Calculus II 08.04 Trigonometric Substitution

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8.4 Trigonometric Substitution

• Use trigonometric substitution to solve an integral.
• Use integrals to model and solve real-life applications.

Trigonometric substitution is useful in solving radical integrals with the form

\(\sqrt{a^{2}-u^{2}}\), \(\sqrt{a^{2}+u^{2}}\), and \(\sqrt{u^{2}-a^{2}}\)

by eliminating the radicals. The Pythagorean Trigonometric identities, listed below, are used to remove the radicals.

$$ \cos^{2} \theta = 1 - \sin^{2} \theta $$

$$ \sec^{2} \theta = 1 + \tan^{2} \theta $$

$$ \tan^{2} \theta = \sec^{2} - 1 $$

For example, for \( a > 0\), let \(u=a \sin \theta \), where \(-\pi/2 \leqslant x \leqslant \pi /2\). This produces

\(\sqrt{a^{2}-u^{2}}\)	\(= \sqrt{a^{2}-a^{2}\sin^{2} \theta}\)
	\(= \sqrt{a^{2}(1-\sin^{2} \theta)}\)
	\(= \sqrt{a^{2}\cos^{2} \theta}= a \cos \theta\)

Note that \(\cos x \geqslant 0\), because \(-\pi/2 \leqslant \theta \leqslant \pi /2\).

Trigonometric Substitution \( a > 0 \).
Figure 8.4.1

The restrictions on \(\theta\) ensures the function is one-to-one. These are the same intervals over which the arcsine, arctangent, and arcsecant are defined.

Example 8.4.1 Trigonometric Substitution with \(u= a \sin \theta\)

Figure 8.4.2

Find $$ \int \frac{dx}{x^{2}\sqrt{9-x^{2}}} $$ Solution Note that no basic integration rule discussed so far applies. To apply trigonometric substitution note that $$ \sqrt{9-x^{2}} \text{ has the form } \sqrt{a^{2}-u^{2}} $$ For the substitution let \(x=a \sin \theta = 3 \sin \theta \). Using differentiation and Figure 8.4.2 produces $$ dx = 3 \cos \theta \: d\theta \text{, } \sqrt{9-x^{2}} =3 \cos \theta \text{, and } x^{2}=9 \sin^{2} \theta .$$ Now trigonometric substitution can be applied. $$ \int \frac{dx}{x^{2}\sqrt{9-x^{2}}} $$ $$= \int \frac{ 3 \cos \theta \: d\theta}{(9 \sin^{2} \theta)(3 \cos \theta)} $$ Substitute $$= \frac{1}{9} \int \frac{ d \theta}{\sin^{2} \theta} $$ Simplify $$= -\frac{1}{9} \int \csc^{2} \theta \:d\theta $$ Trigonometric identity $$= -\frac{1}{9} \cot \theta + C $$ Apply Cosecant Rule $$= -\frac{1}{9} \left( \frac{\sqrt{9-x^{2}}}{x} \right) + C \:\:\:\: $$ Substitute for \(\cot \theta \) $$= - \frac{\sqrt{9-x^{2}}}{9x} + C$$

Example 8.4.2 Trigonometric Substitution with \(u= a \tan \theta\)

Figure 8.4.3

Find $$ \int \frac{dx}{\sqrt{4x^{2}+1}} $$ Solution Section 5.8 describes how inverse hyperbolic functions are used to evaluate integrals with the form $$ \int \frac{du}{\sqrt{u^{2} \pm a^{2}}} \text{, } \int \frac{du}{a^{2} - u^{2}} \text{, and } \int \frac{du}{u\sqrt{a^{2} \pm u^{2}}} .$$ Inverse hyperbolic functions and trigonometric substitution can be combined to solve this problem. Let \(u=2x\), \(a=1\), and \(2x= \tan \theta\), as shown in Figure 8.4.3. This produces $$ dx = \frac{1}{2} \sec^{2} \theta \: d\theta \text{ and } \sqrt{4x^{2}+1} = \sec \theta .$$ Trigonometric substitution produces $$ \int \frac{1}{\sqrt{4x^{2}+1}}\:dx $$ $$= \frac{1}{2} \int \frac{\sec^{2} \theta \: d\theta}{\sec \theta} $$ Substitute $$= \frac{1}{2} \int \sec \theta \: d\theta $$ Simplify $$= \frac{1}{2} \ln | \sec \theta + \tan \theta | + C $$ Apply Secant Rule and Integrate $$= \frac{1}{2} \ln | \sqrt{4x^{2}+1} + 2x | + C \:\:\:\:$$ Back substitute

Example 8.4.3 Trigonometric Substitution with Rational Powers

Figure 8.4.4

Trigonometric substitution can cover integrals involving expressions such as \((a^{2}-u^{2})^{n/2}\) by writing the expression as $$(a^{2} \pm u^{2})^{n/2} = (\sqrt{a^{2} \pm u^{2}})^{n} ,$$ Find $$ \int \frac{dx}{(x^{2}+1)^{3/2}} $$ Solution Begin by writing \((x^{2}+1)^{3/2}\) as $$ ( \sqrt{x^{2} + 1} )^{3} .$$ Let \(a=1\) and \(u=x=\tan \theta\), as shown in Figure 8.4.4. Using $$ dx=\sec^{2} \theta\:d\theta \text{ and }\sqrt{x^{2}+1}=\sec \theta $$ trigonometric substitution and be applied. $$ \int \frac{dx}{(x^{2}+1)^{3/2}} $$ $$= \int \frac{dx}{(\sqrt{x^{2} + 1} )^{3}} \:\:\:\: $$ Rewrite denominator $$= \int \frac{ \sec^{2} \theta\:d\theta}{ \sec^{3} \theta} \:\:\:\: $$ Substitute $$= \int \frac{ d\theta}{ \sec \theta } $$ Simplify $$= \int \cos \theta\: d\theta $$ Trigonometric identity $$= \sin \theta + C $$ Apply Cosine Rule and integrate $$= \frac{x}{\sqrt{x^{2}+1}} + C $$ Back-substitute

Example 8.4.4 Converting Integration Limits

Figure 8.4.5

Evaluate $$ \int_{\sqrt{3}}^{2} \frac{ \sqrt{x^{2}-3} }{ x }\:dx $$ Solution Section 4.5, Examples 4.5.8 and 4.5.9 describes how to determine integration limits for \(\theta\) that avoid converting back to \(x\). Because \( \sqrt{x^{2}-3} \)has the form \( \sqrt{x^{2}-a^{2}} \) consider a triangle where \(u=x\), \(a=\sqrt{3}\), and \(x=\sqrt{3} \sec \theta\) as shown in Figure 8.4.5. This produces $$ dx=\sqrt{3} \sec \theta \tan \theta \:d\theta \text{ and } \sqrt{x^{2}-3}=\sqrt{3} \tan \theta $$ To determine the upper and lower integration limits, use the substitution \( x=\sqrt{3} \sec \theta \) Lower Limit Upper Limit When \(x=\sqrt{3}\), \(\sec \theta = 1\) When \(x=2\), \( \sec \theta = \frac{2}{\sqrt{3}} \) and \(\theta=0\). and \( \theta = \frac{\pi}{6} \). $$ \int_{\sqrt{3}}^{2} \frac{ \sqrt{x^{2}-3} }{ x }\:dx $$ $$= \int_{0}^{\pi/6} \frac{(\sqrt{3}\tan \theta)(\sqrt{3}\sec \theta \tan \theta)d\theta}{\sqrt{3}\sec \theta} $$ $$= \int_{0}^{\pi/6} \sqrt{3}\tan^{2} \theta\: d\theta $$ $$= \sqrt{3} \int_{0}^{\pi/6} (\sec^{2} \theta - 1)\: d\theta $$ $$= \sqrt{3} \left[ \tan \theta - \theta \right]_{0}^{\pi/6} $$ $$= \sqrt{3} \left( \frac{1}{\sqrt{3}} - \frac{\pi}{6} \right) $$ $$= 1 - \frac{\sqrt{3}\pi}{6}\approx 0.0931 $$

Example 8.4.5 Choosing an Interval for \(\theta\)

Find

$$ \int_{-2}^{-\sqrt{3}} \frac{ \sqrt{x^{2}-3} }{ x }\:dx $$

Solution When using trigonometric substitution to evaluate definite integrals make sure the values for \(\theta\) lie in ranges described in the first three guidelines described in this section. This definite integral is the same as in Example 8.4.4 except that the range is negative, \([-2, -\sqrt{3}]\). This makes \(u < -a\), which is covered in guideline 3. Choosing \(\theta\) such that \( \pi /2 < \theta \leqslant \pi\). The integral evaluates as follows

$$ \int_{-2}^{-\sqrt{3}} \frac{ \sqrt{x^{2}-3} }{ x }\:dx $$	$$= \int_{5 \pi /6}^{\pi} \frac{(-\sqrt{3}\tan \theta)(\sqrt{3}\sec \theta \tan \theta)d\theta}{\sqrt{3}\sec \theta}$$
	$$= \int_{5 \pi /6}^{\pi} -\sqrt{3}\tan^{2} \theta\: d\theta $$
	$$= -\sqrt{3} \int_{0}^{\pi/6} (\sec^{2} \theta - 1)\: d\theta $$
	$$= -\sqrt{3} \left[ \tan \theta - \theta \right]_{5 \pi /6}^{\pi} $$
	$$= -\sqrt{3} \left[ (0-\pi) - \left( -\frac{1}{\sqrt{3}} - \frac{5\pi}{6} \right)\right] $$
	$$= -1 + \frac{\sqrt{3}\pi}{6}\approx 0.0931 $$

Theorem 8.4.1 Special Integration Formulas where \((a > 0)\)

Trigonometric substitution can be used to complete the square. For instance, try finding the integral

$$ \int \sqrt{x^{2}-2x}\:dx $$

Complete the square and write the integral as

$$ \int \sqrt{(x-1)^{2}-1^{2}}\:dx $$

The integrand now has the form

$$ \sqrt{u^{2}-a^{2}}$$

with \(u=x-1\) and \(a=1\). Trigonometric substitution can now be used to find the integral.

Theorem 8.4.1 describes these formulas.

Theorem 8.4.1 Special Integration Formulas where \((a > 0)\)
$$ \int \sqrt{a^{2}-u^{2}}\:du $$	$$= \frac{1}{2} \left( a^{2} \arcsin \frac{u}{a} + u \sqrt{a^{2}-u^{2}} \right) + C $$
$$ \int \sqrt{u^{2}-a^{2}}\:du $$	$$= \frac{1}{2} \left( u \sqrt{u^{2}-a^{2}} - a^{2} \ln | u + \sqrt{u^{2}-a^{2}}| \right) + C, \: u > a $$
$$ \int \sqrt{u^{2}+a^{2}}\:du $$	$$= \frac{1}{2} \left( u \sqrt{u^{2}+a^{2}} + a^{2} \ln | u + \sqrt{u^{2}+a^{2}}| \right) + C $$

Example 8.4.6 Finding an Arc Length

The curve's arc length is \(\approx 1.148\). Figure 8.4.6

Find the arc length for the graph \(f(x)= \frac{1}{2} x^{2}\) from \(x=0\) to \(x=1\) as shown in Figure 8.4.6. Solution Refer to the arc length formula in Section 7.4 and apply. \(S\) $$= \int_{0}^{1} \sqrt{1+[f'(x)]^{2}}\:dx $$ Formula for arc length $$= \int_{0}^{1} \sqrt{1+(x)^{2}}\:dx $$ \(f'(x) = x\) $$= \int_{0}^{\pi/4} \sec^{3} \theta \:d\theta $$ Let \(a=1\) and \(x= \tan \theta \) $$= \frac{1}{2} \bigg[ \sec \theta \tan \theta + \ln | \sec \theta + \tan \theta | \bigg]_{0}^{\pi/4} $$ Example 8.2.5 $$= \frac{1}{2} \left[ \sqrt{2} + \ln( \sqrt{2} +1)\right] \approx 1.148$$

Example 8.4.7 Comparing two Fluid Forces

Barrel is almost full with 0.2 cubic feet empty. Figure 8.4.7 Figure 8.4.8

A sealed oil barrel of oil is floating in seawater as shown in Figures 8.4.7 and 8.4.8. The oil weighs 48 pounds per cubic foot. The seawater weighs 64 pounds per cubic foot. The barrel is lying on its side with the top 0.2 feet empty. Compare the fluid forces against one end from the inside and from the outside. Solution Place the coordinate system with the origin at the side's center as shown in Figure 8.4.8. This makes the equation for the surface area $$ x^{2} + y^{2} = 1.$$ To find the fluid force pressing out from the inside, integrate between -1 and 0.8 where \(w=48\). \(F\) $$ = w \int_{c}^{d} h(y)L(y)\: dy $$ Section 7.7 \(F_{\text{inside}}\) $$ = 48 \int_{-1}^{0.8} (0.8-y)(2)\sqrt{1-y^{2}}\: dy $$ $$ = 76.8 \int_{-1}^{0.8} \sqrt{1-y^{2}}\: dy - 96 \int_{-1}^{0.8} y\sqrt{1-y^{2}}\: dy $$ To find the fluid force pressing in from the outside, integrate between -1 and 0.4 where \(w=64\). \(F_{\text{outside}}\) $$ = 64 \int_{-1}^{0.4} (0.4-y)(2)\sqrt{1-y^{2}}\: dy $$ $$ = 51.2 \int_{-1}^{0.4} \sqrt{1-y^{2}}\: dy - 128 \int_{-1}^{0.4} y\sqrt{1-y^{2}}\: dy $$ Integrate $$ \int \sqrt{1-y^{2}}\: dy $$ then apply the variables. Let \(y= \sin u\), \(u= \arcsin y\), and \(\frac{dx}{du} = \cos u\). $$ \int \sqrt{1-y^{2}}\: dy $$ $$= \int \sqrt{-y^{2}+ 1}\: dy $$ Rewrite to ease integration $$= \int \cos u \sqrt{-\sin^{2}y + 1}\: du $$ Trigonometric substitution $$= \int \cos^{2} u \: du $$ Simplify $$= \frac{\cos^{2} u \sin u}{2} + \frac{1}{2} \int 1 \: du $$ Apply reduction formula $$= \frac{\cos^{2} u \sin u}{2} + \frac{u}{2} $$ $$= \frac{\arcsin y }{2} + \frac{y\sqrt{-y^{2}+ 1}}{2} + C\:\:\:\: $$ Back substitution The answer is \( F_{\text{inside}} \approx 121.3 \text{pounds} \) and \( F_{\text{outside}} \approx 93.03 \text{pounds} \)

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Parent Article: Calculus II 08 Integration Techniques