0-pi-sec-2-x-1-tan-2-x-dx-

Question Number 106964 by Ar Brandon last updated on 08/Aug/20

\int_{0}^{π} \frac{\sec^{2} x}{\sqrt{1 - \tan^{2} x}} dx

Commented by Her_Majesty last updated on 08/Aug/20

I b e l i e v e t h i s i n t e g r a l i s d i v e r g e n t

m u s t s l e e p n o w t o k e e p m y m a j e s t i c b e a u t y

w i l l t r y t o s h o w l a t e r

Commented by Dwaipayan Shikari last updated on 08/Aug/20

probably

Commented by Dwaipayan Shikari last updated on 08/Aug/20

Commented by Ar Brandon last updated on 08/Aug/20

��Alright her majesty ��

Answered by bemath last updated on 08/Aug/20

^{@ b e m a t h @}

l e t u = \tan x \to {\begin{cases} d u = \sec^{2} x d x \\ \to {\begin{cases} x = 0 \to u = 0 \\ x = π \to u = 0 \end{cases} \end{cases}

I = \underset{0}{\int^{0}} \frac{d u}{\sqrt{1 - u^{2}}} = 0

Answered by 1549442205PVT last updated on 08/Aug/20

I = \int_{0}^{π} \frac{d (tanx)}{\sqrt{1 - \tan^{2} x}} \underset{tanx = u}{=} \int_{0}^{1} \frac{du}{\sqrt{1 - u^{2}}} = \sin^{- 1} u ∣_{0}^{0}

= 0 - 0 = 0

Answered by Dwaipayan Shikari last updated on 08/Aug/20

\int_{0}^{π} \frac{d t}{\sqrt{1 - t^{2}}} t a n x = t, {s e c}^{2} x = \frac{d t}{d x}

{[{s i n}^{- 1} (t a n x)]}_{0}^{π} = 0

Commented by Ar Brandon last updated on 08/Aug/20

Thank y^{'} all . I really needed to verify something .

And {you}^{'} ve been of great help .

Answered by Ar Brandon last updated on 08/Aug/20

I = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{\sqrt{1 - \tan^{2} x}} dx + \int_{\frac{π}{2}}^{π} \frac{\sec^{2} x}{\sqrt{1 - \tan^{2} x}} dx

= \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{\sqrt{1 - \tan^{2} x}} dx - \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{\sqrt{1 - \tan^{2} x}} dx = 0

Answered by Her_Majesty last updated on 08/Aug/20

y = \frac{{s e c}^{2} x}{\sqrt{1 - {t a n}^{2} x}} = \frac{1}{∣ c o s x ∣ \sqrt{c o s 2 x}}

f o r 0 ⩽ x ⩽ π y i s d e f i n e d f o r 0 ⩽ x < \frac{π}{4} a n d

\frac{3 π}{4} < y ⩽ π \Rightarrow t h e i n t e g r a l o n l y e x i s t s i f t h e

i m a g i n a r y / c o m p l e x p a r t s c a n c e l o u t

b u t {l e t}^{'} s t r y

\underset{0}{\int^{π}} \frac{d x}{∣ c o s x ∣ \sqrt{c o s 2 x}} = 2 \underset{0}{\int^{π / 2}} \frac{d x}{c o s x \sqrt{c o s 2 x}}

w i t h t = t a n x w e g e t

\underset{0}{\int^{π / 2}} \frac{d x}{c o s x \sqrt{c o s 2 x}} = \underset{0}{\int^{+ \infty}} \frac{d t}{\sqrt{1 - t^{2}}} = {[a r c s i n (t)]}_{0}^{+ \infty} =

= \underset{t \to + \infty}{l i m a r c s i n} (t) w h i c h i s \frac{π}{2} - i \infty (t r y s o m e

v a l u e s l i k e t = 10^{10}, 10^{20} \dots

\Rightarrow t h e i n t e g r a l i s d i v e r g e n t

Commented by Ar Brandon last updated on 08/Aug/20

OK Sir, thanks