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Trigonometric Substitution for Integrals with Square Roots

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For the integral a2x2\displaystyle \sqrt{a^2-x^2}, make the trigonometric substitution x=asinθx=a\sin\theta and find the differential dx=acosθdθdx=a\cos\theta\,d\theta.

Trigonometric substitution is a powerful technique often used to simplify integrals, especially when dealing with expressions involving square roots. In this problem, the substitution of x with a sine function of theta is a common method to transform the integral. This approach is particularly helpful when the integral has the form of a square root of a squared term minus another squared term, since it allows the square root to be handled by trigonometric identities. In this context, the trigonometric identity involving sine and cosine simplifies the expression inside the square root, turning it into a manageable form. By substituting x = a sin theta, the differential dx is transformed into a cos theta d theta, indicating how these substitutions work together to simplify the integration process. The idea is to swap a difficult algebraic problem for an easier trigonometric problem.

This technique can be generalized depending on the form of the integral. For instance, using tangent or secant substitutions may be appropriate in other cases, following similar methods. Recognizing the patterns of which trigonometric identity and substitution to use can significantly reduce the complexity of solving integrals of this nature. The mastery of this method is crucial for anyone delving deeper into calculus, especially in topics involving integration of complex algebraic forms.

Posted by grwgreg 21 days ago

Related Problems

Simplify and integrate the expression (x2+9)3/2(x^2 + 9)^{3/2} using trigonometric substitution where x=3tan(θ)x = 3\tan(\theta).

Evaluate the integral dtt2+9\displaystyle \int \frac{dt}{t^2 + 9} using trigonometric substitution.

Evaluate the integral of the form dxx2a2 \displaystyle \int \frac{dx}{\sqrt{x^2 - a^2}} by making the substitution x=asecθx = a \sec\theta.

Convert x=secθx = \sec\theta back in terms of xx using a right triangle and basic SOHCAHTOA.