calculate-0-pi-2-dt-1-a-cos-2-t-

Question Number 32712 by caravan msup abdo. last updated on 31/Mar/18

c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d t}{1 + a {c o s}^{2} t} .

Commented by abdo imad last updated on 03/Apr/18

l e t p u t F (a) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + a {c o s}^{2} t}

F (a) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + a \frac{1 + c o s (2 t)}{2}} = \int_{0}^{\frac{π}{2}} \frac{2 d t}{2 + a + a c o s (2 t)}

=_{2 t = u} \int_{0}^{π} \frac{2}{2 + a + a c o s (u)} \frac{d u}{2}

= \int_{0}^{π} \frac{d u}{2 + a + a c o s u} b u t t h e c h . t a n (\frac{u}{2}) = x g i v e

F (a) = \int_{0}^{+ \infty} \frac{1}{2 + a + a \frac{1 - x^{2}}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}}

= \int_{0}^{\infty} \frac{2 d x}{(2 + a) (1 + x^{2}) + a (1 - x^{2})}

= \int_{0}^{\infty} \frac{2 d x}{2 + a + (2 + a) x^{2} + a - {a x}^{2}}

= \int_{0}^{\infty} \frac{2 d x}{2 + 2 a + 2 x^{2}} = \int_{0}^{\infty} \frac{d x}{1 + a + x^{2}}

c a s e 1 1 + a > 0 \Rightarrow F (a) =_{x = \sqrt{1 + a} u} \int_{0}^{\infty} \frac{\sqrt{1 + a} d u}{(1 + a) (1 + u^{2})}

= \frac{π}{2 \sqrt{1 + a}}

c a s e 2 1 + a < 0 \Rightarrow F (a) = \int_{0}^{\infty} \frac{d x}{x^{2} - {(\sqrt{- (1 + a)})}^{2}}

= \int_{0}^{\infty} \frac{d x}{(x - α) (x + α)} (α = \sqrt{- 1 - a})

= \frac{1}{2 α} \int_{0}^{\infty} (\frac{1}{x - α} - \frac{1}{x + α}) d x = \frac{1}{2 α} {[l n ∣ \frac{x - α}{x + α} ∣]}_{0}^{+ \infty} = 0