Question

∫5sec (π/4−x) tan (π/4−x) dx....does anyone know how to solve this....?

Answer 1

\[put \frac{ \pi }{ 4 }-x=t,-dx=dt\]
\[\int\limits \sec x \tan x dx=\sec x+c\]

Answer 2

can u explain more?

Answer 3

\[I=5\int\limits \sec \left( \frac{ \pi }{4 }-x \right)\tan \left( \frac{ \pi }{4 }-x \right)dx\]
\[put \frac{ \pi }{ 4 }-x=t,-dx=dt,dx=-dt\]
\[I=-5\int\limits \sec t \tan t dt=-5\sec t+c\]
\[=-5\sec \left( \frac{ \pi }{ 4}-x \right)+c\]