Question

Integration of (sin4x-cos2x)/cos x dx

Answer 1

Step-by-step explanation:

We have,

$\int \frac{ \sin(4x) - \cos(2x) }{ \cos(x) } dx$

$= \int \frac{ \sin(4x) }{ \cos(x) } dx - \int \frac{ \cos(2x) }{ \cos(x) } dx$

$= \int \frac{4 \sin(x) \cos(x) \cos(2x) }{ \cos(x) } dx - \int \frac{2 \cos ^{2} (x) - 1 }{ \cos(x) } dx$

$=4 \int \sin(x) (2 \cos^{2} (x) - 1)dx - \int (2 \cos(x) - \sec(x)) dx$

In the first integral putting cos(x) = t,

$= - 4 \int (2t^{2} - 1)dx - 2\int \cos(x) dx - \int \sec(x) dx$

$= - \frac{8 { \cos}^{3} (x)}{3} + \cos(x) - 2 \sin(x) - ln( | \sec(x) + \tan(x) | ) + c$