[tex] \frac{1}{2}∫ - \frac{1}{2} { \cos}^{2}xdx \: \times \: ∫ \frac{1}{cosx}dx[/tex]Soal gampang in.... Question from @14299

August 2022 1 4 Report

[tex] \frac{1}{2}∫ - \frac{1}{2} { \cos}^{2}xdx \: \times \: ∫ \frac{1}{cosx}dx[/tex]
Soal gampang ini

Jawaban:

$= - \frac{1}{4} ( \frac{ \sin(2x) }{4} + \frac{1}{2}x) \int \frac{1}{ \cos(x) } \: dx \\ = ( - \frac{1}{16} \sin(2x) - \frac{1}{8} x)( ln( \sec(x) + \tan(x) ) ) + c$

Penjelasan dengan langkah-langkah:

$= \frac{1}{2} \times - \frac{1}{2} \int { \cos }^{2} (x) \times \int \frac{1}{ \cos(x) } \\ = - \frac{1}{4} \int { \cos }^{2} (x) \times \int \frac{1}{ \cos(x) } \\ - \frac{1}{4} \int \frac{ \cos(2x) + 1 }{2} \: dx \: \int \frac{1}{ \cos(x) } \: dx \\ = - \frac{1}{4} (\int \frac{ \cos(2x) }{2} \: dx + \int \frac{1}{2} \: dx ) \int \frac{1}{ \cos(x) } dx \\ = - \frac{1}{4} ( \frac{ \sin(2x) }{4} + \frac{1}{2}x) \int \frac{1}{ \cos(x) } \: dx$

$\int \frac{1}{ \cos(x) } \: dx\\ = \int { \sec }(x) \: dx \\ = \int \frac{ { \sec {}^{2} (x) } + \sec(x) \tan(x) }{ \sec(x) + \tan(x) } \: dx \\ =$

u= sec x + tan x

du= sec²x+ sec x tan x dx

$\int \frac{du}{u} \\ = \int \frac{1}{u} du \\ = ln(u) + c \\ = ln( \sec(x) + \tan(x) ) + c$

maka:

$= - \frac{1}{4} ( \frac{ \sin(2x) }{4} + \frac{1}{2}x) \int \frac{1}{ \cos(x) } \: dx \\ = ( - \frac{1}{16} \sin(2x) - \frac{1}{8} x)( ln( \sec(x) + \tan(x) ) ) + c$

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