Trigonometric substitution

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]

In the case of a definite integral, this method of integration by substitution uses the substitution to change the interval of integration. Alternatively, the antiderivative of the integrand may be applied to the original interval.

Case I: Integrands containing a² − x²

Let ${\displaystyle x=a\sin \theta ,}$ and use the identity ${\displaystyle 1-\sin ^{2}\theta =\cos ^{2}\theta .}$

Examples of Case I

Geometric construction for Case I

Example 1

In the integral

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

we may use

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

Then, $${\displaystyle {\begin{aligned}\int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}&=\int {\frac {a\cos \theta \,d\theta }{\sqrt {a^{2}-a^{2}\sin ^{2}\theta }}}\\[6pt]&=\int {\frac {a\cos \theta \,d\theta }{\sqrt {a^{2}(1-\sin ^{2}\theta )}}}\\[6pt]&=\int {\frac {a\cos \theta \,d\theta }{\sqrt {a^{2}\cos ^{2}\theta }}}\\[6pt]&=\int d\theta \\[6pt]&=\theta +C\\[6pt]&=\arcsin {\frac {x}{a}}+C.\end{aligned}}}$$

The above step requires that ${\displaystyle a>0}$ and ${\displaystyle \cos \theta >0.}$ We can choose ${\displaystyle a}$ to be the principal root of ${\displaystyle a^{2},}$ and impose the restriction ${\displaystyle -\pi /2<\theta <\pi /2}$ by using the inverse sine function.

For a definite integral, one must figure out how the bounds of integration change. For example, as ${\displaystyle x}$ goes from ${\displaystyle 0}$ to ${\displaystyle a/2,}$ then ${\displaystyle \sin \theta }$ goes from ${\displaystyle 0}$ to ${\displaystyle 1/2,}$ so ${\displaystyle \theta }$ goes from ${\displaystyle 0}$ to ${\displaystyle \pi /6.}$ Then,

$${\displaystyle \int _{0}^{a/2}{\frac {dx}{\sqrt {a^{2}-x^{2}}}}=\int _{0}^{\pi /6}d\theta ={\frac {\pi }{6}}.}$$

Some care is needed when picking the bounds. Because integration above requires that ${\displaystyle -\pi /2<\theta <\pi /2}$ , ${\displaystyle \theta }$ can only go from ${\displaystyle 0}$ to ${\displaystyle \pi /6.}$ Neglecting this restriction, one might have picked ${\displaystyle \theta }$ to go from ${\displaystyle \pi }$ to ${\displaystyle 5\pi /6,}$ 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

$${\displaystyle \int _{0}^{a/2}{\frac {dx}{\sqrt {a^{2}-x^{2}}}}=\arcsin \left({\frac {x}{a}}\right){\Biggl |}_{0}^{a/2}=\arcsin \left({\frac {1}{2}}\right)-\arcsin(0)={\frac {\pi }{6}}}$$ as before.

Example 2

The integral

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

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

Then, $${\displaystyle {\begin{aligned}\int {\sqrt {a^{2}-x^{2}}}\,dx&=\int {\sqrt {a^{2}-a^{2}\sin ^{2}\theta }}\,(a\cos \theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}(1-\sin ^{2}\theta )}}\,(a\cos \theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}(\cos ^{2}\theta )}}\,(a\cos \theta )\,d\theta \\[6pt]&=\int (a\cos \theta )(a\cos \theta )\,d\theta \\[6pt]&=a^{2}\int \cos ^{2}\theta \,d\theta \\[6pt]&=a^{2}\int \left({\frac {1+\cos 2\theta }{2}}\right)\,d\theta \\[6pt]&={\frac {a^{2}}{2}}\left(\theta +{\frac {1}{2}}\sin 2\theta \right)+C\\[6pt]&={\frac {a^{2}}{2}}(\theta +\sin \theta \cos \theta )+C\\[6pt]&={\frac {a^{2}}{2}}\left(\arcsin {\frac {x}{a}}+{\frac {x}{a}}{\sqrt {1-{\frac {x^{2}}{a^{2}}}}}\right)+C\\[6pt]&={\frac {a^{2}}{2}}\arcsin {\frac {x}{a}}+{\frac {x}{2}}{\sqrt {a^{2}-x^{2}}}+C.\end{aligned}}}$$

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

For example, the definite integral

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

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

Because ${\displaystyle \arcsin(1/{2})=\pi /6}$ and ${\displaystyle \arcsin(-1/2)=-\pi /6,}$ $${\displaystyle {\begin{aligned}\int _{-1}^{1}{\sqrt {4-x^{2}}}\,dx&=\int _{-\pi /6}^{\pi /6}{\sqrt {4-4\sin ^{2}\theta }}\,(2\cos \theta )\,d\theta \\[6pt]&=\int _{-\pi /6}^{\pi /6}{\sqrt {4(1-\sin ^{2}\theta )}}\,(2\cos \theta )\,d\theta \\[6pt]&=\int _{-\pi /6}^{\pi /6}{\sqrt {4(\cos ^{2}\theta )}}\,(2\cos \theta )\,d\theta \\[6pt]&=\int _{-\pi /6}^{\pi /6}(2\cos \theta )(2\cos \theta )\,d\theta \\[6pt]&=4\int _{-\pi /6}^{\pi /6}\cos ^{2}\theta \,d\theta \\[6pt]&=4\int _{-\pi /6}^{\pi /6}\left({\frac {1+\cos 2\theta }{2}}\right)\,d\theta \\[6pt]&=2\left[\theta +{\frac {1}{2}}\sin 2\theta \right]_{-\pi /6}^{\pi /6}=[2\theta +\sin 2\theta ]{\Biggl |}_{-\pi /6}^{\pi /6}\\[6pt]&=\left({\frac {\pi }{3}}+\sin {\frac {\pi }{3}}\right)-\left(-{\frac {\pi }{3}}+\sin \left(-{\frac {\pi }{3}}\right)\right)={\frac {2\pi }{3}}+{\sqrt {3}}.\end{aligned}}}$$

On the other hand, direct application of the boundary terms to the previously obtained formula for the antiderivative yields $${\displaystyle {\begin{aligned}\int _{-1}^{1}{\sqrt {4-x^{2}}}\,dx&=\left[{\frac {2^{2}}{2}}\arcsin {\frac {x}{2}}+{\frac {x}{2}}{\sqrt {2^{2}-x^{2}}}\right]_{-1}^{1}\\[6pt]&=\left(2\arcsin {\frac {1}{2}}+{\frac {1}{2}}{\sqrt {4-1}}\right)-\left(2\arcsin \left(-{\frac {1}{2}}\right)+{\frac {-1}{2}}{\sqrt {4-1}}\right)\\[6pt]&=\left(2\cdot {\frac {\pi }{6}}+{\frac {\sqrt {3}}{2}}\right)-\left(2\cdot \left(-{\frac {\pi }{6}}\right)-{\frac {\sqrt {3}}{2}}\right)\\[6pt]&={\frac {2\pi }{3}}+{\sqrt {3}}\end{aligned}}}$$ as before.

Case II: Integrands containing a² + x²

Let ${\displaystyle x=a\tan \theta ,}$ and use the identity ${\displaystyle 1+\tan ^{2}\theta =\sec ^{2}\theta .}$

Examples of Case II

Geometric construction for Case II

Example 1

In the integral

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

we may write

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

so that the integral becomes

$${\displaystyle {\begin{aligned}\int {\frac {dx}{a^{2}+x^{2}}}&=\int {\frac {a\sec ^{2}\theta \,d\theta }{a^{2}+a^{2}\tan ^{2}\theta }}\\[6pt]&=\int {\frac {a\sec ^{2}\theta \,d\theta }{a^{2}(1+\tan ^{2}\theta )}}\\[6pt]&=\int {\frac {a\sec ^{2}\theta \,d\theta }{a^{2}\sec ^{2}\theta }}\\[6pt]&=\int {\frac {d\theta }{a}}\\[6pt]&={\frac {\theta }{a}}+C\\[6pt]&={\frac {1}{a}}\arctan {\frac {x}{a}}+C,\end{aligned}}}$$

provided ${\displaystyle a\neq 0.}$

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

For example, the definite integral

$${\displaystyle \int _{0}^{1}{\frac {4\,dx}{1+x^{2}}}\,}$$

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

Since ${\displaystyle \arctan 0=0}$ and ${\displaystyle \arctan 1=\pi /4,}$ $${\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {4\,dx}{1+x^{2}}}&=4\int _{0}^{1}{\frac {dx}{1+x^{2}}}\\[6pt]&=4\int _{0}^{\pi /4}{\frac {\sec ^{2}\theta \,d\theta }{1+\tan ^{2}\theta }}\\[6pt]&=4\int _{0}^{\pi /4}{\frac {\sec ^{2}\theta \,d\theta }{\sec ^{2}\theta }}\\[6pt]&=4\int _{0}^{\pi /4}d\theta \\[6pt]&=(4\theta ){\Bigg |}_{0}^{\pi /4}=4\left({\frac {\pi }{4}}-0\right)=\pi .\end{aligned}}}$$

Meanwhile, direct application of the boundary terms to the formula for the antiderivative yields $${\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {4\,dx}{1+x^{2}}}\,&=4\int _{0}^{1}{\frac {dx}{1+x^{2}}}\\[6pt]&=4\left[{\frac {1}{1}}\arctan {\frac {x}{1}}\right]_{0}^{1}\\[6pt]&=4(\arctan x){\Bigg |}_{0}^{1}\\[6pt]&=4(\arctan 1-\arctan 0)\\[6pt]&=4\left({\frac {\pi }{4}}-0\right)=\pi ,\end{aligned}}}$$ same as before.

Example 2

The integral

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

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

where ${\displaystyle a>0}$ so that ${\displaystyle {\sqrt {a^{2}}}=a,}$ and ${\displaystyle -{\frac {\pi }{2}}<\theta <{\frac {\pi }{2}}}$ by the range of arctangent, so that ${\displaystyle \sec \theta >0}$ and ${\displaystyle {\sqrt {\sec ^{2}\theta }}=\sec \theta .}$

Then, $${\displaystyle {\begin{aligned}\int {\sqrt {a^{2}+x^{2}}}\,dx&=\int {\sqrt {a^{2}+a^{2}\tan ^{2}\theta }}\,(a\sec ^{2}\theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}(1+\tan ^{2}\theta )}}\,(a\sec ^{2}\theta )\,d\theta \\[6pt]&=\int {\sqrt {a^{2}\sec ^{2}\theta }}\,(a\sec ^{2}\theta )\,d\theta \\[6pt]&=\int (a\sec \theta )(a\sec ^{2}\theta )\,d\theta \\[6pt]&=a^{2}\int \sec ^{3}\theta \,d\theta .\\[6pt]\end{aligned}}}$$ The integral of secant cubed may be evaluated using integration by parts. As a result, $${\displaystyle {\begin{aligned}\int {\sqrt {a^{2}+x^{2}}}\,dx&={\frac {a^{2}}{2}}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)+C\\[6pt]&={\frac {a^{2}}{2}}\left({\sqrt {1+{\frac {x^{2}}{a^{2}}}}}\cdot {\frac {x}{a}}+\ln \left|{\sqrt {1+{\frac {x^{2}}{a^{2}}}}}+{\frac {x}{a}}\right|\right)+C\\[6pt]&={\frac {1}{2}}\left(x{\sqrt {a^{2}+x^{2}}}+a^{2}\ln \left|{\frac {x+{\sqrt {a^{2}+x^{2}}}}{a}}\right|\right)+C.\end{aligned}}}$$

Case III: Integrands containing x² − a²

Let ${\displaystyle x=a\sec \theta ,}$ and use the identity ${\displaystyle \sec ^{2}\theta -1=\tan ^{2}\theta .}$

Examples of Case III

Geometric construction for Case III

Integrals such as

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

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

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

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

where ${\displaystyle a>0}$ so that ${\displaystyle {\sqrt {a^{2}}}=a,}$ and ${\displaystyle 0\leq \theta <{\frac {\pi }{2}}}$ by assuming ${\displaystyle x>0,}$ so that ${\displaystyle \tan \theta \geq 0}$ and ${\displaystyle {\sqrt {\tan ^{2}\theta }}=\tan \theta .}$

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

One may evaluate the integral of the secant function by multiplying the numerator and denominator by ${\displaystyle (\sec \theta +\tan \theta )}$ and the integral of secant cubed by parts.^[3] As a result, $${\displaystyle {\begin{aligned}\int {\sqrt {x^{2}-a^{2}}}\,dx&={\frac {a^{2}}{2}}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)-a^{2}\ln |\sec \theta +\tan \theta |+C\\[6pt]&={\frac {a^{2}}{2}}(\sec \theta \tan \theta -\ln |\sec \theta +\tan \theta |)+C\\[6pt]&={\frac {a^{2}}{2}}\left({\frac {x}{a}}\cdot {\sqrt {{\frac {x^{2}}{a^{2}}}-1}}-\ln \left|{\frac {x}{a}}+{\sqrt {{\frac {x^{2}}{a^{2}}}-1}}\right|\right)+C\\[6pt]&={\frac {1}{2}}\left(x{\sqrt {x^{2}-a^{2}}}-a^{2}\ln \left|{\frac {x+{\sqrt {x^{2}-a^{2}}}}{a}}\right|\right)+C.\end{aligned}}}$$

When ${\displaystyle {\frac {\pi }{2}}<\theta \leq \pi ,}$ which happens when ${\displaystyle x<0}$ given the range of arcsecant, ${\displaystyle \tan \theta \leq 0,}$ meaning ${\displaystyle {\sqrt {\tan ^{2}\theta }}=-\tan \theta }$ instead in that case.

Substitutions that eliminate trigonometric functions

Substitution can be used to remove trigonometric functions.

For instance,

$${\displaystyle {\begin{aligned}\int f(\sin(x),\cos(x))\,dx&=\int {\frac {1}{\pm {\sqrt {1-u^{2}}}}}f\left(u,\pm {\sqrt {1-u^{2}}}\right)\,du&&u=\sin(x)\\[6pt]\int f(\sin(x),\cos(x))\,dx&=\int {\frac {1}{\mp {\sqrt {1-u^{2}}}}}f\left(\pm {\sqrt {1-u^{2}}},u\right)\,du&&u=\cos(x)\\[6pt]\int f(\sin(x),\cos(x))\,dx&=\int {\frac {2}{1+u^{2}}}f\left({\frac {2u}{1+u^{2}}},{\frac {1-u^{2}}{1+u^{2}}}\right)\,du&&u=\tan \left({\frac {x}{2}}\right)\\[6pt]\end{aligned}}}$$

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

For example,

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

Hyperbolic substitution

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

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

$${\displaystyle {\begin{aligned}\int {\frac {dx}{\sqrt {a^{2}+x^{2}}}}&=\int {\frac {a\cosh u\,du}{\sqrt {a^{2}+a^{2}\sinh ^{2}u}}}\\[6pt]&=\int {\frac {\cosh {u}\,du}{\sqrt {1+\sinh ^{2}{u}}}}\\[6pt]&=\int {\frac {\cosh {u}}{\cosh u}}\,du\\[6pt]&=u+C\\[6pt]&=\sinh ^{-1}{\frac {x}{a}}+C.\end{aligned}}}$$

If desired, this result may be further transformed using other identities, such as using the relation ${\displaystyle \sinh ^{-1}{z}=\operatorname {arsinh} {z}=\ln(z+{\sqrt {z^{2}+1}})}$: $${\displaystyle {\begin{aligned}\sinh ^{-1}{\frac {x}{a}}+C&=\ln \left({\frac {x}{a}}+{\sqrt {{\frac {x^{2}}{a^{2}}}+1}}\,\right)+C\\[6pt]&=\ln \left({\frac {x+{\sqrt {x^{2}+a^{2}}}}{a}}\,\right)+C.\end{aligned}}}$$

See also

• Mathematics portal

• Weierstrass substitution
• Euler substitution

References

1. Stewart, James (2008). Calculus: Early Transcendentals (6th ed.). Brooks/Cole. ISBN 978-0-495-01166-8.
2. Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th ed.). Addison-Wesley. ISBN 978-0-321-58876-0.
3. Stewart, James (2012). "Section 7.2: Trigonometric Integrals". Calculus - Early Transcendentals. United States: Cengage Learning. pp. 475–6. ISBN 978-0-538-49790-9.
4. Boyadzhiev, Khristo N. "Hyperbolic Substitutions for Integrals" (PDF). Archived from the original (PDF) on 26 February 2020. Retrieved 4 March 2013.