Question

evaluate integration of sin²x / 1+cosx dx​

Answer 1

Answer:

Step-by-step explanation:

∫sin^2 x dx /1+cosx

∫1-cos^2 x dx /1+cos x

∫(1-cosx) dx

x-sinx+c

Answer 2

Answer:

$solving \: \: the \: \: given \: \: function = > \\ \\ f(x) = \frac{ {sin}^{2} (x)}{1 + cosx} \\ \\ f(x) = \frac{1 - {cos}^{2}(x) }{1 + cos(x)} \\ \\ f(x) = \frac{(1 + \cos(x ))(1 - cosx)}{1 + cosx} \\ \\ f(x) = 1 - cosx \\ \\ now \: \: integrate \: \: this \: \: function = > \\ \\ integration \: = x - sinx + c$

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