# Find an anti derivative(or integral)of the function by the method of inspection $\sin2x-4e^{3x}$

Toolbox:
• $\frac{d}{dx}[F(x)+c]=f(x),$
• $\int f(x)\;dx=F(x)+c$
We know that $\frac{d}{dx}(\cos 2x)=-2\sin 2x.$

$\Rightarrow \frac{-1}{2}\frac{d}{dx}(\cos2x)=\sin2x.$

$\sin2x=\frac{d}{dx}\big(\frac{-1}{2}\cos2x\big)$-----(1)

We know that the $\frac{d}{dx}(e^{3x})=3e^{3x}.$

$\Rightarrow e^{3x}=\frac{1}{3}\frac{d}{dx}(e^{3x}).$

$4e^{3x}=\frac{4}{3}\frac{d}{dx}(e^{3x})$--------(2)

Hence the anti derivative of $\sin2x-4e^{3x}$ can be obtained by combining equ(1) and equ(2).

The antiderivative is $\frac{-1}{2}\cos2x-\frac{4}{3}e^{3x}.$
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