Consider the following statements, where C is a constant of integration:
(A) \(\int e^x\sin x\,dx = \displaystyle\frac{e^x}{2}\left(\sin x-\cos x\right)+C\)
(B) \(\int e^x\cos x\,dx = \displaystyle\frac{e^x}{2}\left(\cos x-\sin x\right)+C\)
(C) \(\int e^{-x}\sin x\,dx = \displaystyle -\frac{e^{-x}}{2}\left(\sin x+\cos x\right)+C\)
(D) \(\int e^x\left(\sin x+\cos x\right)dx = e^x\sin x+C\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C) and (D) only
Let \(I = \int e^x\sin x\,dx\) and \(J = \int e^x\cos x\,dx\). Integration by parts with the trigonometric function as the first function gives
\( I = e^x\sin x-J \)
\( J = e^x\cos x+I \)
Solving the two equations:
\( \displaystyle I = \frac{e^x}{2}\left(\sin x-\cos x\right),\quad J = \frac{e^x}{2}\left(\sin x+\cos x\right) \)
In the same way, with \(K = \int e^{-x}\sin x\,dx\), by parts twice gives
\( K = -e^{-x}\sin x-e^{-x}\cos x-K \)
\( \displaystyle K = -\frac{e^{-x}}{2}\left(\sin x+\cos x\right) \)
Also \(e^x\left(\sin x+\cos x\right)\) is of the form \(e^x\left[f(x)+f'(x)\right]\) with \(f(x) = \sin x\).
So \(\int e^x\cos x\,dx = \frac{e^x}{2}\left(\sin x+\cos x\right)+C\), and the other three results hold.
Checking the statements
(A) \(\int e^x\sin x\,dx = \frac{e^x}{2}\left(\sin x-\cos x\right)+C\) — Correct. Its derivative is \(\frac{e^x}{2}\left(\sin x-\cos x+\cos x+\sin x\right) = e^x\sin x\).
(B) \(\int e^x\cos x\,dx = \frac{e^x}{2}\left(\cos x-\sin x\right)+C\) — Incorrect. The bracket is \(\sin x+\cos x\); the given result has derivative \(-e^x\sin x\).
(C) \(\int e^{-x}\sin x\,dx = -\frac{e^{-x}}{2}\left(\sin x+\cos x\right)+C\) — Correct. Its derivative is \(\frac{e^{-x}}{2}\left(\sin x+\cos x-\cos x+\sin x\right) = e^{-x}\sin x\).
(D) \(\int e^x\left(\sin x+\cos x\right)dx = e^x\sin x+C\) — Correct. The derivative of \(e^x\sin x\) is \(e^x\sin x+e^x\cos x\).