# What is the antiderivative of sin3x?

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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.

Here is a similar problem:

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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

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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.

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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