Question

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

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Answer 1

We split the integral as,

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{\sin x\ dx}{\sqrt{cos(2x)}}+\int\dfrac{\cos x\ dx}{\sqrt{cos(2x)}} }$

Since $\small\cos(2x)=2\cos^2x-1=1-2\sin^2x,$

$\small\text{ \displaystyle\longrightarrow I=\int\dfrac{\sin x\ dx}{\sqrt{2\cos^2x-1}}+\int\dfrac{\cos x\ dx}{\sqrt{1-2\sin^2x}} }$

Consider,

$\small\text{ \displaystyle\longrightarrow I_1=\int\dfrac{\sin x\ dx}{\sqrt{2\cos^2x-1}} }$

$\small\text{ \displaystyle\longrightarrow I_1=-\int\dfrac{-\sin x\ dx}{\sqrt{\left(\sqrt2\,\cos x\right)^2-1^2}} }$

We have,

• $\small\text{ \displaystyle\int\dfrac{dx}{\sqrt{(ax+b)^2-c^2}}=\dfrac{1}{a}\log\left|(ax+b)+\sqrt{(ax+b)^2-c^2}\right|+C }$

Then,

$\small\displaystyle\longrightarrow I_1=-\dfrac{1}{\sqrt2}\,\log\left|\sqrt2\,\cos x+\sqrt{2\cos^2x-1}\right|+C_1$

$\small\displaystyle\longrightarrow I_1=-\dfrac{1}{\sqrt2}\,\log\left|\sqrt2\,\cos x+\sqrt{\cos(2x)}\right|+C_1$

Consider,

$\small\text{ \displaystyle\longrightarrow I_2=\int\dfrac{\cos x\ dx}{\sqrt{1-2\sin^2x}} }$

$\small\text{ \displaystyle\longrightarrow I_2=\int\dfrac{\cos x\ dx}{\sqrt{1^2-\left(\sqrt2\,\sin x\right)^2}} }$

We have,

• $\small\text{ \displaystyle\int\dfrac{dx}{\sqrt{c^2-(ax+b)^2}}=\dfrac{1}{a}\sin^{-1}\left(\dfrac{ax+b}{c}\right)+C }$

Then,

$\small\displaystyle\longrightarrow I_2=\dfrac{1}{\sqrt2}\sin^{-1}\left(\sqrt2\,\sin x\right)+C_2$

Hence,

$\small\displaystyle\longrightarrow I=\dfrac{1}{\sqrt2}\sin^{-1}\left(\sqrt2\,\sin x\right)-\dfrac{1}{\sqrt2}\log\left|\sqrt2\,\cos x+\sqrt{\cos(2x)}\right|+C_1+C_2$

Taking $\smallC_1+C_2=C,$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{\sqrt2}\left[\sin^{-1}\left(\sqrt2\,\sin x\right)-\log\left|\sqrt2\,\cos x+\sqrt{\cos(2x)}\right|\right]+C}} }$

Answer 2

Solution−

Given integral is

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

can be rewritten as

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

We know,

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

So, above integrals can be rewritten as

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

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

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

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

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

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

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

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

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

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

Additional Information:-

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