Question

integration of 1/sinx cosx​

Answer 1

$\huge \boxed{ \red{\mathfrak{solution}}}$

To find :

$\boxed{ \bold{\int \: \frac{1}{ sinx \: cosx} dx }}$

In place of 1 we write

$\boxed{ \bold{{sin}^{2} x + {cos}^{2} x = 1}}$

$\leadsto \int \frac{ {sin}^{2}x + {cos}^{2} x }{sinx \: cosx}dx \\ \\ \leadsto \: \int \frac{ {sin}^{2}x }{sinx \: cosx} dx + \int \: \frac{ {cos}^{2}x }{sinx \: cosx} dx \\ \\ \leadsto \: \int \: \cancel \frac{ {sin}^{2}x }{sinx \: \: \: cosx} dx + \int \cancel \frac{ {cos}^{2}x }{cosx \: \: sinx} dx \\ \\ \leadsto \int \: \frac{sinx}{cosx} dx + \int \: \frac{cosx}{sinx} dx \\ \\ \leadsto \int \: tanx \: dx + \int \: cot \: dx \\ \\ \leadsto \: log( secx) + log(sinx ) + c \:$

Formula used :

$\boxed{ \green{\bold{\int \: cot \: dx = log(sinx) + c}}}$

$\boxed{ \purple{\bold{\int \: tanx \: dx = log(secx) + c}}}$

Answer 2

Question :-

Integrate it ,

$\large \int \sf{ \frac{1}{ \sin x \cos } dx}$

Answer :-

$\large \boxed{\int \sf{ \frac{1}{ \sin x \cos } dx} = log( \tan x) + c}$

Formula used :-

$\star \: \sf{ \frac{1}{ \cos x} = \sec x} \\ \\ \star \: \frac{ \sin x}{ \cos x} \: = \tan x$

Explanation :-

We have ,

$\rightarrow \large \int \sf{ \frac{1}{ \sin x \cos } dx} \\ \\ \large \mathfrak{we \: can \: write \: it} \\ \\ \rightarrow \large \int \sf{ \frac{1}{ \sin x \cos \times \frac{ \cos x}{ \cos x} } dx} \\ \: \\ \\ \rightarrow \large \int \sf{ \frac{1}{ \frac{ \sin x}{ \cos x} \times { \cos}^{2} x } dx} \\ \\ \rightarrow \large \int \sf{ \frac{1}{ \tan x \: \: { \cos}^{2} x} dx} \\ \\ \rightarrow \large \int \: \sf{ \frac{ { \sec}^{2} x}{ \tan x} dx} \\ \\ \sf{let \: \tan x = t \: ....(1)} \\ \\ \mathfrak{differentiate \: with \: respect \: to \: } \sf{x} \\ \\ \implies \large \sf{ { \sec}^{2} x \: dx = dt} \\ \\ \rightarrow \large \int \sf{ \frac{1}{t} dt} \\ \\ \because \sf{ \int \large \frac{1}{x} dx = log(x) + c} \\ \\ \therefore \\ \rightarrow \large \sf{ log(t) + c} \\ \\ \sf{ from \: (1)} \\ \\ \rightarrow \large \sf{ log( \tan x) + c}$

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