Question

integrate tan x/(log cos x)^2​

Answer 1

Here

we used substitution method

•let

log cosx= t

we get

-tanx dx= dt

•put tanx dx =- dt

and

log cosx = t

in Given integration :

Formula used :

$\bold{\boxed{\frac{d}{dx} cos \: x = - sinx \: }}$

and

$\boxed{ \bold{\int \: {x}^{y} = } \frac{ {x}^{y + 1} }{y + 1} }$

soln refers to the attachment

Answer 2

Question :-

Integrate it .

$\rightarrow \sf{ \int \: \frac{ \red{ \tan(x)} }{ { (\:\green{log \cos(x) } \:) }^{2} } } dx$

Answer :-

$\rightarrow \boxed{\sf{ \red{\frac{1}{ log \cos(x) } } + c}} \\ \:$

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Explanation :-

We used here substitution method of integration .

Firstly let,

$\rightarrow \sf{ \red{ log \cos(x) } \: = k} \: \: .......(1)\\ \\ \sf{ \pink{differentiate \: with \: respect \: to \: x}} \\ \\ \rightarrow \pink{\sf{ \frac{1}{ \cos(x) } \frac{d}{dx} \cos(x)} = \green{\frac{dk}{dx} }} \\ \\ \rightarrow \sf{ \green{ - \frac{ \sin(x) }{ \cos(x) } }= \red{ \frac{dk}{dx} } }\\ \\ \rightarrow \sf{ \pink{ - \tan(x) \: dx \: }= dk} \\ \\ \rightarrow \boxed{ \sf{ \green{\tan(x) \: dx \: = - dk}}}$

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Now put tan(x) dx = - dk in question,

$\rightarrow \sf{ \red{ - \int \: } \green{ \frac{dk}{ {k}^{2} }} } \\ \\ \rightarrow \sf{ \pink{- \int \: } \red{ {k}^{ - 2} \: dk} }\\ \\ \rightarrow \sf{ - \green{ \bigg( \frac{ {k}^{ - 2 + 1} }{ - 2 + 1} \bigg)} }+ c \\ \\ \rightarrow \sf{ \pink{{k}^{ - 1} } + c} \\ \\ \rightarrow \sf{ \red{\frac{1}{k} + c}}$

From (1)

$\rightarrow \boxed{\sf{ \red{\frac{1}{ \green{log \cos(x)} }} + c}}$

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