Question

integrate.......2-3sinx/cos^2x​

Answer 1

$\huge\mathcal{Given}\\=\int\frac{2-3sinx}{cos^2x}dx\\=\int\frac{2}{cos^2x}dx-\int\frac{3sinx}{cos^2x}dx\\Let\: \bold{I_1=\int\frac{2}{cos^2x}dx}\\\bold{I_2=\int\frac{3sinx}{cos^2x}dx}\\Evaluate\:I_1\\=>\int\frac{2}{cos^2x}dx=2\int\frac{1}{cos^2x}dx\\=2\int sec^2x dx\\\bold{I_1=2Tanx + C}\\Hope\:it's\:familiar\:that\:\int sec^2xdx=Tanx+C\\Now\:I_2\\=3\int\frac{sinx}{cos^2x}dx\\put\:cosx=r\\-\:sinx.dx=dr\\sinx.dx=-dr\\I_2\:becomes\:=3\int\frac{-dr}{r^2}\\=-3\frac{r^{-2+1}}{-2+1}+K\\=-3\frac{-1}{r}+K\\\bold{I_2=\frac{3}{r}=\frac{3}{cosx}+K=3sec x+K}\\Now\:combine\:I_1\:and\:I_2\\=2\int\frac{1}{cos^2x}dx-3\int\frac{sinx}{cos^2x}dx\\=>2Tanx+C\:-\:\frac{(3)}{cosx}+K\\=>\huge{2Tanx-3secx+L}\\=>\frac{2sinx}{cosx}-\frac{3}{cosx}+L\\=>\huge\frac{2sinx-3}{cosx}+L$

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