Question

integration of 1/cos(x-a)sin(x-b)

Answer 1

Final Answer :
$\sec( a - b) ln( \frac{ \sin(x - b) }{ \cos(x - b) } )$

Steps:
1) Multiply numerator and denominator by cos(a-b) .
2) Write cos(a-b) = cos((x-b) -(x-a)) in numerator.
3)Now, expand the terms and see magic.
We will get terns which are easily integrated.

Ex. Integration of cot(x-b) wrt. to x is
$- ln( \cos(x - b) )$

Answer 2

Answer:

= $\frac{1}{cos (b-a)}$  $log | \frac{sin (x-b)}{cos(x-a)} |$ $+C$

Given

∫1/cos(x-a)sin(x-b)

To Find

Integration of $\int\ { 1/cos(x-a)sin(x-b)} \, dx$

Solution

$=\int\ \frac{1}{cos (x-a) sin (x-b)} \, dx$

$=\frac{1}{cos(b-a)}$ $\int\ \frac{1}{cos (x-a) sin (x-b)} \, dx$

$=\frac{1}{cos (b-a)}$ $\int\ \frac{cos(b-a)}{cos (x-a) sin (x-b)} \, dx$

$=\frac{1}{cos (b-a)}$ $\int\ {\frac{cos (x-b) cos (x-a) + sin(x-b) sin(x-a)}{cos(x-a) sin(x-b)} } \, dx$

$=\frac{1}{cos (b-a)}$$\int\ {\frac{cos(x-b)}{sin(x-b)} + \frac{sin(x-a)}{cos (x-a)} } \, dx$

$=\frac{1}{cos (b-a)}$$\int\{cot(x-b)} \,$$+ \int\ {tan(x-a)} \, dx$

$=\frac{1}{cos (b-a)}$  $[log| sin (x-b)| -log |cos(x-a)|]$

$=\frac{1}{cos (b-a)}$  $log |\frac{sin (x-b)}{cos(x-a)} | + C$

∴Final answer is   $\frac{1}{cos (b-a)}$ $log |\frac{sin (x-b)}{cos(x-a)} | + C$