Question

Evaluate (cos 2x+2sin^2x)/ cis^2x dx

Answer 1

Step-by-step explanation:

$\tt \large \underline{ solution} \\ \\ \\ \tt cosx = {cos}^{2} x - {sin}^{2} x \\ \\ \\ \tt {cos}^{2} x = 1 - {sin}^{2} x \\ \\ \\ \tt cos2x = 1 - {sin}^{2} x - {sin}^{2} x \\ \\ \\ \tt : \implies 1 - 2 {sin}^{2} x \\ \\ \\ \tt \large \int \frac{ {cos}^{2}x + 2 {sin}^{2} x }{ {cos}^{2} x} \: \: dx \\ \\ \\ \tt :\implies \large\int \frac{1 - \cancel{2 {sin}^{2}x} + \cancel{2 {sin}^{2} x} }{ {cos}^{2}x } \\ \\ \\ \tt :\implies \large \int \frac{1}{ {cos}^{2} x} dx = \large \int {sec}^{2} x \: \: dx \\ \\ \\ : \implies \tt\large tanx \: + c$