Question

If $\int e^{3x}cos4xdx=e^{3x}(Asin4x+Bcos4x)+c$, then

• A. 4A = 3B
• B. 2A = 3
• C. 3A = 4B
• D. 4A + 3B = 1

Answer 1

Answer: (C), (D)

The correct options are
C 3A = 4B
D 4A + 3B = 1

Using $\int e^{ax}\cos bxdx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+\sin bx)$
Then
$\int e^{3x}\cos 4xdx=\frac{e^{3x}}{25}\left(3\cos 4x+4\sin 4x\right)$
We get $A=\frac{4}{25}$ and $B=\frac{3}{25}$
$\therefore 3A=4B,4A+3B=1$