[tex] \frac{1}{2}∫ - \frac{1}{2} { \cos}^{2}xdx \: \times \: ∫ \frac{1}{cosx}dx[/tex]
Soal gampang ini
Jawaban:
[tex]= - \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[/tex]
Penjelasan dengan langkah-langkah:
[tex] = \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[/tex]
[tex] \int \frac{1}{ \cos(x) } \: dx\\ = \int { \sec }(x) \: dx \\ = \int \frac{ { \sec {}^{2} (x) } + \sec(x) \tan(x) }{ \sec(x) + \tan(x) } \: dx \\ = [/tex]
u= sec x + tan x
du= sec²x+ sec x tan x dx
[tex] \int \frac{du}{u} \\ = \int \frac{1}{u} du \\ = ln(u) + c \\ = ln( \sec(x) + \tan(x) ) + c[/tex]
maka:
[tex]= - \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[/tex]
[answer.2.content]