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 *a*2 − *x*2 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 *a*2 + *x*2, let
> 
> 
> {\displaystyle x=a\tan \theta }
> and use the identity
> 
> 
> {\displaystyle 1+\tan ^{2}\theta =\sec ^{2}\theta .}
> 
> 
> 

> **प्रतिस्थापन 3.** If the integrand contains *x*2 − *a*2, let
> 
> 
> {\displaystyle x=a\sec \theta }
> and use the identity
> 
> 
> {\displaystyle \sec ^{2}\theta -1=\tan ^{2}\theta .}

उदाहरण
------

### Integrals containing *a*2 − *x*2

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 *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

{\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

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.

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