Question

Evaluate the following :

$\sf \: \int \: \frac{dx}{secx + cosecx}$
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Answer 1

Question :-

Evaluate : $\rm{ \int \frac{dx}{ \sec x + \cos \sec x} }$

Solution :-

$\small\rm:\longmapsto{ \int \frac{dx}{ \frac{1}{ \cos x} + \frac{1}{ \sin x} } } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \small\rm:\longmapsto{ \frac{1}{2} \times \int \frac{2 \sin x \cos x}{ \sin x + \cos x}dx } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \small\rm:\longmapsto{ \frac{1}{2} \int \frac{(1 + \sin 2x) - 1}{ \sin x + \cos x} dx } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\\ \\ \small\rm:\longmapsto{ \frac{1}{2} \int \frac{( \sin x + \cos x) {}^{2} - 1 }{ \sin x + \cos x} dx } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \small\rm:\longmapsto{ \frac{1}{2} \int ( \sin x + \cos x)dx - \frac{1}{2} \int \frac{dx}{ \sin x + \cos x} } \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \small\rm:\longmapsto{ \frac{1}{2 } \int( \sin x + \cos x)dx - \frac{1}{2} \int \frac{dx}{ \sqrt{2} \sin \bigg( x + \frac{\pi}{4} \bigg) } } \: \: \: \: \: \\ \\ \small\rm:\longmapsto{ \frac{1}{2} \int( \sin x + \cos x)dx - \frac{1}{2 \sqrt{2} } \int \csc \bigg(x + \frac{\pi}{4} \bigg)dx } \\ \\ \\ \tiny\rm:\longmapsto \pink{{ \boxed{ \rm{ \frac{1}{2} ( \sin x - \cos x) - \frac{1}{2 \sqrt{2} }ln \bigg | \csc \bigg(x + \frac{\pi}{4} \bigg) - \cot \bigg(x + \frac{\pi}{4} \bigg) \bigg| + c}}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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Answer 2

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

Given integral is

$\red{\rm :\longmapsto\:\displaystyle\int \frac{dx}{secx + cosecx}}$

Let assume that,

$\red{\rm :\longmapsto\:I \: = \: \displaystyle\int \frac{dx}{secx + cosecx}}$

${\rm :\longmapsto\:I \: = \: \displaystyle\int \frac{dx}{\dfrac{1}{cosx} + \dfrac{1}{sinx} }}$

${\rm :\longmapsto\:I \: = \: \displaystyle\int \frac{dx}{\dfrac{sinx + cosx}{sinx \: cosx}}}$

$\rm :\longmapsto\:I = \displaystyle\int \frac{sinxcosx \: dx}{sinx + cosx}$

On multiply and divide by 2, we get

$\rm :\longmapsto\:I =\dfrac{1}{2} \displaystyle\int \frac{2sinxcosx \: dx}{sinx + cosx}$

$\rm :\longmapsto\:I =\dfrac{1}{2} \displaystyle\int \frac{1 + 2sinxcosx - 1}{sinx + cosx} \: dx$

$\rm :\longmapsto\:I =\dfrac{1}{2} \displaystyle\int \frac{ {sin}^{2}x + {cos}^{2}x + 2sinxcosx - 1}{sinx + cosx} \: dx$

$\rm :\longmapsto\:I =\dfrac{1}{2} \displaystyle\int \frac{ {(sinx + cosx)}^{2} - 1}{sinx + cosx} \: dx$

$\rm :\longmapsto\:I =\dfrac{1}{2} \displaystyle\int \frac{ {(sinx + cosx)}^{2}}{sinx + cosx} \: dx - \frac{1}{2} \displaystyle\int \frac{1}{sinx + cosx}dx$

$\rm :\longmapsto\:I =\dfrac{1}{2} \displaystyle\int [sinx + cosx] \: dx - \frac{1}{2} \displaystyle\int \frac{1}{sinx + cosx}dx$

Let assume that,

$\red{\rm :\longmapsto\:I = I_1 - I_2} - - - (1)$

where,

$\rm :\longmapsto\:I_1 = \dfrac{1}{2}\displaystyle\int [sinx + cosx] \: dx$

and

$\rm :\longmapsto\:I_2 = \dfrac{1}{2}\displaystyle\int \frac{1}{sinx + cosx} \: dx$

Consider,

$\rm :\longmapsto\:I_2 = \dfrac{1}{2}\displaystyle\int \frac{1}{sinx + cosx} \: dx$

$\rm :\longmapsto\:I_2 = \dfrac{1}{2 \sqrt{2} }\displaystyle\int \frac{1}{\dfrac{1}{ \sqrt{2} } sinx + \dfrac{1}{ \sqrt{2} } cosx} \: dx$

$\rm :\longmapsto\:I_2 = \dfrac{1}{2 \sqrt{2} }\displaystyle\int \frac{1}{cos\dfrac{\pi}{ 4 } sinx + sin\dfrac{\pi}{ 4 } cosx} \: dx$

We know

$\boxed{ \bf{ \: sin(x + y) = sinxcosy + sinycosx}}$

So, using this

$\rm :\longmapsto\:I_2 = \dfrac{1}{2 \sqrt{2} }\displaystyle\int \frac{1}{sin\bigg(x + \dfrac{\pi}{ 4 } \bigg)} \: dx$

$\rm :\longmapsto\:I_2 = \dfrac{1}{2 \sqrt{2} }\displaystyle\int cosec\bigg(x + \dfrac{\pi}{ 4 }\bigg) \: dx$

We know,

$\boxed{ \bf{ \: \displaystyle\int cosecx \: dx \: = \: log \: tan \frac{x}{2} + c}}$

$\rm :\longmapsto\:I_2 = \dfrac{1}{2 \sqrt{2} } \: log \: tan\bigg(\dfrac{x}{2} + \dfrac{\pi}{ 8 }\bigg) \: + c_2 - - - (2)$

Now Consider,

$\rm :\longmapsto\:I_1 = \dfrac{1}{2}\displaystyle\int [sinx + cosx] \: dx$

$\rm :\longmapsto\:I_1 = - cosx + sinx + c_1 - - - (2)$

Hence, On substituting the values of equation (2) and (3) in equation (1), we get

$\rm :\longmapsto\:I= -\dfrac{1}{2 \sqrt{2} } \: log \: tan\bigg(\dfrac{x}{2} + \dfrac{\pi}{8}\bigg)-c_2 +\dfrac{1}{2}\bigg[ - cosx + sinx\bigg] + c_1$

$\rm :\longmapsto\:I= -\:\dfrac{1}{2 \sqrt{2} } \: log \: tan\bigg(\dfrac{x}{2} + \dfrac{\pi}{8}\bigg) + \dfrac{1}{2}\bigg[ - cosx + sinx\bigg] +c$

Additional Information :-

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