Find the

Question:

$\int \frac{2-3 \sin x}{\cos ^{2} x} d x$

Solution:

$\int \frac{2-3 \sin x}{\cos ^{2} x} d x$

$=\int\left(\frac{2}{\cos ^{2} x}-\frac{3 \sin x}{\cos ^{2} x}\right) d x$

$=\int 2 \sec ^{2} x d x-3 \int \tan x \sec x d x$

$=2 \tan x-3 \sec x+\mathrm{C}$

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