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Answer 1 · 2017-01-09T15:13:34

$\int \frac{4 - 5 sin (x)}{{cos}^{2} (x)} d x = 4 tan (x) - 5 sec (x) + C$ $int(4 - 5sin(x))/(cos^2(x))dx = 4tan(x) - 5sec(x) + C$

Given: $\int \frac{4 - 5 sin (x)}{{cos}^{2} (x)} d x$ $int(4 - 5sin(x))/(cos^2(x))dx$

Use the identities $\frac{1}{cos (x)} = sec (x) and tan (x) = \frac{sin (x)}{cos (x)}$ $1/cos(x) = sec(x) and tan(x) = sin(x)/cos(x)$ :

$\int (4 {sec}^{2} (x) - 5 tan (x) sec (x)) d x$ $int(4sec^2(x) - 5tan(x)sec(x))dx$

Break the integral into two integrals:

$4 \int {sec}^{2} (x) d x - 5 \int tan (x) sec (x) d x$ $4intsec^2(x)dx - 5inttan(x)sec(x)dx$

Both of these are well known:

$4 tan (x) - 5 sec (x) + C$ $4tan(x) - 5sec(x) + C$