To find the integral of sin 3x we need to eliminate 3x.

Let us equate t = 3x

dt/dx = 3

=> dx = dt/3

Now substitute this in the given expression:

Int [ sin 3x dx] = Int [ (1/3)*sin t dt]

=> (1/3) (-cos t) + C

substitute...

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To find the integral of sin 3x we need to eliminate 3x.

Let us equate t = 3x

dt/dx = 3

=> dx = dt/3

Now substitute this in the given expression:

Int [ sin 3x dx] = Int [ (1/3)*sin t dt]

=> (1/3) (-cos t) + C

substitute back t = 3x

=> (1/3) (-cos 3x) + C

**Therefore integral of sin 3x is (1/3) (-cos 3x) + C.**

### Videos

Let f(x) = sin3x.

We need to find the anti-derivative of f(x).

Then, we need to find the integral of f(x).

==> intg f(x) = intg sin3x dx

Let u = 3x ==> du = 3 dx

==> intg f(x) = intg sinu * du/3

= intg sin(u)/3 du

= (1/3) intg sinu du

= (1/3) * - cos(u) + C

Now we will substitute with u = 3x

**==> intg f(x) = - (1/3) cos(3x) + C **

** = -cos(3x)/ 3 + C**