Question

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

Answer 1

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

Given function is $\sin 2x-4e^{3x}$
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
${\displaystyle \frac{d}{dx}}\left[{\displaystyle \frac{-1}{2}}\cos 2x-{\displaystyle \frac{4}{3}}{e}^{3x}\right]=\sin 2x-4e^{3x}$
Hence by inspection method we find that anti derivative of $\sin 2x-4e^{3x}$is
$\left({\displaystyle \frac{-1}{2}}\cos 2x-{\displaystyle \frac{4}{3}}{e}^{3x}\right)$

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.