Question

Which value of constant of integration makes the value of integral of $\sin 3x\cos 5x$ equal to zero at $x=0$

• A. $\frac{3}{16}$
• B. $-\frac{3}{16}$
• C. $-\frac{5}{6}$
• D. $\frac{1}{8}$

Answer 1

The correct option is B $-\frac{3}{16}$

$I=\int \sin 3x\cos 5xdx$
$=\frac{1}{2}\int (\sin 8x-\sin 2x)dx$
$I=\frac{-\cos 8x}{16}+\frac{\cos 2x}{4}+C$
Now, at $x=0$,given $I=0$
$0=\frac{-1}{16}+\frac{1}{4}+C$
$\Rightarrow C=-\frac{3}{16}$