Question

$\red{ \bf \: evaluate \: this \: : }\\ \\ \pink{\leadsto \: }\orange{ \displaystyle \int \sf \: \frac{sin \: x + cos \: x}{ \sqrt{cos2x} } \: \: dx}$
​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle \int \sf \: \frac{sin \: x + cos \: x}{ \sqrt{cos2x} } \: \: dx$

can be rewritten as

$\rm \: = \:\displaystyle \int \sf \: \frac{sin \: x }{ \sqrt{cos2x} } \: \: dx + \displaystyle \int \sf \: \frac{ cos \: x}{ \sqrt{cos2x} } \: \: dx$

We know,

$\boxed{ \bf{ \: cos2x = 1 - {2sin}^{2}x = {2cos}^{2}x - 1}}$

So, above integrals can be rewritten as

$\rm \: = \red{ \:\displaystyle \int \sf \: \frac{sin \: x }{ \sqrt{ {2cos}^{2}x - 1 } } \: \: dx} + \purple{\displaystyle \int \sf \: \frac{ cos \: x}{ \sqrt{ {2sin}^{2}x - 1 } } \: \: dx}$

In the first integral,

$\red{\rm :\longmapsto\:Put \: cosx = u}$

$\red{\rm :\longmapsto\: - sinx \: dx \: = \: du}$

$\red{\rm :\longmapsto\: sinx \: dx \: = \: - \: du}$

In the second integral,

$\purple{\rm :\longmapsto\:Put \: sinx = v}$

$\purple{\rm :\longmapsto\: \: cosx \: dx = dv}$

So, above integrals can be reduced to

$\rm \: = \: - \red{\displaystyle \int \sf \: \frac{du}{ \sqrt{ {2u}^{2} - 1} } } \: + \: \purple{\displaystyle \int \sf \: \frac{dv}{ \sqrt{1 - {2v}^{2} } } }$

$\rm \: = \: - \red{\dfrac{1}{ \sqrt{2} } \displaystyle \int \sf \: \frac{du}{ \sqrt{ {u}^{2} - \dfrac{1}{2} } } } \: + \: \purple{\dfrac{1}{ \sqrt{2} } \displaystyle \int \sf \: \frac{dv}{ \sqrt{\dfrac{1}{2} - {v}^{2} } } }$

$\rm \: = \: - \red{\dfrac{1}{ \sqrt{2} } \displaystyle \int \sf \: \frac{du}{ \sqrt{ {u}^{2} - \dfrac{1}{ {( \sqrt{2}) }^{2} } } } } \: + \: \purple{\dfrac{1}{ \sqrt{2} } \displaystyle \int \sf \: \frac{dv}{ \sqrt{\dfrac{1}{ {( \sqrt{2} )}^{2} } - {v}^{2} } } }$

We know,

$\boxed{ \bf{ \: \displaystyle \int \sf \: \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} + c }}$

and

$\boxed{ \bf{ \: \displaystyle \int \sf \: \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log | \: x + \sqrt{ {x}^{2} - {a}^{2} } \: | + c }}$

$\rm \: = \:\dfrac{1}{ \sqrt{2} }\bigg[ - log\bigg |u + \sqrt{ {u}^{2} - \dfrac{1}{2} } \bigg| + {sin}^{ - 1}\dfrac{v}{\dfrac{1}{ \sqrt{2} } }\bigg] + c$

$\rm \: = \:\dfrac{1}{ \sqrt{2} }\bigg[ - log\bigg |u + \sqrt{\dfrac{ {2u}^{2} - 1}{2} } \bigg| + {sin}^{ - 1}( \sqrt{2} v) \bigg] + c$

$\rm \: = \:\dfrac{1}{ \sqrt{2} }\bigg[ - log\bigg |\dfrac{ \sqrt{2} u + \sqrt{ {2u}^{2} - 1} }{ \sqrt{2} } \bigg| + {sin}^{ - 1}( \sqrt{2} v) \bigg] + c$

$\rm \: = \:\dfrac{1}{ \sqrt{2} }\bigg[ - log\bigg |\sqrt{2} u + \sqrt{ {2u}^{2} - 1}\bigg| + log \sqrt{2} + {sin}^{ - 1}( \sqrt{2} v) \bigg] + c$

$\rm \: = \:\dfrac{1}{ \sqrt{2} }\bigg[ - log\bigg |\sqrt{2} u + \sqrt{ {2u}^{2} - 1}\bigg| + {sin}^{ - 1}( \sqrt{2} v) \bigg] + d$

$\rm \: = \:\dfrac{1}{ \sqrt{2} }\bigg[ - log\bigg |\sqrt{2}cosx + \sqrt{ {2cos}^{2}x - 1}\bigg| + {sin}^{ - 1}( \sqrt{2} sinx) \bigg] + d$

$\rm \: = \:\dfrac{1}{ \sqrt{2} }\bigg[ - log\bigg |\sqrt{2}cosx + \sqrt{ cos2x}\bigg| + {sin}^{ - 1}( \sqrt{2} sinx) \bigg] + d$

Additional Information

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