Trigonometric substitution

Let $x\; =\; a\; \backslash sin\; \backslash theta,$ and use the identity $1-\backslash sin^2\; \backslash theta\; =\; \backslash cos^2\; \backslash theta.$

File:Trig Sub Triangle 1.png

In the integral

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

we may use

$$x=a\backslash sin\; \backslash theta,\backslash quad\; dx=a\backslash cos\backslash theta\backslash ,\; d\backslash theta,\; \backslash quad\; \backslash theta=\backslash arcsin\backslash frac\{x\}\{a\}.$$

Then,

$$\backslash begin\{align\}$$

\int\frac{dx}{\sqrt{a^2-x^2}} &= \int\frac{a\cos\theta \,d\theta}{\sqrt{a^2-a^2\sin^2 \theta}} \\[6pt]

&= \int\frac{a\cos\theta\,d\theta}{\sqrt{a^2(1 - \sin^2 \theta )}} \\[6pt]

&= \int\frac{a\cos\theta\,d\theta}{\sqrt{a^2\cos^2\theta}} \\[6pt]

&= \int d\theta \\[6pt]

&= \theta + C \\[6pt]

&= \arcsin\frac{x}{a}+C.

\end{align}

The above step requires that $a\; >\; 0$ and $\backslash cos\; \backslash theta\; >\; 0.$ We can choose $a$ to be the principal root of $a^2,$ and impose the restriction 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 $\backslash sin\; \backslash theta$ goes from $0$ to $1/2,$ so $\backslash theta$ goes from $0$ to $\backslash pi\; /\; 6.$ Then,

$$\backslash int\_0^\{a/2\}\backslash frac\{dx\}\{\backslash sqrt\{a^2-x^2\}\}=\backslash int\_0^\{\backslash pi/6\}\; d\backslash theta\; =\; \backslash frac\{\backslash pi\}\{6\}.$$

Some care is needed when picking the bounds. Because integration above requires that , $\backslash theta$ can only go from $0$ to $\backslash pi\; /\; 6.$ Neglecting this restriction, one might have picked $\backslash theta$ to go from $\backslash pi$ to $5\backslash pi\; /6,$ 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

$$\backslash int\_\{0\}^\{a/2\}\; \backslash frac\{dx\}\{\backslash sqrt\{a^2\; -\; x^2\}\}\; =\; \backslash arcsin\; \backslash left(\; \backslash frac\{x\}\{a\}\; \backslash right)\; \backslash Biggl|\_\{0\}^\{a/2\}$$

= \arcsin \left ( \frac{1}{2}\right) - \arcsin (0) = \frac{\pi}{6} as before.

The integral

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

may be evaluated by letting $x=a\backslash sin\; \backslash theta,\backslash ,\; dx=a\backslash cos\backslash theta\backslash ,\; d\backslash theta,\backslash ,\; \backslash theta=\backslash arcsin\backslash dfrac\{x\}\{a\},$ where $a\; >\; 0$ so that $\backslash sqrt\{a^2\}=a,$ and $-\backslash pi/2\; \backslash le\; \backslash theta\; \backslash le\; \backslash pi/2$ by the range of arcsine, so that $\backslash cos\; \backslash theta\; \backslash ge\; 0$ and $\backslash sqrt\{\backslash cos^2\; \backslash theta\}\; =\; \backslash cos\; \backslash theta.$

Then,

$$\backslash begin\{align\}$$

\int\sqrt{a^2-x^2}\,dx &= \int\sqrt{a^2-a^2\sin^2\theta}\,(a\cos\theta) \,d\theta \\[6pt]

&= \int\sqrt{a^2(1-\sin^2\theta)}\,(a\cos\theta) \,d\theta \\[6pt]

&= \int\sqrt{a^2(\cos^2\theta)}\,(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{1+\cos 2\theta}{2}\right)\,d\theta \\[6pt]

&= \frac{a^2}{2} \left(\theta+\frac{1}{2}\sin 2\theta \right) + C \\[6pt]

&= \frac{a^2}{2}(\theta+\sin\theta\cos\theta) + C \\[6pt]

&= \frac{a^2}{2}\left(\arcsin\frac{x}{a}+\frac{x}{a}\sqrt{1-\frac{x^2}{a^2}}\right) + C \\[6pt]

&= \frac{a^2}{2}\arcsin\frac{x}{a}+\frac{x}{2}\sqrt{a^2-x^2}+C.

\end{align}

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

For example, the definite integral

$$\backslash int\_\{-1\}^1\backslash sqrt\{4-x^2\}\backslash ,dx,$$

may be evaluated by substituting $x\; =\; 2\backslash sin\backslash theta,\; \backslash ,dx\; =\; 2\backslash cos\backslash theta\backslash ,d\backslash theta,$ with the bounds determined using $\backslash theta\; =\; \backslash arcsin\backslash dfrac\{x\}\{2\}.$

Because $\backslash arcsin(1/\{2\})\; =\; \backslash pi/6$ and $\backslash arcsin(-1/2)\; =\; -\backslash pi/6,$

$$\backslash begin\{align\}$$

\int_{-1}^1\sqrt{4-x^2}\,dx &= \int_{-\pi/6}^{\pi/6}\sqrt{4-4\sin^2\theta}\,(2\cos\theta) \,d\theta \\[6pt]

&= \int_{-\pi/6}^{\pi/6}\sqrt{4(1-\sin^2\theta)}\,(2\cos\theta) \,d\theta \\[6pt]

&= \int_{-\pi/6}^{\pi/6}\sqrt{4(\cos^2\theta)}\,(2\cos\theta) \,d\theta \\[6pt]

&= \int_{-\pi/6}^{\pi/6}(2\cos\theta)(2\cos\theta) \,d\theta \\[6pt]

&= 4\int_{-\pi/6}^{\pi/6}\cos^2\theta\,d\theta \\[6pt]

&= 4\int_{-\pi/6}^{\pi/6}\left(\frac{1+\cos 2\theta}{2}\right)\,d\theta \\[6pt]

&= 2 \left[\theta+\frac{1}{2} \sin 2\theta \right]^{\pi/6}_{-\pi/6}

= [2\theta+\sin 2\theta] \Biggl |^{\pi/6}_{-\pi/6} \\[6pt]

&= \left(\frac{\pi}{3}+\sin\frac{\pi}{3}\right)-\left(-\frac{\pi}{3}+\sin\left(-\frac{\pi}{3}\right)\right)

= \frac{2\pi}{3}+\sqrt{3}.

\end{align}

On the other hand, direct application of the boundary terms to the previously obtained formula for the antiderivative yields

$$\backslash begin\{align\}$$

\int_{-1}^1\sqrt{4-x^2}\,dx &= \left[ \frac{2^2}{2}\arcsin\frac{x}{2}+\frac{x}{2}\sqrt{2^2-x^2} \right]_{-1}^{1}\\[6pt]

&= \left( 2 \arcsin \frac{1}{2} + \frac{1}{2}\sqrt{4-1}\right) - \left( 2 \arcsin \left(-\frac{1}{2}\right) + \frac{-1}{2}\sqrt{4-1}\right)\\[6pt]

&= \left( 2 \cdot \frac{\pi}{6} + \frac{\sqrt{3}}{2}\right) - \left( 2\cdot \left(-\frac{\pi}{6}\right) - \frac{\sqrt 3}{2}\right)\\[6pt]

&= \frac{2\pi}{3} + \sqrt{3}

\end{align}

as before.

Let $x\; =\; a\; \backslash tan\; \backslash theta,$ and use the identity $1+\backslash tan^2\; \backslash theta\; =\; \backslash sec^2\; \backslash theta.$

File:Trig Sub Triangle 2.png

In the integral

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

we may write

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

so that the integral becomes

$$\backslash begin\{align\}$$

\int\frac{dx}{a^2+x^2} &= \int\frac{a\sec^2\theta\, d\theta}{a^2 + a^2\tan^2\theta} \\[6pt]

&= \int\frac{a\sec^2\theta\, d\theta}{a^2(1+\tan^2\theta)} \\[6pt]

&= \int\frac{a\sec^2\theta\, d\theta}{a^2\sec^2\theta} \\[6pt]

&= \int\frac{d\theta}{a} \\[6pt]

&= \frac{\theta}{a}+C \\[6pt]

&= \frac{1}{a} \arctan \frac{x}{a} + C,

\end{align}

provided $a\; \backslash neq\; 0.$

For a definite integral, the bounds change once the substitution is performed and are determined using the equation $\backslash theta\; =\; \backslash arctan\backslash frac\{x\}\{a\},$ with values in the range Alternatively, apply the boundary terms directly to the formula for the antiderivative.

For example, the definite integral

$$\backslash int\_0^1\backslash frac\{4\backslash ,\; dx\}\{1+x^2\}\backslash ,$$

may be evaluated by substituting $x\; =\; \backslash tan\backslash theta,\; \backslash ,dx\; =\; \backslash sec^2\backslash theta\backslash ,d\backslash theta,$ with the bounds determined using $\backslash theta\; =\; \backslash arctan\; x.$

Since $\backslash arctan\; 0\; =\; 0$ and $\backslash arctan\; 1\; =\; \backslash pi/4,$

$$\backslash begin\{align\}$$

\int_0^1\frac{4\,dx}{1+x^2} &= 4\int_0^1\frac{dx}{1 + x^2} \\[6pt]

&= 4\int_0^{\pi/4}\frac{\sec^2\theta\, d\theta}{1+\tan^2\theta} \\[6pt]

&= 4\int_0^{\pi/4}\frac{\sec^2\theta\, d\theta}{\sec^2\theta} \\[6pt]

&= 4\int_0^{\pi/4}d\theta \\[6pt]

&= (4\theta)\Bigg|^{\pi/4}_0 = 4 \left (\frac{\pi}{4} - 0 \right) = \pi.

\end{align}

Meanwhile, direct application of the boundary terms to the formula for the antiderivative yields

$$\backslash begin\{align\}$$

\int_0^1\frac{4\,dx}{1+x^2}\,

&= 4\int_0^1\frac{dx}{1+x^2} \\[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{\pi}{4} - 0 \right) = \pi,

\end{align}

same as before.

The integral

$$\backslash int\backslash sqrt\{a^2+x^2\}\backslash ,\{dx\}$$

may be evaluated by letting $x=a\backslash tan\backslash theta,\backslash ,\; dx=a\backslash sec^2\backslash theta\backslash ,\; d\backslash theta,\; \backslash ,\; \backslash theta=\backslash arctan\backslash frac\{x\}\{a\},$

where $a\; >\; 0$ so that $\backslash sqrt\{a^2\}=a,$ and by the range of arctangent, so that $\backslash sec\; \backslash theta\; >\; 0$ and $\backslash sqrt\{\backslash sec^2\; \backslash theta\}\; =\; \backslash sec\; \backslash theta.$

Then,

$$\backslash begin\{align\}$$

\int\sqrt{a^2+x^2}\,dx &= \int\sqrt{a^2 + a^2\tan^2\theta}\,(a \sec^2\theta)\, d\theta \\[6pt]

&= \int\sqrt{a^2 (1+\tan^2\theta)}\,(a \sec^2\theta)\, d\theta \\[6pt]

&= \int\sqrt{a^2 \sec^2\theta}\,(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{align}

The integral of secant cubed may be evaluated using integration by parts. As a result,

$$\backslash begin\{align\}$$

\int\sqrt{a^2+x^2}\,dx &= \frac{a^2}{2}(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|)+C \\[6pt]

&= \frac{a^2}{2}\left(\sqrt{1+\frac{x^2}{a^2}}\cdot\frac{x}{a} + \ln\left|\sqrt{1+\frac{x^2}{a^2}}+\frac{x}{a}\right|\right)+C \\[6pt]

&= \frac{1}{2}\left(x\sqrt{a^2+x^2} + a^2\ln\left|\frac{x+\sqrt{a^2+x^2}}{a}\right|\right)+C.

\end{align}

Let $x\; =\; a\; \backslash sec\; \backslash theta,$ and use the identity $\backslash sec^2\; \backslash theta\; -1\; =\; \backslash tan^2\; \backslash theta.$

File:Trig Sub Triangle 3.png

Integrals such as

$$\backslash int\backslash frac\{dx\}\{x^2\; -\; a^2\}$$

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

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

cannot. In this case, an appropriate substitution is:

$$x\; =\; a\; \backslash sec\backslash theta,\backslash ,\; dx\; =\; a\; \backslash sec\backslash theta\backslash tan\backslash theta\backslash ,\; d\backslash theta,\; \backslash ,\; \backslash theta\; =\; \backslash arcsec\backslash frac\{x\}\{a\},$$

where $a\; >\; 0$ so that $\backslash sqrt\{a^2\}=a,$ and by assuming $x\; >\; 0,$ so that $\backslash tan\; \backslash theta\; \backslash ge\; 0$ and $\backslash sqrt\{\backslash tan^2\; \backslash theta\}\; =\; \backslash tan\; \backslash theta.$

Then,

$$\backslash begin\{align\}$$

\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}

One may evaluate the integral of the secant function by multiplying the numerator and denominator by $(\; \backslash sec\; \backslash theta\; +\; \backslash tan\; \backslash theta)$ and the integral of secant cubed by parts.{{Cite book|last=Stewart|first=James|title=Calculus - Early Transcendentals|publisher=Cengage Learning|year=2012|isbn=978-0-538-49790-9|location=United States|pages=475–6|chapter=Section 7.2: Trigonometric Integrals|author-link=James Stewart (mathematician)}} As a result,

$$\backslash begin\{align\}$$

\int\sqrt{x^2-a^2}\,dx &= \frac{a^2}{2}(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|)-a^2\ln|\sec\theta+\tan\theta|+C \\[6pt]

&= \frac{a^2}{2}(\sec\theta \tan\theta - \ln|\sec\theta+\tan\theta|)+C \\[6pt]

&= \frac{a^2}{2}\left(\frac{x}{a}\cdot\sqrt{\frac{x^2}{a^2}-1} - \ln\left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right|\right)+C \\[6pt]

&= \frac{1}{2}\left(x\sqrt{x^2-a^2} - a^2\ln\left|\frac{x+\sqrt{x^2-a^2}}{a}\right|\right)+C.

\end{align}

When which happens when given the range of arcsecant, $\backslash tan\; \backslash theta\; \backslash le\; 0,$ meaning $\backslash sqrt\{\backslash tan^2\; \backslash theta\}\; =\; -\backslash tan\; \backslash theta$ instead in that case.

Substitution can be used to remove trigonometric functions.

For instance,

$$\backslash begin\{align\}$$

\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) \\[6pt]

\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) \\[6pt]

\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 (\frac{x}{2} \right ) \\[6pt]

\end{align}

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

For example,

$$\backslash begin\{align\}$$

\int\frac{4 \cos x}{(1+\cos x)^3}\, dx &= \int\frac2{1+u^2}\frac{4\left(\frac{1-u^2}{1+u^2}\right)}{\left(1+\frac{1-u^2}{1+u^2}\right)^3}\, du = \int (1-u^2)(1+u^2)\, du \\&= \int (1-u^4)\,du = u - \frac{u^5}{5} + C = \tan \frac{x}{2} - \frac{1}{5} \tan^5 \frac{x}{2} + C.

\end{align}

Substitutions of hyperbolic functions can also be used to simplify integrals.{{cite web|last=Boyadzhiev|first=Khristo N.|title=Hyperbolic Substitutions for Integrals|url=http://www2.onu.edu/~m-caragiu.1/bonus_files/HYPERSUB.pdf|access-date=4 March 2013|archive-date=26 February 2020|archive-url=https://web.archive.org/web/20200226040813/http://www2.onu.edu/~m-caragiu.1/bonus_files/HYPERSUB.pdf|url-status=dead}}

For example, to integrate $1/\backslash sqrt\{a^2+x^2\}$, introduce the substitution $x=a\backslash sinh\{u\}$ (and hence $dx=a\backslash cosh\; u\; \backslash ,du$), then use the identity $\backslash cosh^2\; (x)\; -\; \backslash sinh^2\; (x)\; =\; 1$ to find:

$$\backslash begin\{align\}$$

\int \frac{dx}{\sqrt{a^2+x^2}} &= \int \frac{a\cosh u \,du}{\sqrt{a^2+a^2\sinh^2 u}} \\[6pt]

&=\int \frac{\cosh{u} \,du}{\sqrt{1+\sinh^2{u}}} \\[6pt]

&=\int \frac{\cosh{u}}{\cosh u} \,du \\[6pt]

&=u+C \\[6pt]

&=\sinh^{-1}{\frac{x}{a}} + C.

\end{align}

If desired, this result may be further transformed using other identities, such as using the relation $\backslash sinh^\{-1\}\{z\}\; =\; \backslash operatorname\{arsinh\}\{z\}\; =\; \backslash ln(z\; +\; \backslash sqrt\{z^2\; +\; 1\})$:

$$\backslash begin\{align\}$$

\sinh^{-1}{\frac{x}{a}} + C

&=\ln\left(\frac{x}{a} + \sqrt{\frac{x^2}{a^2} + 1}\,\right) + C \\[6pt]

&=\ln\left(\frac{x + \sqrt{x^2+a^2}}{a}\,\right) + C.

\end{align}

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{{Wikiversity|Trigonometric Substitutions}}

{{Wikibooks|Calculus/Integration techniques/Trigonometric Substitution}}

• Integration by substitution
• Weierstrass substitution
• Euler substitution

{{reflist}}

{{DEFAULTSORT:Trigonometric Substitution}}

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Category:Integral calculus

Category:Trigonometry