Question

integrate I=
1/ 1+sinx​

Answer 1

Answer:

$\(\int \frac{1}{1 + sin x}dx\) =\(\tan x- \sec x . dx +c\)$

Step-by-step explanation:

$\frac{1}{(1 +sin x)} \\ \text {Multiplying numerator and denominator by (1 - sin x) } \\we get\\\(\int \frac{1}{1 + sin x}dx\) \\\\= \(\int \frac{1-sin x}{1-sin^{2}x} dx\) \\from trigonometric identity we know that sin^2x + cos^2x = 1 \\cos^2x = 1 - sin^2x \\= \(\int \frac{1-sin x}{cos^{2}x} dx\) \\=\(\int \sec ^{2}x - \tan x \sec xdx\) \\=\(\tan x- \sec x . dx +c\)$

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