Evaluate the integral using the trigonometric substitution. \int\frac{x}{\sqrt{25-x^{2}}}dx

Question:

Evaluate the integral using the trigonometric substitution.

{eq}\int\frac{x}{\sqrt{25-x^{2}}}dx {/eq}

Trigonometric Substitution

The substitution of the trigonometric functions for any other expression is known as the trigonometric substitution. By trigonometric substitution, this question becomes easy to solve.

Answer and Explanation:

Given that,

{eq}I=\int\frac{x}{\sqrt{25-x^{2}}}dx\\ Let,\\ \Rightarrow \ x=5\sin(t)\\ \Rightarrow \ t=\sin^{-1}(\frac{x}{5})\\ \text{Differentiate with respect to t}\\ \Rightarrow \ \frac{dx}{dt}=5\cos(t)\\ \Rightarrow \ dx=5\cos(t)dt\\ Now,\\ \Rightarrow \ I=\int\frac{25\sin(t)\cos(t)}{5\cos(t)}dt\\ \Rightarrow \ I=5\int\sin(t)dt\\ \Rightarrow \ I=-5\cos(t)+C\\ \text{Undo substitution}\\ \Rightarrow \ I=-5\cos(\sin^{-1}(\frac{x}{5}))+C\\ \text{As we know that }\\ \sin(\theta)=''opposit/hypotenuse''=\frac{x}{5} \ and \ \cos(\theta)=''adjacent/hypotenuse''=\frac{\sqrt{25-x^{2}}}{5}\\ Then,\\ \Rightarrow \ I=-\sqrt{25-x^{2}}+C\\ {/eq}


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Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4
512

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