Question

Evaluate the integral.\\ $\int \frac{8 dx}{\sqrt{9 - 64x^2}}$ $\frac{1}{3} \sin^{-1} (\frac{8x}{3}) + C$ $\frac{1}{3} \tan^{-1} (\frac{8x}{3}) + C$ $\sin^{-1} (\frac{8x}{3}) + C$ $\tan^{-1} (\frac{8x}{3}) + C$

          Evaluate the integral.\\
$\int \frac{8 dx}{\sqrt{9 - 64x^2}}$
$\frac{1}{3} \sin^{-1} (\frac{8x}{3}) + C$
$\frac{1}{3} \tan^{-1} (\frac{8x}{3}) + C$
$\sin^{-1} (\frac{8x}{3}) + C$
$\tan^{-1} (\frac{8x}{3}) + C$

Evaluate the integral.

∫(8 dx)/(√(9 - 64x^2))
(1)/(3)sin^-1 ((8x)/(3)) + C
(1)/(3)tan^-1 ((8x)/(3)) + C
sin^-1 ((8x)/(3)) + C
tan^-1 ((8x)/(3)) + C

Added by Mike A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert J Hardin

02/09/2018

Step 1

Step 1: Recognize that the integral is of the form \int \frac{8dx}{\sqrt{9-64x^2}}. Show more…

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Evaluate the integral. int (8dx)/(sqrt(9-64x^(2))) (1)/(3)sin^(-1)((8)/(3)x)+C (1)/(3)tan^(-1)((8)/(3)x)+C sin^(-1)((8)/(3)x)+C tan^(-1)((8)/(3)x)+C Evaluate the integral. 8dx 9-642 sin -1 X + C -1 113 tan + C 1 sin +C 1 tan' + C