Solve this: â ‹Q60 ∫sin 2xsin2x +2 cos2xdx = - Maths - Integrals

Question

Solve this:

â€‹Q60 ∫ sin   2 x sin 2 x   + 2
  cos 2 x d x   =

(1) log ( 1 + cos2 x ) + C (2) log ( 1 + tan2
x) + C

(3) – log (1 + sin2 x) + C (4) – log (1 +
cos2x ) + C

Lovina Kansal · Accepted Answer

Dear student
Let I=∫sin2xsin2x+2cos2xdx=∫sin2x1-cos2x+2cos2xdx=∫sin2xcos2x+1dxTake u=cos2x⇒du=-2cosx sinxdx⇒du=-2sin2x dxSo,I=-∫1udu=-logu+C=-logcos2x+1+C
Regards

Solve this: ​Q60. ∫ sin 2 x sin 2 x + 2 cos 2 x d x = (1) log ( 1 + cos2 x ) + C (2) log ( 1 + tan2 x) + C (3) – log (1 + sin2 x) + C (4) – log (1 + cos2x ) + C

Solve this:

Q60. $\int \frac{\sin 2 x}{\sin^{2} x + 2 \cos^{2} x} d x =$

(1) log ( 1 + cos² x ) + C (2) log ( 1 + tan² x) + C

(3) – log (1 + sin² x) + C (4) – log (1 + cos²x ) + C