0-pi-dt-1-sina-cost-a-0-pi-2-

Question Number 165741 by metamorfose last updated on 07/Feb/22

\int_{0}^{π} \frac{d t}{1 - s i n a . c o s t} = ? ? ?, a \in] 0, \frac{π}{2} [

Answered by MJS_new last updated on 08/Feb/22

a \in] 0; \frac{π}{2} [\Rightarrow 0 < \sin a < 1 \Rightarrow let \sin a = A; 0 < A < 1

\int \frac{d t}{1 - A \cos t} =

[u = \tan \frac{t}{2} \Rightarrow d t = \frac{2 d u}{u^{2} + 1}]

= 2 \int \frac{d u}{(1 + A) u^{2} + (1 - A)} =

[let B = 1 + A \land C = 1 - A \Rightarrow B, C > 0]

= 2 \int \frac{d u}{{B u}^{2} + C} =

[u = \frac{\sqrt{C}}{\sqrt{B}} v \to d u = \frac{\sqrt{C}}{\sqrt{B}} d v]

= \frac{2}{\sqrt{B C}} \int \frac{d v}{v^{2} + 1} = \frac{2}{\sqrt{B C}} \arctan v =

= \frac{2}{\sqrt{1 - A^{2}}} \arctan \frac{\sqrt{1 + A} u}{\sqrt{1 - A}} =

= \frac{2}{\cos a} \arctan \frac{\sqrt{1 + \sin a} \tan \frac{t}{2}}{\sqrt{1 - \sin a}} + C

\Rightarrow

answer is \lim_{t \to π^{-}} (\frac{2}{\cos a} \arctan \frac{\sqrt{1 + \sin a} \tan \frac{t}{2}}{\sqrt{1 - \sin a}}) = \frac{π}{\cos a}

Commented by metamorfose last updated on 09/Feb/22

t h n x s i r

Answered by Mathspace last updated on 08/Feb/22

l e t s i n a = α \Rightarrow

I = \int_{0}^{π} \frac{d t}{1 - α c o s t}

=_{t a n (\frac{t}{2}) = y} \int_{0}^{\infty} \frac{2 d y}{(1 + y^{2}) (1 - α \frac{1 - y^{2}}{1 + y^{2}})}

= 2 \int_{0}^{\infty} \frac{d y}{1 + y^{2} - α + α y^{2}}

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

= \frac{2}{1 - α} \int_{0}^{\infty} \frac{d y}{1 + \frac{1 + α}{1 - α} y^{2}}

=_{z = \sqrt{\frac{1 + α}{1 - α}} y} \frac{2}{1 - α} \int_{0}^{\infty} \frac{d z}{\sqrt{\frac{1 + α}{1 - α}} (1 + z^{2})}

= \frac{2}{1 - α} \frac{\sqrt{1 - α}}{\sqrt{1 + α}} \times \frac{π}{2}

= \frac{π}{\sqrt{1 - α^{2}}} = \frac{π}{\sqrt{1 - {s i n}^{2} α}} = \frac{π}{∣ c o s α ∣}

o r 0 < α < \frac{π}{2} \Rightarrow I = \frac{π}{c o s α}

Commented by metamorfose last updated on 09/Feb/22

t h n x s i r

Commented by Mathspace last updated on 08/Feb/22

I = \frac{π}{c o s a}