How do you integrate $sin (3 x) d x$ $sin(3x) dx$ ?

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How do you integrate $sin (3 x) d x$ $sin(3x) dx$ ?

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Answer 1 · 2016-10-18T15:50:12

The answer is $- \frac{cos 3 x}{3} + C$ $-(cos3x)/3+C$

Explanation:

The integral of $sin θ$ $sintheta$ is $- cos θ$ $-costheta$
You can use the substitution, $u = 3 x$ $u=3x$ the $d u = 3 d x$ $du=3dx$
$\int sin 3 x d x = \int \frac{sin u d u}{3} = - \frac{cos u}{3} = - \frac{cos 3 x}{3} + C$ $intsin3xdx=int(sinudu)/3=-cosu/3=-(cos3x)/3+C$

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How do you integrate $sin (3 x) d x$ $sin(3x) dx$ ?

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