Question

Question 14 5 pts $$ \int \sin^3 x dx = $$ $$ -\frac{\cos^3 x}{3} + \cos x + C $$ $$ \frac{\cos^3 x}{3} - \cos x + C $$ $$ \frac{\sin^4 x}{4} + C $$ $$ \frac{\cos^3 x}{3} - x + C $$

Name: question 14 5 pts int sin3 x dx fraccos3 x3 cos x c fraccos3 x3 cos x c fracsin4 x4 c fraccos3 x3 x c 39469
Uploaded: 2025-03-07T23:33:21-08:00
Duration: 2 min 24 s
Channel: Eduard Sanchez
Description: question 14 5 pts int sin3 x dx fraccos3 x3 cos x c fraccos3 x3 cos x c fracsin4 x4 c fraccos3 x3 x c 39469

          Question 14
5 pts
$$ \int \sin^3 x dx = $$
$$ -\frac{\cos^3 x}{3} + \cos x + C $$
$$ \frac{\cos^3 x}{3} - \cos x + C $$
$$ \frac{\sin^4 x}{4} + C $$
$$ \frac{\cos^3 x}{3} - x + C $$

$question 14 5 pts int sin3 x dx fraccos3 x3 cos x c fraccos3 x3 cos x c fracsin4 x4 c fraccos3 x3 x c 39469$

Added by Michael D.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Eduard Sanchez

03/07/2025

Step 1

We can rewrite $$ \sin^3 x $$ as $$ \sin^2 x \cdot \sin x $$. Using the identity $$ \sin^2 x = 1 - \cos^2 x $$, we get: $$ \int \sin^3 x dx = \int (1 - \cos^2 x) \sin x dx $$ Show more…

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Question 14 5 pts $$ \int \sin^3 x dx = $$ $$ -\frac{\cos^3 x}{3} + \cos x + C $$ $$ \frac{\cos^3 x}{3} - \cos x + C $$ $$ \frac{\sin^4 x}{4} + C $$ $$ \frac{\cos^3 x}{3} - x + C $$