Question

Evaluate ∫ 1 dx./(sinx. cosx)

Answer 1

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

Given integral is

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

can be rewritten as

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

$\rm \:  =  \: 2 \: \displaystyle\int\rm \frac{dx}{\: sin2x}$

$\rm \:  =  \: 2 \: \displaystyle\int\rm cosec2x \: dx$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int\rm cosecx \: dx = log |cosecx \: - \: cotx| + c}}}$

So, using this identity, we get

$\rm \:  =  \: 2 \: \times \dfrac{1}{2} \: log |cosec2x \: - \: cot2x| + c$

$\rm \:  =  \: log |cosec2x \: - \: cot2x| + c$

Hence,

$\boxed{\tt{ \rm \: \displaystyle\int\rm \frac{dx}{sinx \: cosx} =  \: log |cosec2x \: - \: cot2x| + c}}$

ALTERNATIVE METHOD

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

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

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

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

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

$\rm \:  =  \: \displaystyle\int\rm tanx \: dx \: + \: \displaystyle\int\rm cotx \: dx$

$\rm \:  =  \: - log |cosx| + log |sinx| + c$

$\rm \:  =  \: log\bigg |\dfrac{sinx}{cosx} \bigg| + c$

$\rm \:  =  \: log |tanx| + c$

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LEARN MORE

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

Answer 2

$\large\underline{\sf{Solution-}} \\ \\ Given \: \: integral \: \: is \\ \\ \rm :\longmapsto\:\displaystyle\int\rm \frac{dx}{sinx \: cosx} \\ \\ can \: \: be \: \: rewritten \: \: as \\ \\ \rm \: = \: 2 \: \displaystyle\int\rm \frac{dx}{2 \: sinx \: cosx} \\ \\ \rm \: = \: 2 \:\displaystyle\int\rm \frac{dx}{\: sin2x} \\ \\ \rm \: = \: 2 \: \displaystyle\int\rm cosec2x \: </p><p>\\ \\ We \: \: know \: , \\ \\\begin{gathered}\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int\rm cosecx \: dx = log |cosecx \: - \: cotx| + c}}} \\ \end{gathered} \\ \\ So \: \: , using \: \: this \: \: identity \: \: , \: \: we \: \: get \\ \\ \rm \: = \: 2 \: \times \dfrac{1}{2} \: log |cosec2x \: - \: cot2x| + c \\ \\ \rm \: = \: log |cosec2x \: - \: cot2x| + c \\ \\ </p><p>Hence \: , \\ \\ \begin{gathered}\boxed{\tt{ \rm \: \displaystyle\int\rm \frac{dx}{sinx \: cosx} = \: log |cosec2x \: - \: cot2x| + c}} \\ \end{gathered}$

ALTERNATIVE METHOD

$\rm :\longmapsto\:\displaystyle\int\rm \frac{dx}{sinx \: cosx} \\ \\ \rm \: = \: \displaystyle\int\rm \frac{1}{sinx \: cosx} \: dx = ∫ \\ \\ \rm \: = \: \displaystyle\int\rm \frac{ {sin}^{2}x + {cos}^{2}x}{sinx \: cosx} \: dx = ∫ \\ \\ \rm \: = \: \displaystyle\int\rm \frac{ {sin}^{2}x}{sinx \: cosx} \: dx \: + \: \displaystyle\int\rm \frac{ {cos}^{2}x}{sinx \: cosx} \: dx = ∫ \\ \\ </p><p>\rm \: = \: \displaystyle\int\rm \frac{ {sin}x}{\: cosx} \: dx \: + \: \displaystyle\int\rm \frac{ {cos}x}{sinx \: } \: dx = ∫ \\ \\ \rm \: = \: \displaystyle\int\rm tanx \: dx \: + \: \displaystyle\int\rm cotx \: dx \\ \\ \begin{gathered}\rm \: = \: - log |cosx| + log |sinx| + c \\ \end{gathered} \\ \\\begin{gathered}\rm \: = \: log\bigg |\dfrac{sinx}{cosx} \bigg| + c \\ \end{gathered} \\ \\ </p><p>\begin{gathered}\rm \: = \: log |tanx| + c \\ \end{gathered}$

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