Question

Question 9 Evaluate $\int \sin x \cos^2 x \,dx$. $\circ \frac{\cos^3 x}{3} + C$ $\circ \frac{\cos^3 x}{3}$ $\circ -\frac{\cos^3 x}{3}$ $\circ -\frac{\cos^3 x}{3} + C$

Name: question 9 evaluate int sin x cos2 x dx circ fraccos3 x3 c circ fraccos3 x3 circ fraccos3 x3 circ fraccos3 x3 c 79105
Uploaded: 2021-09-16T00:43:54-08:00
Duration: 8 min 51 s
Channel: Sandip Ranjan
Description: question 9 evaluate int sin x cos2 x dx circ fraccos3 x3 c circ fraccos3 x3 circ fraccos3 x3 circ fraccos3 x3 c 79105

          Question 9
Evaluate $\int \sin x \cos^2 x \,dx$.
$\circ \frac{\cos^3 x}{3} + C$
$\circ \frac{\cos^3 x}{3}$
$\circ -\frac{\cos^3 x}{3}$
$\circ -\frac{\cos^3 x}{3} + C$

$question 9 evaluate int sin x cos2 x dx circ fraccos3 x3 c circ fraccos3 x3 circ fraccos3 x3 circ fraccos3 x3 c 79105$

Added by Joanne C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sandip Ranjan

09/16/2021

Step 1

We can use a substitution method to solve this integral. Let $u = \cos x$. Then, the differential $du$ with respect to $x$ is $du = \frac{d}{dx}(\cos x) \,dx = -\sin x \,dx$. From this, we can express $\sin x \,dx$ as $-\,du$. Show more…

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Question 9 Evaluate $\int \sin x \cos^2 x \,dx$. $\circ \frac{\cos^3 x}{3} + C$ $\circ \frac{\cos^3 x}{3}$ $\circ -\frac{\cos^3 x}{3}$ $\circ -\frac{\cos^3 x}{3} + C$