What is the antiderivative of sin3x?

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justaguide's profile pic

Posted on

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.

Here is a similar problem:

hala718's profile pic

Posted on

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

neela's profile pic

Posted on

To find the antiderivative of sin3x.

f(x) = sin3x.

Let 3x = t.

Differentiating 3x = t, we get 3dx = dt.

Therefore dx = dt/3.

Therefore Int f(x)dx = Int sint dt/3.

Int f(x) dx = (-cost)/3 +C....(1).

Now replace t = 3x in (1) and get:

Int f(x) dx = (-cos3x)/3 + C.

giorgiana1976's profile pic

Posted on

To find the antiderivative of sin 3x, we'll have to determine the indefinite integral of sin 3x.

Int sin 3x dx

We'll substitute 3x by t.

3x = t

We'll differentiate both sides:

3dx = dt

We'll divide by 3:

dx = dt/3

We'll re-write the integral of the function , changing the variable x to t:

Int sin t*(dt/3) = (1/3)*Int sin t dt

(1/3)*Int sin t dt = - cos t/3 +C

We'll substitute t by 3x:

Int sin 3x dx =  - (cos 3x)/3 +C

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