# त्रिकोणमितीय प्रतिस्थापन

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 a2 − x2 हो तो ,

${\displaystyle x=a\sin \theta }$

रखें और यह सर्वसमिका प्रयोग करें-

${\displaystyle 1-\sin ^{2}\theta =\cos ^{2}\theta .}$

प्रतिस्थापन 2. If the integrand contains a2 + x2, let

${\displaystyle x=a\tan \theta }$

and use the identity

${\displaystyle 1+\tan ^{2}\theta =\sec ^{2}\theta .}$

प्रतिस्थापन 3. If the integrand contains x2 − a2, let

${\displaystyle x=a\sec \theta }$

and use the identity

${\displaystyle \sec ^{2}\theta -1=\tan ^{2}\theta .}$

## उदाहरण

### Integrals containing a2 − x2

In the integral

${\displaystyle \int {\frac {\mathrm {d} x}{\sqrt {a^{2}-x^{2}}}}}$

we may use

${\displaystyle x=a\sin(\theta ),\quad \mathrm {d} x=a\cos(\theta )\,\mathrm {d} \theta ,\quad \theta =\arcsin \left({\frac {x}{a}}\right)}$
{\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 a2; 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

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

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 a2 + x2

In the integral

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

we may write

${\displaystyle x=a\tan(\theta ),\quad \mathrm {d} x=a\sec ^{2}(\theta )\,\mathrm {d} \theta ,\quad \theta =\arctan \left({\tfrac {x}{a}}\right)}$

so that the integral becomes

{\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).

## सन्दर्भ

1. Stewart, James (2008). Calculus: Early Transcendentals (6th संस्करण). Brooks/Cole. आई॰ऍस॰बी॰ऍन॰ 0-495-01166-5.
2. Thomas, George B.; Weir, Maurice D.; Hass, Joel (2010). Thomas' Calculus: Early Transcendentals (12th संस्करण). Addison-Wesley. आई॰ऍस॰बी॰ऍन॰ 0-321-58876-2.