Question

If $\int e^{3x}\sin 7xdx=\frac{e^{3x}}{a}[b\sin cx+d\cos cx]+C,$ then the value of $a+b+c+d$ is
$($where $C$ is constant of integration$)$

• A. 61.0
• B. 61.00
• C. 61

Answer 1

Answer: (A), (B), (C)

We know
$\int e^{ax}\sin bx=\frac{e^{ax}}{a^2+b^2}[a\sin bx-b\cos bx]+C\Rightarrow \int e^{3x}\sin 7xdx=\frac{e^{3x}}{3^2+7^2}[3\sin 7x-7\cos 7x]+C\therefore a=49+9=58,b=3,c=7,d=-7$
Hence,
$a+b+c+d=61$