In mathematics , trigonometric substitution is the substitution of trigonometric functions for other expressions. One may use the trigonometric identities to simplify certain integrals containing radical expressions :[1] [2]
Substitution 1. If the integrand contains a 2 − x 2 , let
x
=
a
sin
(
θ
)
{\displaystyle x=a\sin(\theta )}
and use the identity
1
−
sin
2
(
θ
)
=
cos
2
(
θ
)
.
{\displaystyle 1-\sin ^{2}(\theta )=\cos ^{2}(\theta ).}
Substitution 2. If the integrand contains a 2 + x 2 , let
x
=
a
tan
(
θ
)
{\displaystyle x=a\tan(\theta )}
and use the identity
1
+
tan
2
(
θ
)
=
sec
2
(
θ
)
.
{\displaystyle 1+\tan ^{2}(\theta )=\sec ^{2}(\theta ).}
Substitution 3. If the integrand contains x 2 − a 2 , let
x
=
a
sec
(
θ
)
{\displaystyle x=a\sec(\theta )}
and use the identity
sec
2
(
θ
)
−
1
=
tan
2
(
θ
)
.
{\displaystyle \sec ^{2}(\theta )-1=\tan ^{2}(\theta ).}
Examples
Integrals containing a 2 − x 2
In the integral
∫
d
x
a
2
−
x
2
{\displaystyle \int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}}
we may use
x
=
a
sin
(
θ
)
,
d
x
=
a
cos
(
θ
)
d
θ
,
θ
=
arcsin
(
x
a
)
{\displaystyle x=a\sin(\theta ),\quad dx=a\cos(\theta )\,d\theta ,\quad \theta =\arcsin \left({\frac {x}{a}}\right)}
∫
d
x
a
2
−
x
2
=
∫
a
cos
(
θ
)
d
θ
a
2
−
a
2
sin
2
(
θ
)
=
∫
a
cos
(
θ
)
d
θ
a
2
(
1
−
sin
2
(
θ
)
)
=
∫
a
cos
(
θ
)
d
θ
a
2
cos
2
(
θ
)
=
∫
d
θ
=
θ
+
C
=
arcsin
(
x
a
)
+
C
{\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 )}}}\\&=\int {\frac {a\cos(\theta )\,d\theta }{\sqrt {a^{2}(1-\sin ^{2}(\theta ))}}}\\&=\int {\frac {a\cos(\theta )\,d\theta }{\sqrt {a^{2}\cos ^{2}(\theta )}}}\\&=\int d\theta \\&=\theta +C\\&=\arcsin \left({\tfrac {x}{a}}\right)+C\end{aligned}}}
Note that the above step requires that a > 0 and cos(θ) > 0; we can choose the a to be the positive square root of a 2 ; and we impose the restriction on θ to be −π/2 < θ < π/2 by using the arcsin 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 sin(θ) goes from 0 to 1/2, so θ goes from 0 to π/6. Then we have
∫
0
a
2
d
x
a
2
−
x
2
=
∫
0
π
6
d
θ
=
π
6
.
{\displaystyle \int _{0}^{\frac {a}{2}}{\frac {dx}{\sqrt {a^{2}-x^{2}}}}=\int _{0}^{\frac {\pi }{6}}d\theta ={\tfrac {\pi }{6}}.}
Some care is needed when picking the bounds. The integration above requires that −π/2 < θ < π/2, so θ going from 0 to π/6 is the only choice. If we had missed this restriction, we might have picked θ to go from π to 5π/6, which would give us the negative of the result.
Integrals containing a 2 + x 2
In the integral
∫
d
x
a
2
+
x
2
{\displaystyle \int {\frac {dx}{a^{2}+x^{2}}}}
we may write
x
=
a
tan
(
θ
)
,
d
x
=
a
sec
2
(
θ
)
d
θ
,
θ
=
arctan
(
x
a
)
{\displaystyle x=a\tan(\theta ),\quad dx=a\sec ^{2}(\theta )\,d\theta ,\quad \theta =\arctan \left({\tfrac {x}{a}}\right)}
so that the integral becomes
∫
d
x
a
2
+
x
2
=
∫
a
sec
2
(
θ
)
d
θ
a
2
+
a
2
tan
2
(
θ
)
=
∫
a
sec
2
(
θ
)
d
θ
a
2
(
1
+
tan
2
(
θ
)
)
=
∫
a
sec
2
(
θ
)
d
θ
a
2
sec
2
(
θ
)
=
∫
d
θ
a
=
θ
a
+
C
=
1
a
arctan
(
x
a
)
+
C
{\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 )}}\\&=\int {\frac {a\sec ^{2}(\theta )\,d\theta }{a^{2}(1+\tan ^{2}(\theta ))}}\\&=\int {\frac {a\sec ^{2}(\theta )\,d\theta }{a^{2}\sec ^{2}(\theta )}}\\&=\int {\frac {d\theta }{a}}\\&={\tfrac {\theta }{a}}+C\\&={\tfrac {1}{a}}\arctan \left({\tfrac {x}{a}}\right)+C\end{aligned}}}
(provided a ≠ 0).
Integrals containing x 2 − a 2
Integrals like
∫
d
x
x
2
−
a
2
{\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}}
should be done by partial fractions rather than trigonometric substitutions. However, the integral
∫
x
2
−
a
2
d
x
{\displaystyle \int {\sqrt {x^{2}-a^{2}}}\,dx}
can be done by substitution:
x
=
a
sec
(
θ
)
,
d
x
=
a
sec
(
θ
)
tan
(
θ
)
d
θ
,
θ
=
arcsec
(
x
a
)
{\displaystyle x=a\sec(\theta ),\quad dx=a\sec(\theta )\tan(\theta )\,d\theta ,\quad \theta =\operatorname {arcsec} \left({\tfrac {x}{a}}\right)}
∫
x
2
−
a
2
d
x
=
∫
a
2
sec
2
(
θ
)
−
a
2
⋅
a
sec
(
θ
)
tan
(
θ
)
d
θ
=
∫
a
2
(
sec
2
(
θ
)
−
1
)
⋅
a
sec
(
θ
)
tan
(
θ
)
d
θ
=
∫
a
2
tan
2
(
θ
)
⋅
a
sec
(
θ
)
tan
(
θ
)
d
θ
=
∫
a
2
sec
(
θ
)
tan
2
(
θ
)
d
θ
=
a
2
∫
sec
(
θ
)
(
sec
2
(
θ
)
−
1
)
d
θ
=
a
2
∫
(
sec
3
(
θ
)
−
sec
(
θ
)
)
d
θ
.
{\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}}}
We can then solve this using the formula for the integral of secant cubed .
Substitutions that eliminate trigonometric functions
Substitution can be used to remove trigonometric functions. In particular, see Tangent half-angle substitution .
For instance,
∫
f
(
sin
(
x
)
,
cos
(
x
)
)
d
x
=
∫
1
±
1
−
u
2
f
(
u
,
±
1
−
u
2
)
d
u
u
=
sin
(
x
)
∫
f
(
sin
(
x
)
,
cos
(
x
)
)
d
x
=
∫
1
∓
1
−
u
2
f
(
±
1
−
u
2
,
u
)
d
u
u
=
cos
(
x
)
∫
f
(
sin
(
x
)
,
cos
(
x
)
)
d
x
=
∫
2
1
+
u
2
f
(
2
u
1
+
u
2
,
1
−
u
2
1
+
u
2
)
d
u
u
=
tan
(
x
2
)
∫
cos
x
(
1
+
cos
x
)
3
d
x
=
∫
2
1
+
u
2
1
−
u
2
1
+
u
2
(
1
+
1
−
u
2
1
+
u
2
)
3
d
u
=
∫
1
−
u
2
1
+
u
2
d
u
{\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)\\\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)\\\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({\tfrac {x}{2}}\right)\\\int {\frac {\cos x}{(1+\cos x)^{3}}}\,dx&=\int {\frac {2}{1+u^{2}}}{\frac {\frac {1-u^{2}}{1+u^{2}}}{\left(1+{\frac {1-u^{2}}{1+u^{2}}}\right)^{3}}}\,du=\int {\frac {1-u^{2}}{1+u^{2}}}\,du\end{aligned}}}
Hyperbolic functions
Substitutions of hyperbolic functions can also be used to simplify integrals.[3]
In the integral
∫
1
a
2
+
x
2
d
x
{\displaystyle \int {\frac {1}{\sqrt {a^{2}+x^{2}}}}\,dx}
, make the substitution
x
=
a
sinh
u
{\displaystyle x=a\sinh {u}}
,
d
x
=
a
cosh
u
d
u
{\displaystyle dx=a\cosh {u}\,du}
.
Then, using the identities
cosh
2
(
x
)
−
sinh
2
(
x
)
=
1
{\displaystyle \cosh ^{2}(x)-\sinh ^{2}(x)=1}
and
sinh
−
1
x
=
ln
(
x
+
x
2
+
1
)
{\displaystyle \sinh ^{-1}{x}=\ln(x+{\sqrt {x^{2}+1}})}
,
∫
1
a
2
+
x
2
d
x
=
∫
a
cosh
u
a
2
+
a
2
sinh
2
u
d
u
=
∫
a
cosh
u
a
1
+
sinh
2
u
d
u
=
∫
a
cosh
u
a
cosh
u
d
u
=
u
+
C
=
sinh
−
1
x
a
+
C
=
ln
(
x
2
a
2
+
1
+
x
a
)
+
C
=
ln
(
x
2
+
a
2
+
x
a
)
+
C
{\displaystyle {\begin{aligned}\int {\frac {1}{\sqrt {a^{2}+x^{2}}}}\,dx&=\int {\frac {a\cosh {u}}{\sqrt {a^{2}+a^{2}\sinh ^{2}{u}}}}\,du\\&=\int {\frac {a\cosh {u}}{a{\sqrt {1+\sinh ^{2}{u}}}}}\,du\\&=\int {\frac {a\cosh {u}}{a\cosh {u}}}\,du\\&=u+C\\&=\sinh ^{-1}{\frac {x}{a}}+C\\&=\ln \left({\sqrt {{\frac {x^{2}}{a^{2}}}+1}}+{\frac {x}{a}}\right)+C\\&=\ln \left({\frac {{\sqrt {x^{2}+a^{2}}}+x}{a}}\right)+C\end{aligned}}}
See also
References