Question

integration of (√1+sinx)dx​

Answer 1

I = ∫√1-sinxdx

=∫√1-sinx X (1+sinx/1+sinx) dx

=∫(1^2 - sin^2x / 1+sinx). dx

=∫(cosx/√1+sinx)dx

let 1+sinx = t.

then cosx dx = dt

I=∫(1/√x)dt

= (t^1/2 / 1/2) + C

= 2√1+sinx + C

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