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inegration of (sinx + cosx)/3+sin2x dx......................................

inegration of (sinx + cosx)/3+sin2x dx......................................
 

Grade:12

2 Answers

Tech Xposed
37 Points
5 years ago
The answer is:-
  1. bring the denominator in (sin x – cosx)^2 format
  2. then take sin x- cos x= t
  3. therfore sin x + cos x dx = dt
  4. then the question will be converted into dt/a^2 – t^2 format then we can apply the formula.
\int \frac{sinx+cosx}{3+sin2x} =\int \frac{sinx+cosx}{4-1+sin2x} =\int \frac{sinx+cosx}{4-(1-sin2x)} =\int \frac{sinx+cosx}{4-(sin^{2}x+cos^{2}x-sin2x)} =\int \frac{sinx+cosx}{4-(sinx-cosx)^{2}} let sin x- cos x =t => (cos x +sin x )dx = dt =\int \frac{dt}{4-t^2} = \frac{1}{2*2} ln\frac{2+t}{2-t} = \frac{1}{2*2} ln\frac{2+sinx-cosx}{2-sinx+cosx}
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

bring the denominator in (sin x – cosx)^2 format
then take sin x- cos x= t
therfore sin x + cos x dx = dt
then the question will be converted into dt/a^2 – t^2 format then we can apply the formula

Thanks and Regards


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