how to do 8,10 problems
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Hello student,
Please find the answer to your question below
I = \int \frac{sin2x}{acos^{2}x+bsin^{2}x}dx
I = \int \frac{sin2x}{a(1-sin^{2}x)+bsin^{2}x}dx
I = \int \frac{sin2x}{a+(b-a)sin^{2}x}dx
a+(b-a)sin^{2}x = t
(b-a)2sinxcosxdx = dt
(b-a)sin2x = dt
I = \int \frac{dt}{(b-a)t}
I = \frac{log(t)}{(b-a)} + c
I = \frac{log(a+(b-a)sin^{2}x)}{(b-a)} + c