calculate-0-pi-2-sin-x-dx-cos-2-x-a-2-sin-2-x-dx-

Question Number 40140 by maxmathsup by imad last updated on 16/Jul/18

c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{s i n (x) d x}{{c o s}^{2} x + a^{2} {s i n}^{2} x} d x

Commented by math khazana by abdo last updated on 20/Jul/18

l e t I = \int_{0}^{\frac{π}{2}} \frac{s i n x}{{c o s}^{2} x + a^{2} {s i n}^{2} x} d x c h a n g e m e n t

c o s x = t g i v e

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

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

(a \neq o)

c a s e 1 1 - a^{2} > 0 \Leftrightarrow∣ a ∣< 1 w e d o t h e c h a n g e m e n t

\sqrt{\frac{1 - a^{2}}{a^{2}}} t = u \Rightarrow

I = \frac{1}{a^{2}} \int_{0}^{\sqrt{\frac{1 - a^{2}}{a^{2}}}} \frac{1}{1 + u^{2}} \sqrt{\frac{a^{2}}{1 - a^{2}}} d u

= \frac{∣ a ∣}{a^{2} \sqrt{1 - a^{2}}} \int_{0}^{\frac{\sqrt{1 - a^{2}}}{∣ a ∣}} \frac{d u}{1 + u^{2}}

= \frac{ξ (a)}{a \sqrt{1 - a^{2}}} a r c t a n (\frac{\sqrt{1 - a^{2}}}{∣ a ∣})

ξ (a) = 1 i f a > 0 a n d ξ (a) = - 1 i f a < 0

c a s e 2 1 - a^{2} < 0 \Rightarrow∣ a ∣> 1 \Rightarrow

I = \frac{1}{a^{2}} \int_{0}^{1} \frac{d t}{1 - \frac{a^{2} - 1}{a^{2}} t^{2}} w e d o t h e c h a n g e m e n t

\sqrt{\frac{a^{2} - 1}{a^{2}}} t = u \Rightarrow

I = \frac{1}{a^{2}} \int_{0}^{\frac{\sqrt{a^{2} - 1}}{∣ a ∣}} \frac{1}{1 - u^{2}} \frac{∣ a ∣}{\sqrt{a^{2} - 1}} d u

= \frac{ξ (a)}{\sqrt{a^{2} - 1}} \frac{1}{2} \int_{0}^{\frac{\sqrt{a^{2} - 1}}{∣ a ∣}} (\frac{1}{1 + u} + \frac{1}{1 - u}) d u

= \frac{ξ (a)}{2 \sqrt{a^{2} - 1}} {[l n ∣ \frac{1 + u}{1 - u} ∣]}_{0}^{\frac{\sqrt{a^{2} - 1}}{∣ a ∣}}

= \frac{ξ (a)}{2 \sqrt{a^{2} - 1}} l n (\frac{1 + \frac{\sqrt{a^{2} - 1}}{∣ a ∣}}{1 - \frac{\sqrt{a^{2} - 1}}{∣ a ∣}})

= \frac{ξ (a)}{2 \sqrt{a^{2} - 1}} l n (\frac{∣ a ∣ + \sqrt{a^{2} - 1}}{∣ a ∣ - \sqrt{a^{2} - 1}}) .

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jul/18

t = c o s x d t = - s i n x d x

\int_{1}^{0} \frac{- d t}{t^{2} + a^{2} (1 - t^{2})}

\int_{0}^{1} \frac{d t}{a^{2} + t^{2} (1 - a^{2})}

= \frac{1}{1 - a^{2}} \int_{0}^{1} \frac{d t}{\frac{a^{2}}{1 - a^{2}} + t^{2}}

= \frac{1}{1 - a^{2}} \times \frac{\sqrt{1 - a^{2}}}{a} ∣ {t a n}^{- 1} (\frac{t}{\sqrt{\frac{a^{2}}{1 - a^{2}}}}) ∣_{0}^{1}

= \frac{\sqrt{1 - a^{2}}}{(a - a^{3})} ∣ {t a n}^{- 1} (\frac{t \sqrt{1 - a^{2}}}{a}) ∣_{0}^{1}

= \frac{\sqrt{1 - a^{2}}}{a - a^{3}} {t a n}^{- 1} (\frac{\sqrt{1 - a^{2}}}{a})

Commented by math khazana by abdo last updated on 20/Jul/18

s i r T a n m a y y o u m u s t s t u d y t h e c a s e s ∣ a ∣> 1

a n d ∣ a ∣< 1 \dots .