Trigonometric Substitution in Integration
Like other substitutions in calculus, trigonometric substitutions provide a method for evaluating an integral by reducing it to a simpler one. Trigonometric substitutions take advantage of patterns in the integrand that resemble common trigonometric relations and are most often useful for integrals of radical or rational functions that may not be simply evaluated by other methods. These substitutions are often used in conjunction with the basic trigonometric relations and occasionally product-to-sum identities as well as other integration techniques including integration by parts and \(u\)-substitutions.
Contents
Introduction
Trigonometric substitutions are a specific type of \(u\)-substitutions and rely heavily upon techniques developed for those.
They use the key relations \(\sin^2x + \cos^2x = 1\), \(\tan^2x + 1 = \sec^2x\), and \(\cot^2x + 1 = \csc^2x\) to manipulate an integral into a simpler form. The derivatives of trigonometric functions are also necessary to determine the best way to simplify the expression.
| \( f(x) \quad \rightarrow \)\(\hspace{10mm}\) | \( f'(x) \quad \rightarrow \)\(\hspace{10mm}\) | \( f'(x) \) rewritten |
| \( \sin \theta \) | \(\cos\theta\) | \(\pm\sqrt{1 - \sin^2\theta}\) |
| \( \cos \theta \) | \(-\sin\theta\) | \(\mp\sqrt{1-\cos^2\theta}\) |
| \( \tan \theta \) | \(\sec^2\theta\) | \(1 + \tan^2\theta\) |
| \( \sec \theta \) | \(\sec\theta\tan\theta\) | \(\pm\sec\theta\sqrt{\sec^2\theta - 1}\) |
| \( \csc \theta \) | \(-\csc\theta\cot\theta\) | \(\mp\csc\theta\sqrt{\csc^2\theta - 1 }\) |
| \(\cot\theta\) | \(-\csc^2\theta\) | \(-1-\cot^2\theta\) |
Where applicable, the sign of the square root may be determined by the value of \(\theta\).
One of the most common and direct applications is the integration of the reciprocal of a quadratic function.
Evaluate
\[ \int \frac{dx}{x^2+1}. \]
Note from the above table that \(x = \tan\theta\) satisfies the equation \( \tfrac{dx}{d\theta} = 1 + x^2\). It then makes some intuitive sense to say
\[d\theta = \frac{dx}{1 + x^2}.\]
It then follows that
\[\int \frac{dx}{x^2 + 1} = \int d\theta = \int 1 \, d\theta = \theta + C,\]
where \(C\) is the constant of integration. Recalling \(x = \tan\theta\), the "\(\theta\)-substitution" \(\theta = \arctan x\) shows that the integral is
\[\int \frac{dx}{x^2 + 1} = \theta + C = \arctan x + C.\ _\square\]
In general, tangent or cotangent substitutions help a lot with the integration of rational functions, especially those with denominators of even degree. This concept is further explored below in the section on rational functions.
Trigonometric substitutions also help integrate certain types of radical functions, especially those involving square roots of quadratic functions. In fact, this technique may provide a verification of the well-known formula for the area of a circle.
Determine the area of a circle of radius \(r\) centered at the origin.
The equation of the circle is \(y^2 + x^2 = r^2\), and it has symmetries across the \(x\)- and \(y\)-axes. Therefore, the area of the circle is four times the area of the portion of the circle in Quadrant I, i.e. that region bounded by \(y = 0\), \(x = 0\) and \(y = \sqrt{r^2 - x^2}\). Thus, the integral in question is
\[4 \int_0^r \sqrt{r^2-x^2} \, dx.\]
The above table shows that, in Quadrant I, \(x = r\sin\theta\) satisfies the equation \(\tfrac{dx}{d\theta} = r\sqrt{1 - x^2}\). It then makes some intuitive sense to say
\[dx = r\sqrt{1 - x^2} \, d\theta.\]
Therefore, the "\(\theta\)-substitution" \(\theta = r\arcsin x\) yields
\[\begin{align*} 4 \int_0^r \sqrt{r^2 - x^2} \, dx &= 4 \int_0^{\pi/2} \sqrt{r^2 - r^2\sin^2\theta} \cdot \left(r\sqrt{1 - \sin^2\theta}\right) \, d\theta \\ &= 4r^2 \int_0^{\pi/2} 1 - \sin^2\theta \, d\theta \\ &= 4r^2 \int_0^{\pi/2} \cos^2\theta \, d\theta \\ &= 4r^2 \int_0^{\pi/2} \frac{1 + \cos(2\theta)}{2} \, d\theta \\ &= 4r^2 \left[ \frac{\theta}{2} + \frac{\sin(2\theta)}{4} \right]_0^{\pi/2} \\ &= 4r^2 \left(\frac{\pi}{4} - 0\right) \\ &= \boxed{\pi r^2}. \end{align*}\]
The double angle formula, which follows from the product-to-sum formulas, was necessary to finish the calculation. \(_\square\)
Powers of Trigonometric Functions
For any positive integer \(n\), the functions \(\cos^nx\) and \(\sin^nx\) may be expressed as a sum of multiples of elements of \(\{1, \, \cos x, \, \cos(2x), \, \dots, \, \cos(nx)\}\) and \(\{1, \, \sin x, \, \sin(2x), \, \dots, \, \sin(nx)\},\) respectively. For instance, the double-angle formulas yield
\[\cos^2x = \frac{1}{2} + \frac{1}{2} \cos(2x)\ \text{ and }\ \sin^2x = \frac{1}{2} - \frac{1}{2} \cos(2x).\]
By expressing, a typical \(u\)-substitution like \(u = nx\) may be applied to help integrate an expression.
Evaluate \(\displaystyle \int \sin^3x \, dx.\)
Note that
\[\begin{align} \sin(3x) &= \cos x \sin(2x) + \cos(2x) \sin x \\ &= 2\cos^2 x \sin x + \big(1 - 2\sin^2 x\big) \sin x \\ &= 3 \sin x - 4\sin^3 x. \end{align}\]
It follows that \(\sin^3 x = \frac{3}{4} \sin x - \frac{1}{4} \sin(3x)\). Therefore,
\[\begin{align*} \int \sin^3x \, dx &= \int \frac{3}{4} \sin x - \frac{1}{4} \sin(3x) \, dx \\ &= \frac{1}{12} \cos(3x) - \frac{3}{4} \cos x + C, \end{align*}\]
where \(C\) is the constant of integration. \(_\square\)
Evaluate the indefinite integral
\[ \int (1 + \cos x)^3 \, dx.\]
Clarification: \(C\) denotes the arbitrary constant of integration.
Radical Functions
Many of the relations between trigonometric are second-order, so the inverse relations may involve square roots. As such, integrals involving square roots may be simplified by the use of trigonometric substitutions. In particular, expressions involving square roots of quadratic functions may benefit from cosine or secant substitutions. Sine substitutions work in the same scenarios as cosine ones, and cosecant substitutions work in the same scenarios as secant ones.
Given an expression of the form \(\sqrt{ax^2 + bx + c}\), consider the sign of \(a\) and \(d = b^2 - 4ac\). If \(d\) is positive, then a trigonometric substitution might help. For positive \(a\), a secant substitution could help after completing the square; for negative \(a\), a cosine substitution could help after completing the square. \((\)If \(d\) is negative, then a tangent or hyperbolic trigonometric substitution might help.\()\)
Such a substitution may help because it can remove the radical from the expression through the usage of trigonometric identities.
Evaluate \[\int \frac{1}{\sqrt{2x - x^2}} \, dx.\]
The substitution \(\cos\theta = x - 1\) simplifies the integral: \[\begin{align*} \int \frac{1}{\sqrt{2x - x^2}} \, dx &= \int \frac{1}{\sqrt{1 - (x - 1)^2}} \, dx \\ &= \int \frac{1}{\sqrt{1 - \cos^2\theta}} \cdot (- \sin \theta) \, d\theta \\ &= \int \frac{- \sin\theta}{\sin\theta} \, d\theta \\ &= - \theta + C \\ &= - \arccos(x - 1) + C, \end{align*}\] where \(C\) is the constant of integration. \(_\square\)
Evaluate \[\int \frac{1}{\sqrt{4x^2 - 1}} \, dx.\]
The substitution \(\sec\theta = 2x\) simplifies the integral: \[\begin{align*} \int \frac{1}{\sqrt{4x^2 - 1}} \, dx &= \int \frac{1}{\sqrt{\sec^2\theta - 1}} \cdot \big(\tfrac{1}{2} \sec\theta \tan\theta\big) \, d\theta \\ &= \int \frac{\sec\theta \tan\theta}{2\tan\theta} \, d\theta \\ &= \int \tfrac{1}{2} \sec\theta d\theta \\ &= \tfrac{1}{2} \ln(\tan\theta + \sec\theta) + C\\ &= \tfrac{1}{2} \ln\big(2x + \sqrt{4x^2 - 1}\big) + C, \end{align*}\] where \(C\) is the constant of integration. \(_\square\)
Evaluate the definite integral
\[ \int_0^1 x \sqrt{1 - x^4} \, dx.\]
\[ \int_0^2 x \sqrt{1-(x-1)^2} \, dx = \, ? \]
Give your answer to 2 decimal places.
Rational Functions
Rational functions involving quadratic polynomials in the denominator are often susceptible to a substitution involving the tangent function, and these tangent substitutions \((\)together with knowledge of \(\ln)\) are the cornerstone of partial fraction decomposition. For each quadratic polynomial with no real roots, the idea is to complete the square and replace the square portion with a tangent function.
The idea behind this substitution is to "cancel out" part of the denominator with the differential term \((dx\) in terms of \(d\theta)\) in order to integrate a smaller expression. When applied properly, something will cancel out, since \(\tfrac{dx}{d\theta} = 1 + x^2,\) where \(x = \tan\theta\).
Evaluate
\[\int_0^1\frac{2}{t^2+3} \, dt.\]
The "\(\theta\)-substitution" \(t = \sqrt{3} \tan\theta\) simplifies the integral:
\[\begin{align*} \int_0^1\frac{2}{t^2+3} \, dt &= \int_0^{\pi/6} \frac{2}{3\tan^2\theta + 3} \cdot \big(\sqrt{3} \sec^2\theta\big) \, d\theta \\ &= \int_0^{\pi/6} \frac{2\sqrt{3}}{3} \, d\theta \\ &= \frac{\sqrt{3}\pi}{9}. \end{align*}\]
Note that the limits change in accordance with the \(\theta\)-substitution; that is, \(\sqrt{3} \tan\tfrac{\pi}{6} = \sqrt{3} \cdot \tfrac{1}{\sqrt{3}} = 1\). \(_\square\)
Evaluate the improper integral
\[ \int_0^\infty \frac{1}{e^x + e^{-x}} \, dx.\]
The Half-angle Substitution
Recall the half-angle formulas
\[ \cos\left(\tfrac{\theta}{2}\right) = \sqrt{\tfrac{1}{2} (1 + \cos\theta)}\ \text{ and }\ \sin\left(\tfrac{\theta}{2}\right) = \sqrt{\tfrac{1}{2} ( 1 - \cos\theta)}.\]
These are helpful to know for simplifying many expressions involving trigonometric substitutions. Even more helpful is the tangent half-angle formula
\[ \tan\left(\tfrac{\theta}{2}\right) = \frac{\sin\left(\tfrac{\theta}{2}\right)}{\cos\left(\tfrac{\theta}{2}\right)} = \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} = \sqrt{\frac{(1 - \cos\theta)^2}{1 - \cos^2\theta}} = \frac{1 - \cos\theta}{\sin\theta}. \]
The half-angle tangent substitution has the further property that, if \(t = \tan\left(\tfrac{\theta}{2}\right)\),
\[\begin{array} &\sin\theta=\frac{2t}{1+t^2}, &\cos\theta=\frac{1-t^2}{1+t^2}, &\tan\theta=\frac{2t}{1-t^2}.\end{array}\]
For many rational functions involving trigonometric functions, such a substitution can make the expression seem much more integrable.
Find the value of \(\displaystyle \int_0^\frac{\pi}{2}\frac{dx}{2+\cos x}\, dx.\)
The integral is equal to
\[\int_0^\frac{\pi}{2}\frac{dx}{2+\cos x}\, dx = \int_0^1\frac{\frac{2dt}{1+t^2}}{2+\frac{1-t^2}{1+t^2}}=\int_0^1\frac{2}{t^2+3} \, dt.\]
This integral can be finished using the methods outlined in the previous section, and it was evaluated to be equal to \(\frac{\pi\sqrt{3}}{9}.\ _\square\)
\[\int_0^1\frac{x^4\left(1-x^2\right)^5}{\left(1+x^2\right)^{10}}\, dx=A\]
Given the above, find \(\frac{1}{A}.\)