Question

integrate dx/cosx-sinx

Answer 1

Hi Mate!!

cos ( x ) - sin ( x ) = √2 { Cos ( 45 + x ) }

1 / √2 { log ( sec ( 45 + x ) + Tan ( 45 + x) }

Answer 2

Answer:

$I=\dfrac{1}{\sqrt{2}}\ln|\tan(\frac{x}{2}-\frac{\pi}{8})|+C$

Step-by-step explanation:

Given: $I=\int \dfrac{dx}{\cos x-\sin x}$

Multiply and divide by $\sqrt{2}$ at denominator

$I=\dfrac{1}{\sqrt{2}}\int \dfrac{dx}{\dfrac{1}{\sqrt{2}}\cos x-\dfrac{1}{\sqrt{2}}\sin x}$

$I=\dfrac{1}{\sqrt{2}}\int \dfrac{dx}{\cos\frac{\pi}{4}\cos x-\sin\frac{\pi}{4}\sin x}$

$\because \cos A\cos B-\sin A\sin B=\cos(A+B)$

$I=\dfrac{1}{\sqrt{2}}\int \dfrac{dx}{\cos(x+\frac{\pi}{4})}$

$I=\dfrac{1}{\sqrt{2}}\int \sec(x+\frac{\pi}{4})dx$

$I=\dfrac{1}{\sqrt{2}}\ln|\tan(x+\frac{\pi}{4})+\sec(x+\frac{\pi}{4})|+C$

$\because \tan x=\dfrac{\sin x}{\cos x},\sec x=\dfrac{1}{\cos x}$

$I=\dfrac{1}{\sqrt{2}}\ln|\tan(\frac{x}{2}-\frac{\pi}{8})|+C$