) = \frac+\sqrt. \end

On the other hand, direct application of the boundary terms to the previously obtained formula for the antiderivative yields $\begin\int_^1\sqrt\,dx &= \left[\frac{2^2}{2}\arcsin\frac{x}{2}+\frac{x}{2}\sqrt{2^2-x^2} \right]_^\\[6pt]&= \left(2 \arcsin \frac + \frac\sqrt\right) - \left(2 \arcsin \left(-\frac\right) + \frac\sqrt\right)\\[6pt]&= \left(2 \cdot \frac + \frac\right) - \left(2\cdot \left(-\frac\right) - \frac\right)\\[6pt]&= \frac + \sqrt\end$ as before.

Case II: Integrands containing a² + x²

Let

x=a\tan\theta,

and use the identity

1+\tan²\theta=\sec²\theta.

Examples of Case II

Example 1

In the integral

$\int\frac$

we may write

$x=a\tan\theta,\quad dx=a\sec^2\theta\, d\theta, \quad \theta=\arctan\frac,$

so that the integral becomes

$\begin \int\frac &= \int\frac \\[6pt] &= \int\frac \\[6pt] &= \int\frac \\[6pt] &= \int\frac \\[6pt] &= \frac+C \\[6pt] &= \frac \arctan \frac + C, \end$

provided

a ≠ 0.

For a definite integral, the bounds change once the substitution is performed and are determined using the equation

\theta=\arctan

	x
	a

with values in the range

-	\pi
	2

<\theta<

	\pi
	2

Alternatively, apply the boundary terms directly to the formula for the antiderivative.

For example, the definite integral

$\int_0^1\frac\,$

may be evaluated by substituting

x=\tan\theta,dx=\sec^{2\thetad\theta,}

with the bounds determined using

\theta=\arctanx.

Since

\arctan0=0

and

\arctan1=\pi/4,

\begin \int_0^1\frac &= 4\int_0^1\frac \\[6pt] &= 4\int_0^\frac \\[6pt] &= 4\int_0^\frac \\[6pt] &= 4\int_0^d\theta \\[6pt] &= (4\theta)\Bigg|^_0 = 4 \left (\frac - 0 \right) = \pi. \end

Meanwhile, direct application of the boundary terms to the formula for the antiderivative yields $\begin\int_0^1\frac\,&= 4\int_0^1\frac \\[6pt]&= 4\left[\frac{1}{1} \arctan \frac{x}{1} \right]^1_0 \\[6pt]&= 4(\arctan x)\Bigg|^1_0 \\[6pt]&= 4(\arctan 1 - \arctan 0) \\[6pt]&= 4 \left (\frac - 0 \right) = \pi,\end$ same as before.

Example 2

The integral

$\int\sqrt\,$

may be evaluated by letting

x=a\tan\theta,dx=a\sec^2\thetad\theta,\theta=\arctan

	x
	a

where

a>0

so that

\sqrt{a^2}=a,

and

-	\pi	<\theta<
	2

	\pi
	2

by the range of arctangent, so that

\sec\theta>0

and

\sqrt{\sec²\theta}=\sec\theta.

Then, $\begin \int\sqrt\,dx &= \int\sqrt\,(a \sec^2\theta)\, d\theta \\[6pt] &= \int\sqrt\,(a \sec^2\theta)\, d\theta \\[6pt] &= \int\sqrt\,(a \sec^2\theta)\, d\theta \\[6pt] &= \int(a \sec\theta)(a \sec^2\theta)\, d\theta \\[6pt] &= a^2\int \sec^3\theta\, d\theta. \\[6pt] \end$ The integral of secant cubed may be evaluated using integration by parts. As a result, $\begin \int\sqrt\,dx &= \frac(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|)+C \\[6pt] &= \frac\left(\sqrt\cdot\frac + \ln\left|\sqrt+\frac\right|\right)+C \\[6pt] &= \frac\left(x\sqrt + a^2\ln\left|\frac\right|\right)+C. \end$

Case III: Integrands containing x² − a²

Let

x=a\sec\theta,

and use the identity

\sec²\theta-1=\tan²\theta.

Examples of Case III

Integrals such as

$\int\frac$

can also be evaluated by partial fractions rather than trigonometric substitutions. However, the integral

$\int\sqrt\, dx$

cannot. In this case, an appropriate substitution is: $x = a \sec\theta,\, dx = a \sec\theta\tan\theta\, d\theta, \, \theta = \arcsec\frac,$

where

a>0

so that

\sqrt{a^2}=a,

and

0\le\theta<

	\pi
	2

by assuming

x>0,

so that

\tan\theta\ge0

and

\sqrt{\tan²\theta}=\tan\theta.

Then, $\begin \int\sqrt\, dx &= \int\sqrt \cdot a \sec\theta\tan\theta\, d\theta \\ &= \int\sqrt \cdot a \sec\theta\tan\theta\, d\theta \\ &= \int\sqrt \cdot a \sec\theta\tan\theta\, d\theta \\ &= \int a^2 \sec\theta\tan^2\theta\, d\theta \\ &= a^2 \int (\sec\theta)(\sec^2\theta - 1)\, d\theta \\ &= a^2 \int (\sec^3\theta - \sec\theta)\, d\theta. \end$

One may evaluate the integral of the secant function by multiplying the numerator and denominator by

(\sec\theta+\tan\theta)

and the integral of secant cubed by parts.^[3] As a result,

\begin \int\sqrt\,dx &= \frac(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|)-a^2\ln|\sec\theta+\tan\theta|+C \\[6pt] &= \frac(\sec\theta \tan\theta - \ln|\sec\theta+\tan\theta|)+C \\[6pt] &= \frac\left(\frac\cdot\sqrt - \ln\left|\frac+\sqrt\right|\right)+C \\[6pt] &= \frac\left(x\sqrt - a^2\ln\left|\frac\right|\right)+C. \end

When

	\pi
	2

<\theta\le\pi,

which happens when

x<0

given the range of arcsecant,

\tan\theta\le0,

meaning

\sqrt{\tan²\theta}=-\tan\theta

instead in that case.

Substitutions that eliminate trigonometric functions

Substitution can be used to remove trigonometric functions.

For instance,

$\begin\int f(\sin(x), \cos(x))\, dx &=\int\frac1 f\left(u,\pm\sqrt\right)\, du && u=\sin (x) \\[6pt]\int f(\sin(x), \cos(x))\, dx &=\int\frac f\left(\pm\sqrt,u\right)\, du && u=\cos (x) \\[6pt]\int f(\sin(x), \cos(x))\, dx &=\int\frac2 f \left(\frac,\frac\right)\, du && u=\tan\left (\frac \right) \\[6pt]\end$

The last substitution is known as the Weierstrass substitution, which makes use of tangent half-angle formulas.

For example,

$\begin\int\frac\, dx &= \int\frac2\frac\, du = \int (1-u^2)(1+u^2)\, du \\&= \int (1-u^4)\,du = u - \frac + C = \tan \frac - \frac \tan^5 \frac + C.\end$

Hyperbolic substitution

Substitutions of hyperbolic functions can also be used to simplify integrals.^[4]

For example, to integrate

1/\sqrt{a^2+x^2}

, introduce the substitution

x=a\sinh{u}

(and hence

dx=a\coshudu

), then use the identity

\cosh²(x)-\sinh²(x)=1

to find:

$\begin\int \frac &= \int \frac \\[6pt]&=\int \frac \\[6pt]&=\int \frac \,du \\[6pt]&=u+C \\[6pt]&=\sinh^ + C.\end$

If desired, this result may be further transformed using other identities, such as using the relation

\sinh^-1{z}=\operatorname{arsinh}{z}=ln(z+\sqrt{z²+1})

\begin\sinh^ + C&=\ln\left(\frac + \sqrt\,\right) + C \\[6pt]&=\ln\left(\frac\,\right) + C.\end

Notes and References

Book: Stewart, James . James Stewart (mathematician) . Calculus: Early Transcendentals . . 6th . 2008 . 978-0-495-01166-8 . registration .
Book: Thomas . George B. . Weir . Maurice D. . Hass . Joel . Joel Hass . George B. Thomas . Thomas' Calculus: Early Transcendentals . . 2010 . 12th . 978-0-321-58876-0.
Book: Stewart, James. Calculus - Early Transcendentals. Cengage Learning. 2012. 978-0-538-49790-9. United States. 475–6. Section 7.2: Trigonometric Integrals. James Stewart (mathematician).
Web site: Boyadzhiev. Khristo N.. Hyperbolic Substitutions for Integrals. 4 March 2013. 26 February 2020. https://web.archive.org/web/20200226040813/http://www2.onu.edu/~m-caragiu.1/bonus_files/HYPERSUB.pdf. dead.

Trigonometric substitution explained

Case I: Integrands containing a2 − x2

Examples of Case I

Example 1

Example 2

Case II: Integrands containing a2 + x2

Examples of Case II

Example 1

Example 2

Case III: Integrands containing x2 − a2

Examples of Case III

Substitutions that eliminate trigonometric functions

Hyperbolic substitution

See also

Notes and References

Case I: Integrands containing a² − x²

Case II: Integrands containing a² + x²

Case III: Integrands containing x² − a²