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Given \int \sqrt{a^2 - u^2}du= \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2}\sin^{-1}\frac{u}{a} +...

Question:

Given {eq}\int \sqrt{a^2 - u^2}du= \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2}\sin^{-1}\frac{u}{a} + C. {/eq}

Using this form and an appropriate substitution, we obtain that {eq}\int \sqrt{1 - 4x^2}dx= {/eq} _____

Integrals:

In the given problem on indefinite integrals, we will make use of the given standard formula to evaluate the given integral. Here, we will compare the given formula to the given integral and perform few substitutions, in order to obtain the answer.

Answer and Explanation:


Given {eq}\int \sqrt{1 - 4x^2} \ dx ...........(1) {/eq}

and {eq}\int \sqrt{a^2 - u^2} \ du= \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2}\sin^{-1}\frac{u}{a} + C. .............(2) {/eq}

Comparing {eq}(1) {/eq} with left hand side of {eq}(2) {/eq} , we get:

{eq}\begin{align*} \ & a= 1 \ , \ \ u = 2x \end{align*} {/eq}

Now substituting the values of {eq}a {/eq} and {eq}u {/eq} in right hand side of {eq}(2) {/eq}, we get:

{eq}\begin{align*} \ & = \frac{2x}{2}\sqrt{(1)^2 - (2x)^2} + \frac{(1)^2}{2}\sin^{-1}\frac{2x}{1} + C \\ \\ \ & = x \sqrt{1 - 4x^2} + \frac{1}{2}\sin^{-1} (2x) + C \end{align*} {/eq}


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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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