Let I = ∫ d x 1 + 3 s i n 2 x = 1 2 t a n - 1 2 f x + C (where, C is the constant of integration). If f π 4 = 1 , then the fundamental period of y = f x is

Step-by-step solution

Dividing the numerator and denominator by c o s 2 x , we get I = ∫ s e c 2 x s e c 2 x + 3 t a n 2 x d x = ∫ s e c 2 x 1 + 4 t a n 2 x d x Let, tan ⁡ x = t ⇒ s e c 2 x d x = d t ∴ I = ∫ d t 1 + 4 t 2 = 1 2 t a n - 1 2 t + C I = 1 2 t a n - 1 2 tan ⁡ x + C ⇒ f x = tan ⁡ x Hence, the fundamental period is ‘ π ’