Jawab:
[tex]\displaystyle -\arctan (\cos x)+C[/tex]
Penjelasan dengan langkah-langkah:
Ini hanya integral substitusi
[tex]\displaystyle \int \frac{\sin x}{1+\cos^2 x}~dx\\u=\cos x\\du=-\sin x~dx\\=\int \frac{\sin x}{1+u^2}~\frac{du}{-\sin x}\\=-\frac{du}{1+u^2}\\=-\arctan u+C\\=-\arctan (\cos x)+C[/tex]
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Verified answer
Jawab:
[tex]\displaystyle -\arctan (\cos x)+C[/tex]
Penjelasan dengan langkah-langkah:
Ini hanya integral substitusi
[tex]\displaystyle \int \frac{\sin x}{1+\cos^2 x}~dx\\u=\cos x\\du=-\sin x~dx\\=\int \frac{\sin x}{1+u^2}~\frac{du}{-\sin x}\\=-\frac{du}{1+u^2}\\=-\arctan u+C\\=-\arctan (\cos x)+C[/tex]