Question

I: $\int \frac{\cos (x-a)}{\sin x}dx=\cos a\log |\sin x|+x\sin a+c$
II:$\int \frac{\sin (x-a)}{\cos (x-b)}dx=\cos (b-a)\log |\sec (x-b)|+x\sin (b-a)+c$

• A. only I is true
• B. only II is true
• C. both I and II are true
• D. neither I nor II are true

Answer 1

Answer: (C)

both I and II are true

I: $\int \frac{\cos (x-a)}{\sin x}dx$
$=\int \frac{\cos x\cos a+\sin x\sin a}{\sin x}dx$
$=\cos a\int \cot xdx+\sin a\int dx$
$=\cos a\log |\sin x|+x\sin a+c$
II: $\int \frac{\sin (x-a)}{\cos (x-b)}dx$
$=\int \frac{\sin (x-b+b-a)}{\cos (x-b)}dx$
$=\int \frac{\sin (x-b)\cos (b-a)+\cos (x-b)\sin (b-a)}{\cos (x-b)}dx$
$=\cos (b-a)\int \tan (x-b)dx+\sin (b-a)\int dx$
$=\cos (b-a)\log |\sec (x-b)|+x\sin (b-a)+c$