Question

integration dx/(sinx +cosx)​

Answer 1

$\rm \implies \displaystyle \int \rm \frac{dx}{sin \: x + cos \: x}$

$\rm \implies \displaystyle \int \rm \frac{dx}{ \dfrac{2 \: tan \: \frac{x}{2} }{1 + {tan}^{2}\frac{x}{2} } + \dfrac{1 - {tan}^{2} \frac{x}{2}}{1 + {tan}^{2}\frac{x}{2}} }$

$\rm \implies \displaystyle \int \rm \frac{dx}{ \dfrac{2 \: tan \: \frac{x}{2} + 1 - {tan}^{2} \frac{x}{2}}{1 + {tan}^{2}\frac{x}{2} } }$

$\rm \implies \displaystyle \int \rm \dfrac{1 + {tan}^{2}\frac{x}{2} \: }{ {2 \: tan \: \frac{x}{2} + 1 - {tan}^{2} \frac{x}{2}} } \: dx$

$\rm \implies \displaystyle \int \rm \dfrac{ {sec}^{2}\frac{x}{2} \: }{ {2 \: tan \: \frac{x}{2} + 1 - {tan}^{2} \frac{x}{2}} } \: dx$

$\rm \implies \displaystyle - \int \rm \dfrac{ {sec}^{2}\frac{x}{2} \: }{ { {tan}^{2} \frac{x}{2} - 2 \: tan \: \frac{x}{2} - 1 } } \: dx$

$\rm \: \: let \: \: {tan} \frac{x}{2} = m \\ \\ \rm \: \dfrac{1}{2} \times {sec}^{2} \frac{x}{2} \: dx = dm \\ \\ \rm \: {sec}^{2} \frac{x}{2} \: dx =2 \: d m$

$\rm \implies \displaystyle - \int \rm \dfrac{ 2 \: dm}{ { {m}^{2} - 2 m - 1 } }$

$\rm \implies \displaystyle - \int \rm \dfrac{ 2 \: dm}{ { {m}^{2} - 2 m + 1 - 1 - 1 } }$

$\rm \implies \displaystyle - \int \rm \dfrac{ 2 \: dm}{ { {(m - 1)}^{2} - 2} } \\ \\ \rm \implies \displaystyle - 2\int \rm \dfrac{ \: dm}{ { {(m - 1)}^{2} - { (\sqrt{2} )}^{2} } }$

$\rm \implies \rm - 2 \times \dfrac{1}{2 \sqrt{2} } \times log \bigg| \dfrac{m - 1 - \sqrt{2} }{m - 1 + \sqrt{2}} \bigg| + c$

Now , put m = tan (x/2)

$\rm \implies \rm \dfrac{ - 1}{ \sqrt{2} } \times log \bigg| \dfrac{tan \frac{x}{2} - 1 - \sqrt{2} }{tan \frac{x}{2} - 1 + \sqrt{2}} \bigg| + c$

$\rm \implies \rm \dfrac{ - 1}{ \sqrt{2} } \:\:log \bigg| \dfrac{tan \frac{x}{2} - 1 - \sqrt{2} }{tan \frac{x}{2} - 1 + \sqrt{2}} \bigg| + c$

Answer 2

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \:\:\boxed{\pink{\bf\:sinx \: cosy + siny \: cosx = sin(x + y)}}$

$(2).\: \boxed{ \purple{\bf \: \int cosecx \: dx \: = log(tan\dfrac{x}{2}) + c }}$

$\large\underline\purple{\bold{Solution :- }}$

$\bf \longmapsto \: \bf \:\int \dfrac{dx}{sinx + cosx}$

Now,

• Multiply and divide denominator by √(2), we get

$\bf \longmapsto \: \bf \:\dfrac{1}{ \sqrt{2} } \int \dfrac{dx}{\dfrac{1}{ \sqrt{2}} sinx + \dfrac{1}{ \sqrt{2} } cosx}$

$\bf \longmapsto \: \bf \:\dfrac{1}{ \sqrt{2} } \int \dfrac{dx}{sinx \: cos\dfrac{\pi}{4} + cosx \: sin\dfrac{\pi}{4} }$

$\bf \longmapsto \: \bf \:\dfrac{1}{ \sqrt{2} } \int \dfrac{dx}{sin \bigg(x + \dfrac{\pi}{4} \bigg)}$

$\bf \longmapsto \: \bf \:\dfrac{1}{ \sqrt{2} } \int cosec\bigg( x + \dfrac{\pi}{4} \bigg)dx$

$\bf \longmapsto \: \bf \:\dfrac{1}{ \sqrt{2} } log \bigg \{tan\bigg( \dfrac{\pi}{8} + \dfrac{x}{2} \bigg) \bigg \} + c$