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 ,

$x=a\sin \theta$

रखें और यह सर्वसमिका प्रयोग करें-

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

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

$x=a\tan \theta$

and use the identity

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

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

$x=a\sec \theta$

and use the identity

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

उदाहरण

Integrals containing a² − x²

In the integral

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

we may use

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

$\begin{array}{cc}\int \frac{\mathrm{d}x}{\sqrt{a^2-x^2}}& =\int \frac{a\cos (\theta )\mathrm{d}\theta }{\sqrt{a^2-a^2{\sin }^2(\theta )}}\\ & =\int \frac{a\cos (\theta )\mathrm{d}\theta }{\sqrt{a^2(1-{\sin }^2(\theta ))}}\\ & =\int \frac{a\cos (\theta )\mathrm{d}\theta }{\sqrt{a^2{\cos }^2(\theta )}}\\ & =\int \mathrm{d}\theta \\ & =\theta +C\\ & =\arcsin \left(\frac{x}{a}\right)+C\end{array}$

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

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

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 give us the negative of the result.

Integrals containing a² + x²

In the integral

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

we may write

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

so that the integral becomes

$\begin{array}{cc}\int \frac{\mathrm{d}x}{a^2+x^2}& =\int \frac{a{\sec }^2(\theta )\mathrm{d}\theta }{a^2+a^2{\tan }^2(\theta )}\\ & =\int \frac{a{\sec }^2(\theta )\mathrm{d}\theta }{a^2(1+{\tan }^2(\theta ))}\\ & =\int \frac{a{\sec }^2(\theta )\mathrm{d}\theta }{a^2{\sec }^2(\theta )}\\ & =\int \frac{\mathrm{d}\theta }{a}\\ & =\frac{\theta }{a}+C\\ & =\frac{1}{a}\arctan \left(\frac{x}{a}\right)+C\end{array}$

(provided a ≠ 0).

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