गणित के सन्दर्भ में, त्रिकोणमितीय प्रतिस्थापन (Trigonometric substitution) का अर्थ है, गैर-त्रिकोणमितीय फलनों के स्थान पर त्रिकोणमितीय फलनों को स्थापित करना। इनके उपयोग से कुछ समाकल सरल हो जाते हैं।[1] [2]
प्रतिस्थापन 1. यदि समाकल्य (integrand) में a 2 − x 2 हो तो ,
x
=
a
sin
θ
{\displaystyle x=a\sin \theta }
रखें और यह सर्वसमिका प्रयोग करें-
1
−
sin
2
θ
=
cos
2
θ
.
{\displaystyle 1-\sin ^{2}\theta =\cos ^{2}\theta .}
प्रतिस्थापन 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 .}
प्रतिस्थापन 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 .}
Integrals containing a 2 − x 2
संपादित करें
In the integral
∫
d
x
a
2
−
x
2
{\displaystyle \int {\frac {\mathrm {d} x}{\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 \mathrm {d} x=a\cos(\theta )\,\mathrm {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 {\mathrm {d} x}{\sqrt {a^{2}-x^{2}}}}&=\int {\frac {a\cos(\theta )\,\mathrm {d} \theta }{\sqrt {a^{2}-a^{2}\sin ^{2}(\theta )}}}\\&=\int {\frac {a\cos(\theta )\,\mathrm {d} \theta }{\sqrt {a^{2}(1-\sin ^{2}(\theta ))}}}\\&=\int {\frac {a\cos(\theta )\,\mathrm {d} \theta }{\sqrt {a^{2}\cos ^{2}(\theta )}}}\\&=\int \mathrm {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 {\mathrm {d} x}{\sqrt {a^{2}-x^{2}}}}=\int _{0}^{\frac {\pi }{6}}\mathrm {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
संपादित करें
In the integral
∫
d
x
a
2
+
x
2
{\displaystyle \int {\frac {\mathrm {d} x}{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 \mathrm {d} x=a\sec ^{2}(\theta )\,\mathrm {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 {\mathrm {d} x}{a^{2}+x^{2}}}&=\int {\frac {a\sec ^{2}(\theta )\,\mathrm {d} \theta }{a^{2}+a^{2}\tan ^{2}(\theta )}}\\&=\int {\frac {a\sec ^{2}(\theta )\,\mathrm {d} \theta }{a^{2}(1+\tan ^{2}(\theta ))}}\\&=\int {\frac {a\sec ^{2}(\theta )\,\mathrm {d} \theta }{a^{2}\sec ^{2}(\theta )}}\\&=\int {\frac {\mathrm {d} \theta }{a}}\\&={\tfrac {\theta }{a}}+C\\&={\tfrac {1}{a}}\arctan \left({\tfrac {x}{a}}\right)+C\end{aligned}}}
(provided a ≠ 0).
