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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:
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
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.
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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