Question

If P$=\int e^{ax}cosbxdx$ and Q $=\int e^{ax}sinbxdx$, then $ta{n}^{-1}\left(\frac{Q}{P}\right)+ta{n}^{-1}\left(\frac{b}{a}\right)=$

• A. $ax$
• B. $bx$
• C. $\frac{x}{a}$
• D. $\frac{x}{b}$

Answer 1

The correct option is B $bx$

$ta{n}^{-1}\left(\frac{Q}{P}\right)+ta{n}^{-1}\left(\frac{b}{a}\right)$
$=ta{n}^{-1}\left(\frac{a\sin bx-b\cos bx}{a\cos bx+b\sin bx}\right)+ta{n}^{-1}\frac{b}{a}$
$=ta{n}^{-1}\left(\frac{\frac{a\sin bx-b\cos bx}{a\cos bx+b\sin bx}+\frac{b}{a}}{1-\frac{a\sin bx-b\cos bx}{a\cos bx+b\sin bx}\frac{b}{a}}\right)$
$=ta{n}^{-1}\left(\frac{(a^2+b^2)\sin bx}{(a^2+b^2)\cos bx}\right)$
$=ta{n}^{-1}\left(\tan bx\right)=bx$