Question

integrate sin^2x/1+cosx​

Answer 1

$\huge \green{ \mathfrak{solution}}$

$\star \int \: \frac{ {sin}^{2}x }{1 + cosx} dx \\ \\ \star \int \: \frac{1 - {cos}^{2}x }{1 + cosx} dx \\ \\ \star \int \frac{(1 - cosx)(1 + cosx)}{1 + cosx} dx\\ \\ \star \int \frac{(1 - cosx) \cancel{(1 + cosx)}}{ \cancel{1 + cosx}} dx \\ \\ \star \int \: (1 - cosx )\: dx \\ \\ \star \int \: 1dx \: - \int \: cosx \: dx \\ \\ \star \: \boxed{ \bold{x \: - sinx \: + c}}$

Formula:

$\bold{ {sin}^{2} x \: + {cos}^{2} x = 1} \\ \\ \star \bold{ {sin}^{2} x = 1 - {cos}^{2} x}$

$\star \: \bold{{x}^{2} - {y}^{2} = (x - y)(x + y)}$

Answer 2

Answer:

this is the answer with explanation