Trigonometric substitution

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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.

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

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.

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.

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

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.

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, $${\displaystyle {\begin{aligned}\int {\sqrt {a^{2}+x^{2}}}\,dx&={\frac {a^{2}}{2}}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)+C\\[6pt]&={\frac {a^{2}}{2}}\left({\sqrt {1+{\frac {x^{2}}{a^{2}}}}}\cdot {\frac {x}{a}}+\ln \left|{\sqrt {1+{\frac {x^{2}}{a^{2}}}}}+{\frac {x}{a}}\right|\right)+C\\[6pt]&={\frac {1}{2}}\left(x{\sqrt {a^{2}+x^{2}}}+a^{2}\ln \left|{\frac {x+{\sqrt {a^{2}+x^{2}}}}{a}}\right|\right)+C.\end{aligned}}}$$

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

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.

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,

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

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 ${\displaystyle \sinh ^{-1}{z}=\operatorname {arsinh} {z}=\ln(z+{\sqrt {z^{2}+1}})}$: $${\displaystyle {\begin{aligned}\sinh ^{-1}{\frac {x}{a}}+C&=\ln \left({\frac {x}{a}}+{\sqrt {{\frac {x^{2}}{a^{2}}}+1}}\,\right)+C\\[6pt]&=\ln \left({\frac {x+{\sqrt {x^{2}+a^{2}}}}{a}}\,\right)+C.\end{aligned}}}$$

Wikiversity has learning resources about Trigonometric Substitutions

Wikibooks has a book on the topic of: Calculus/Integration techniques/Trigonometric Substitution

• Weierstrass substitution
• Euler substitution