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।

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).

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