What Is The Integration Of 1/sinx+root3cosx Dx?
Answer:
Find Deravative of function (sin x + root (3 cos x))
Differentiate(sin x + root (3 cos x))= cos x- (root(3).sin x)/(2.root(cos x))
=(2.(cox x)^(3/2)- root(3).sin x)/(2.root(cos x))
Multiply and divide with derivative
((2.root(cos x))/(2.(cox x)^(3/2)- root(3).sin x)) integration((1/sinx+root(3cosx))(2.(cox x)^(3/2)- root(3).sin x)/(2.root(cos x)))
Now as Function in denominator has derivative along beside it, therefore by Integrating we get:
= ((2.root(cos x))/(2.(cox x)^(3/2)- root(3).sin x)) Ln (sinx+root(3.cosx))
the above is the desired integral. Here Ln is natural log
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