Soal tentang Integral Trigonometri


\int {\frac{{\sin x}}{{\sin x - \cos x}}} {\mathop{\rm dx}\nolimits}
= \frac{1}{2}\int {\frac{{2\sin x}}{{\sin x - \cos x}}{\mathop{\rm dx}\nolimits} }
= \frac{1}{2}\int {\frac{{(\sin x + \cos x) + (\sin x - \cos x)}}{{\sin x - \cos x}}{\mathop{\rm dx}\nolimits} }
= \frac{1}{2}\int {\frac{{\sin x - \cos x}}{{\sin x - \cos x}} + \frac{{\sin x + \cos x}}{{\sin x - \cos x}}{\mathop{\rm dx}\nolimits} }
= \frac{1}{2}\left( {\int {{\mathop{\rm dx}\nolimits} }  + \int {\frac{{\sin x + \cos x}}{{\sin x - \cos x}}{\mathop{\rm dx}\nolimits} } } \right)

misalkan
u = sinx – cosx
maka du = cosx + sinx dx
sehingga soal menjadi

\frac{1}{2}\left( {x + \int {\frac{{du}}{u}} } \right)
= \frac{1}{2}\left( {{x^2} + \ln |u|} \right) + C
= \frac{1}{2}({x^2} + \ln |\sin x - \cos x|) + C

Selesai.

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