Question

integrate dx/cosx-sinx​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\int\rm \frac{1}{cosx - sinx} \: dx$

can be rewritten as

$\rm \:  =  \: \dfrac{1}{ \sqrt{2} } \displaystyle\int\rm \frac{1}{\dfrac{1}{ \sqrt{2} }cosx - \dfrac{1}{ \sqrt{2} }sinx} \: dx$

We know

$\boxed{ \tt{ \: cos\dfrac{\pi}{4} = sin\dfrac{\pi}{4} = \dfrac{1}{ \sqrt{2} } \: }}$

So, using this

$\rm \:  =  \: \dfrac{1}{ \sqrt{2} }\displaystyle\int\rm \frac{1}{cos\dfrac{\pi}{4}cosx - sin\dfrac{\pi}{4}sinx} \: dx$

We know

$\boxed{ \tt{ \: cosx \: cosy \: - \: sinx \: siny \: = \: cos(x + y) \: }}$

So, using this, we get

$\rm \:  =  \: \dfrac{1}{ \sqrt{2} }\displaystyle\int\rm \frac{1}{cos\bigg(x + \dfrac{\pi}{4} \bigg) } \: dx$

$\rm \:  =  \: \dfrac{1}{ \sqrt{2} }\displaystyle\int\rm sec\bigg(x + \dfrac{\pi}{4}\bigg) \: dx$

We know,

$\boxed{ \tt{ \: \displaystyle\int\rm secx \: dx \: = \: log |secx \: + \: tanx| \: + \: c \: }}$

So, using this, we get

$\rm \:  =  \: \dfrac{1}{ \sqrt{2} }log\bigg |sec\bigg(x + \dfrac{\pi}{4}\bigg) + tan \bigg(x + \dfrac{\pi}{4} \bigg) \bigg| + c$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to know :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$