Question

Find the following integral : $\int \frac{2- 3\ sin\ x}{cos^2 x} \, dx$

Answer 1

Answer:

Step-by-step explanation:

Concept:

$\int{secx.tanx}\:dx=secx+c\\\\\int{sec^2x}\:dx=tanx+c$

The given integral is solved by decomposition method.

In decomposition method, the given integrand(non-integrable function) is decomposed into integrable functions by using algebraic identities, trigonometry identities, etc.

Now,

$\int[\frac{2-3sinx}{cos^2x}]\:dx\\\\=\int[\frac{2}{cos^2x}-3\frac{sinx}{cos^2x}]\:dx\\\\=\int[2\frac{1}{cos^2x}-3\frac{1}{cosx}\frac{sinx}{cosx}]\:dx\\\\=\int[2sec^2x-3secx.tanx]\:dx\\\\=2\int{sec^2x}\:dx-3\int{secx.tanx}\:dx\\\\=2tanx-3secx+c$