Line 33:
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we may use
we may use
:<math>x=a\sin(\theta),\ dx=a\cos(\theta)\,d\theta</math>
:<math>x=a\sin(\theta)\qquad dx=a\cos(\theta)\,d\theta \qquad \theta=\arcsin\left(\frac{x}{a}\right) </math>
:<math>\theta=\arcsin\left(\frac{x}{a}\right)</math>
so that the integral becomes
so that the integral becomes
Trigonometric Substitutions Trigonometric Substitutions Trigonometric Substitutions 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:
If the integrand contains
a
2
−
x
2
{\displaystyle 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 .\,}
If the integrand contains
a
2
+
x
2
{\displaystyle 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 .\,}
If the integrand contains
x
2
−
a
2
{\displaystyle 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
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 )\qquad dx=a\cos(\theta )\,d\theta \qquad \theta =\arcsin \left({\frac {x}{a}}\right)}
so that the integral becomes
∫
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
)
)
{\displaystyle \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 ))}}}}
=
∫
a
cos
(
θ
)
d
θ
a
2
cos
2
(
θ
)
=
∫
d
θ
=
θ
+
C
=
arcsin
(
x
a
)
+
C
{\displaystyle {}=\int {\frac {a\cos(\theta )\,d\theta }{\sqrt {a^{2}\cos ^{2}(\theta )}}}=\int d\theta =\theta +C=\arcsin \left({\frac {x}{a}}\right)+C}
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}^{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. 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 result in the negative of the result.
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
{\displaystyle x=a\tan(\theta ),\ dx=a\sec ^{2}(\theta )\,d\theta }
θ
=
arctan
(
x
a
)
{\displaystyle \theta =\arctan \left({\frac {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}&{}\quad \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}}={\frac {\theta }{a}}+C={\frac {1}{a}}\arctan \left({\frac {x}{a}}\right)+C\end{aligned}}}
(provided a > 0).
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
{\displaystyle x=a\sec(\theta ),\ dx=a\sec(\theta )\tan(\theta )\,d\theta }
θ
=
arcsec
(
x
a
)
{\displaystyle \theta =\operatorname {arcsec} \left({\frac {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}&{}\quad \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. For instance,
∫
f
(
sin
x
,
cos
x
)
d
x
=
∫
1
±
1
−
u
2
f
(
u
,
±
1
−
u
2
)
d
u
,
u
=
sin
x
{\displaystyle \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,\qquad \qquad u=\sin x}
∫
f
(
sin
x
,
cos
x
)
d
x
=
∫
−
1
±
1
−
u
2
f
(
±
1
−
u
2
,
u
)
d
u
u
=
cos
x
{\displaystyle \int f(\sin x,\cos x)\,dx=\int {\frac {-1}{\pm {\sqrt {1-u^{2}}}}}f\left(\pm {\sqrt {1-u^{2}}},u\right)\,du\qquad \qquad u=\cos x}
(but be careful with the signs)
∫
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
{\displaystyle \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\qquad \qquad u=\tan {\frac {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
=
{\displaystyle \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=}
1
4
∫
(
1
−
u
4
)
d
u
=
1
4
(
u
−
1
5
u
5
)
+
C
=
(
1
+
3
cos
x
+
cos
2
x
)
sin
x
5
(
1
+
cos
x
)
3
+
C
{\displaystyle {\frac {1}{4}}\int (1-u^{4})\,du={\frac {1}{4}}\left(u-{\frac {1}{5}}u^{5}\right)+C={\frac {(1+3\cos x+\cos ^{2}x)\sin x}{5(1+\cos x)^{3}}}+C}
See also
