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 a2x2

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

a

to be the principal root of

a2,

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

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,

\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

0

to

\pi/6.

Neglecting this restriction, one might have picked

\theta

to go from

\pi

to

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

a0.

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=\sec2\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,\endsame as before.

Example 2

The integral

\int\sqrt\,

may be evaluated by letting

x=a\tan\theta,dx=a\sec2\thetad\theta,\theta=\arctan

x
a

,

where

a>0

so that

\sqrt{a2}=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] \endThe 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 x2a2

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{a2}=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{a2+x2}

, 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

See also

Notes and References

  1. Book: Stewart, James . James Stewart (mathematician) . Calculus: Early Transcendentals . . 6th . 2008 . 978-0-495-01166-8 . registration .
  2. 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.
  3. 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).
  4. 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.