Question

If $\int \frac{sin(x-a)}{cos(x-b)}dx=Alog|secx(x-b)|+Bx+C$ then $B=$

• A. $cos(b-a)$
• B. $sin(b-a)$
• C. $cos(b+a)$
• D. $sin(b+a)$

Answer 1

Answer: (B)

$sin(b-a)$

$\frac{\sin (x-a)}{\cos (x-b)}=\frac{\sin (x-b+(b-a\left)\right)}{\cos (x-b)}=\tan (x-b)\cos (b-a)+\sin (b-a)$

$\int \frac{\sin (x-a)}{\cos (x-b)}dx=\cos (b-a)\int \tan (x-b)dx+\sin (b-a)\int dx=\cos (b-a)\log |\text{sec}(x-b)|+\sin (b-a)x+C$

$B=\sin (b-a)$