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Evaluate using trig substitution.

{eq}\int_{0}^{a} x^2 \sqrt{a^2 - x^2} {/eq} dx

Question:

Evaluate using trig substitution.

{eq}\int_{0}^{a} x^2 \sqrt{a^2 - x^2} {/eq} dx

Definite Integrals:

In the given problem, we will evaluate the given definite integral using suitable trigonometric substitutions. Using substitution method makes the integral much easier for evaluation. As we make substitutions, we will also change the limits of the integral accordingly.

Answer and Explanation: 1


{eq}\begin{align*} \ &=\int_{0}^{a} x^2 \sqrt{a^2 - x^2} \ dx \end{align*} {/eq}

Put,

{eq}\begin{align*} \ & x = a \ \sin t \end{align*} {/eq}

Differentiating both sides, we get

{eq}\begin{align*} \ & dx = a \ \cos t \ dt \end{align*} {/eq}

Also when

{eq}\begin{align*} \ & x = 0 \ , t=0 \end{align*} {/eq} and {eq}\begin{align*} \ & x = a \ , t= \frac{\pi}{2} \end{align*} {/eq}

Substituting the above values, we get

{eq}\begin{align*} \ &= \int_{0}^{\frac{\pi}{2}} a^2 \sin^2 t \sqrt{a^2 - a^2 \ sin^2 t} \ \ a \cos t \ dt \\ \\ \ &= \int_{0}^{\frac{\pi}{2}} a^3 \sin^2 t \ \cos t \sqrt{a^2(1 - \ sin^2 t)} \ dt \\ \\ \ &= \int_{0}^{\frac{\pi}{2}} a^3 \sin^2 t \ \cos t \sqrt{a^2 \ \cos^2 t} \ dt \ \ \ \ \ \ \ \ \ \ \left[ \ Because \ 1- \sin^2 x= \cos^2 x \right] \\ \\ \ &= \int_{0}^{\frac{\pi}{2}} a^4 \sin^2 t \ \cos^2 t \ dt \end{align*} {/eq}

Multilpy numerator and denominator by 4, we get

{eq}\begin{align*} \ &= \frac{a^4}{4}\int_{0}^{\frac{\pi}{2}} (4 \sin^2 t \cos^2 t) \ dt \\ \\ \ &= \frac{a^4}{4}\int_{0}^{\frac{\pi}{2}} (2 \sin t \cos t)^2 \ dt \\ \\ \ &= \frac{a^4}{4}\int_{0}^{\frac{\pi}{2}} \sin^2 2t \ dt \\ \\ \ &= \frac{a^4}{4}\int_{0}^{\frac{\pi}{2}} \frac{1-\cos 4t}{2} \ dt\ \ \ \ \ \ \ \ \ \ \left[ \ Because \sin^2 2x= \frac{1-\cos 4x}{2} \right] \\ \\ \ &= \frac{a^4}{8}\int_{0}^{\frac{\pi}{2}} (1-\cos 4t) \ dt \\ \\ \ &= \frac{a^4}{8} \left[t-\frac{\sin 4t}{4}\right]_{0}^{\frac{\pi}{2}} \\ \\ \ &= \frac{a^4}{8} \left[\frac{\pi}{2}-\frac{\sin 2\pi}{4} - 0+ 0\right] \\ \\ \ &= \frac{a^4}{8} \left[\frac{\pi}{2}-0- 0+ 0\right] \\ \\ \ &= \frac{a^4 \pi}{16} \end{align*} {/eq}


Learn more about this topic:

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How to Use Trigonometric Substitution to Solve Integrals

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Chapter 13 / Lesson 12
15K

Trigonometric substitutions can be useful by plugging in a function of a variable, thus simplifying the calculation of an integral. Learn how to solve integrals using substitution, tables, by parts, and Riemann Sums through a variety of examples.


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