What is the integral of $\int {sin}^{3} 3 x cos 3 x d x$ $int sin^3 3x cos 3x dx$ ?

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What is the integral of $\int {sin}^{3} 3 x cos 3 x d x$ $int sin^3 3x cos 3x dx$ ?

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Answer 1 · 2016-01-23T14:36:14

$\int {sin}^{3} (3 x) cos (3 x) d x = \frac{{sin}^{4} (3 x)}{12} + c$ $intsin^3(3x)cos(3x)dx=frac{sin^4(3x)}{12}+c$

Explanation:

Take the given equation $\int {sin}^{3} (3 x) cos 3 x d x$ $intsin^3(3x)cos3xdx$
Take the function $sin (3 x) = t$ $sin(3x)=t$
Now differentiate with respect to $t$ $t$ on both sides of this
$\frac{d}{d t} (sin 3 x) = 1 \Rightarrow 3 cos 3 x \cdot \frac{d x}{d t} = 1 \Rightarrow cos 3 x d x = \frac{d t}{3}$ $\frac{d}{dt}(sin3x)=1\implies3cos3x*\frac{dx}{dt}=1\impliescos3xdx=dt/3$

Substituting the given above values for the main equation, we get
$\int \frac{t^{3}}{3} d t$ $intt^3/3dt$
This looks easy. Let's directly integrate.
$\int \frac{t^{3}}{3} d t = \frac{t^{4}}{3 \cdot 4} + c = \frac{t^{4}}{12} + c$ $intt^3/3dt=t^4/(3*4)+c=t^4/12+c$

That's what you want, right? Oh, substitution? That's there up in the answer section.

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What is the integral of $\int {sin}^{3} 3 x cos 3 x d x$ $int sin^3 3x cos 3x dx$ ?

1 Answer

Explanation:

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