Question

Integration of cos^2x / 1+sinx dx=

Answer 1

$\boxed{\int \frac{\cos^2x}{1+\sin x} dx=x+\cos x+c}$

Step-by-step explanation:

$\int \frac{\cos^2x}{1+\sin x} dx$

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

$=\int\frac{(1-\sin x)(1+\sin x)}{(1+\sin x)}dx  (∵ [tex]a^2-b^2=(a+b)(a-b)$)

$=\int (1-\sin x)dx$

$=\int dx-\int sin x dx$

$=x-(-\cos x)+c$  (where, c is a constant)

$=x+\cos x+c$

Hope this answer is helpful,

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Answer 2

INTEGRATION

Q. To integrate cos²x/1+sin x

$\int\limits {\frac{cos^{2} x }{1+sinx} } \, dx \\\\\int\limits {\frac{1-sin^{2}x }{1+sinx} } \, dx \\\\\int\limits{\frac{(1+sinx)(1-sinx)}{1+sinx} } \, dx \\\\\int\limits {(1-sinx)} \, dx \\\\\int\limits {1.} \, dx - \int\limits {sinx.} \, dx \\\\x - (- cosx) + C\\\\ \boxed { \int\limits {\frac{cos^{2} x }{1+sinx} } = x + cosx + C }$

→ It is necessary to add Integration Constant "C" in order to cover the family of function as it is an indefinite integration. There is no need to add "C" if the integration is definite.