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evaluate the following integral∫ sin 2x / a cos^2 (x) + b sin^2 (x) dx

taniska , 10 Years ago
Grade 12
anser 2 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans:
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

Ishu

Last Activity: 6 Years ago

Put a cos²x+bsin²x=t                                                                 so {a.2cosx(-sinx)+b.2sinxcosx}dx=dt                                      (b-a)sin2xdx=dt               i.e. sin 2xdx=dt/b-a                   int.sin2x/a cos²x+bsin²x dx        =int.1/t.dt/b-a               =1/b-aint.1/t dt              = 1/b-a log| a cos²x+b sin²x|+ c

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