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

Tech Xposed
37 Points
4 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
2 years ago
Dear Student,

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