Question

If $sin\left(ax\right)sin\left(bx\right)=cos\left(ax\right)cos\left(bx\right)(a,b\ne 0$ & $a\ne b)$, then $\int \frac{sin\left(ax\right)+cos\left(bx\right)}{cos\left(ax\right)+sin\left(bx\right)}dx$ is

• A. $\frac{1}{a}ln|secax|+C$
• B. $\frac{1}{b}ln|sinbx|+C$
• C. $\frac{1}{a}ln|cosecbx|+C$
• D. $\frac{1}{b}ln|cosax|+C$

Answer 1

The correct option is D $\frac{1}{b}ln|cosax|+C$

$sinaxsinbx=cosaxcosbx(a,b\ne 0,a\ne b)$
$\Rightarrow cos(a+b)x=0$ $\Rightarrow (ax+bx)=(2n+1)\frac{\pi }{2}$
$\Rightarrow bx=(2n+1)\frac{\pi }{2}-ax$
but we take $bx=(4n+1)\frac{\pi }{2}-ax$ only
because at $bx=(4n+3)\frac{\pi }{2}-ax$
denominator becomes zero.
$\int \frac{sin\left(ax\right)+cos\left(bx\right)}{cos\left(ax\right)+sin\left(bx\right)}dx$
$\int tanaxdx=\frac{1}{a}ln|secax|+C$
$\Rightarrow \frac{1}{a}ln|cosecbx|+C$
$\int \frac{sin\left(ax\right)+cos\left(bx\right)}{cos\left(ax\right)+sin\left(bx\right)}dx$
$\int cotbxdx=\frac{1}{b}ln|sinbx|+C$
$\Rightarrow \frac{1}{b}ln|cosax|+C$