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।[१][२]
प्रतिस्थापन 1. यदि समाकल्य (integrand) में a2 − x2 हो तो ,
- <math>x = a \sin \theta</math>
रखें और यह सर्वसमिका प्रयोग करें-
- <math>1-\sin^2 \theta = \cos^2 \theta.</math>
प्रतिस्थापन 2. If the integrand contains a2 + x2, let
- <math>x = a \tan \theta</math>
and use the identity
<math>1+\tan^2 \theta = \sec^2 \theta.</math>
प्रतिस्थापन 3. If the integrand contains x2 − a2, let
- <math>x = a \sec \theta</math>
and use the identity
- <math>\sec^2 \theta -1 = \tan^2 \theta.</math>
उदाहरण
Integrals containing a2 − x2
In the integral
- <math>\int\frac{\mathrm dx}{\sqrt{a^2-x^2}}</math>
we may use
- <math>x=a\sin(\theta),\quad \mathrm dx=a\cos(\theta)\,\mathrm d\theta, \quad \theta=\arcsin\left(\frac{x}{a}\right)</math>
- <math>\begin{align}
\int\frac{\mathrm dx}{\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{align}</math>
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
- <math>\int_0^{\frac{a}{2}}\frac{\mathrm dx}{\sqrt{a^2-x^2}}=\int_0^{\frac{\pi}{6}} \mathrm d\theta = \tfrac{\pi}{6}.</math>
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
- <math>\int\frac{\mathrm dx}साँचा:a^2+x^2</math>
we may write
- <math>x=a\tan(\theta),\quad \mathrm dx=a\sec^2(\theta)\,\mathrm d\theta, \quad \theta=\arctan\left(\tfrac{x}{a}\right)</math>
so that the integral becomes
- <math>\begin{align}
\int\frac{\mathrm dx}साँचा:a^2+x^2 &= \int\frac{a\sec^2(\theta)\,\mathrm d\theta}साँचा:a^2+a^2\tan^2(\theta) \\ &= \int\frac{a\sec^2(\theta)\,\mathrm d\theta}साँचा:a^2(1+\tan^2(\theta)) \\ &= \int \frac{a\sec^2(\theta)\,\mathrm d\theta}साँचा: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{align}</math>
(provided a ≠ 0).