Question

# Find the antiderivative (or integral) of the given function by the method of inspection.$\sin 2x - 4{e^{3x}}$

Hint- Use the method of inspection where directly possible, if not bring it in direct form of inspection.

Given function is $\sin 2x - 4{e^{3x}}$
As we know that derivative of $\cos x$ is$- \sin x$.
Also we know that the derivative of ${e^x}$ is ${e^x}$ itself. And also we are well aware that derivative of a constant comes into its base so keeping all the above points in mind. First term must have $\cos x$ with a base $2$ and the second term must contain ${e^x}$ with a base $3$.
So we have
$\dfrac{d}{{dx}}\left[ {\dfrac{{ - 1}}{2}\cos 2x - \dfrac{4}{3}{e^{3x}}} \right] = \sin 2x - 4{e^{3x}}$
Hence by inspection method we find that anti derivative of $\sin 2x - 4{e^{3x}}$is
$\left( {\dfrac{{ - 1}}{2}\cos 2x - \dfrac{4}{3}{e^{3x}}} \right)$

Note- Integration or anti derivative of a function can be found easily by inspection method only if the function is a common one and readily used. The inspection method is not suitable to find the integral of the function with complex identity. An anti differentiable function $f$ has infinitely many antiderivatives.