Math  /  Calculus

QuestionCompute the given integral. 149x2dx=+C\int \frac{1}{\sqrt{49-x^{2}}} d x=\square+C

Studdy Solution

STEP 1

1. The integral 149x2dx\int \frac{1}{\sqrt{49-x^{2}}} \, dx resembles the standard form of an inverse trigonometric function.
2. We will use the inverse sine function to solve the integral.

STEP 2

1. Identify the standard form of the integral.
2. Apply the inverse trigonometric identity.
3. Solve the integral and add the constant of integration.

STEP 3

Recognize that the integral 149x2dx\int \frac{1}{\sqrt{49-x^{2}}} \, dx is of the form 1a2x2dx\int \frac{1}{\sqrt{a^2-x^2}} \, dx, where a=7a = 7.

STEP 4

Apply the inverse trigonometric identity for integrals of the form 1a2x2dx=arcsin(xa)+C\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C.

STEP 5

Substitute a=7a = 7 into the identity:
149x2dx=arcsin(x7)+C\int \frac{1}{\sqrt{49-x^{2}}} \, dx = \arcsin\left(\frac{x}{7}\right) + C
The solution to the integral is:
arcsin(x7)+C\boxed{\arcsin\left(\frac{x}{7}\right) + C}

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