Question

Evaluate the following integral

$\frac{1}{secx + cosecx}$

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Answer 1

Answer:

The law of conservation of mass states that in a chemical reaction mass is neither created nor destroyed. ... The carbon atom changes from a solid structure to a gas but its mass does not change. Similarly, the law of conservation of energy states that the amount of energy is neither created nor destroyed.

Answer 2

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

Given integral is

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

can be rewritten as

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

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

can be rewritten as

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

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

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

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

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

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

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

$\rm=\dfrac{1}{2}[ - cosx + sinx] - \dfrac{1}{2 \sqrt{2} }\displaystyle\int\rm \frac{dx}{cos\bigg[x - \dfrac{\pi}{4} \bigg] }$

$\rm=\dfrac{1}{2}[ - cosx + sinx] - \dfrac{1}{2 \sqrt{2} }\displaystyle\int\rm sec\bigg[x - \dfrac{\pi}{4} \bigg]dx$

$\rm=\dfrac{1}{2}\bigg[ - cosx + sinx\bigg] - \dfrac{1}{2 \sqrt{2} }log\bigg | sec\bigg[x - \dfrac{\pi}{4} \bigg] + tan\bigg[x - \dfrac{\pi}{4} \bigg]\bigg | + c$

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Formula Used :-

$\boxed{ \tt{ \: cosecx = \frac{1}{sinx} \: }}$

$\boxed{ \tt{ \: secx = \frac{1}{cox} \: }}$

$\boxed{ \tt{ \: \displaystyle\int\rm \: cosx \: dx \: = \: sinx \: + \: c \: }}$

$\boxed{ \tt{ \: \displaystyle\int\rm \: sinx \: dx \: = - \: \: cosx \: + \: c \: }}$

$\boxed{ \tt{ \: cosxcosy + sinxsiny = cos(x - y) \: }}$

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

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More to know :-

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