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

Question Number 86313 by john santu last updated on 28/Mar/20

\underset{0}{\int^{\frac{π}{2}}} \frac{d x}{\sqrt{1 + \tan x}}

Commented by john santu last updated on 28/Mar/20

\frac{\sqrt{\sin x}}{\sqrt{\sin x + \cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x + \cos x}} = 1

h o w ?

Commented by john santu last updated on 28/Mar/20

\sqrt{\cos x} + \sqrt{\sin x} = \sqrt{\sin x + \cos x} ?

Answered by TANMAY PANACEA. last updated on 28/Mar/20

i t h i n k \dots

\int_{0}^{\frac{π}{2}} \frac{d x}{1 + \frac{\sqrt{s i n x}}{\sqrt{c o s x}}} = \int_{0}^{\frac{π}{2}} \frac{\sqrt{c o s x}}{\sqrt{s i n x} + \sqrt{c o s x}} d x

u s i n g \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x

I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{c o s (\frac{π}{2} - x)}}{\sqrt{s i n (\frac{π}{2} - x)} + \sqrt{c o s (\frac{π}{2} - x)}} d x

= \int_{0}^{\frac{π}{2}} \frac{\sqrt{s i n x}}{\sqrt{c o s x} + \sqrt{s i n x}} d x

2 I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{c o s x}}{\sqrt{s i n x} + \sqrt{c o s x}} + \frac{\sqrt{s i n x}}{\sqrt{c o s x} + \sqrt{s i n x}} d x

2 I = \int_{0}^{\frac{π}{2}} d x

I = \frac{π}{2} \times \frac{1}{2} = \frac{π}{4}

Commented by jagoll last updated on 28/Mar/20

king formula

Commented by TANMAY PANACEA. last updated on 28/Mar/20

t h a n k y o u s i r

Commented by john santu last updated on 28/Mar/20

it does mean the equation wrong sir ?

Commented by TawaTawa1 last updated on 28/Mar/20

Sir tanmay, please help with Q 86324 .

Answered by MJS last updated on 28/Mar/20

\int \frac{d x}{\sqrt{1 + \tan x}} =

[t = \sqrt{1 + \tan x} \to d x = {2 \cos}^{2} x \sqrt{1 + \tan x} d t]

= 2 \int \frac{d t}{t^{4} - 2 t^{2} + 2}

and now {it}^{'} s easy

Commented by jagoll last updated on 28/Mar/20

= 2 \int \frac{dt}{{(t^{2} - 1)}^{2} + 1}

let t^{2} - 1 = \tan w \Rightarrow 2 t dt = \sec^{2} w dw

= \int \frac{\sec^{2} w dw}{\sec^{2} w} = w + c

= \tan^{- 1} (t^{2} - 1) + c

= \tan^{- 1} (\tan (x)) + c

it correct sir ?

Commented by MJS last updated on 28/Mar/20

no

w = \arctan (t^{2} - 1) \to d t = \frac{t^{4} - 2 t^{3} + 2}{2 t} d w

\Rightarrow \int \frac{d w}{\sqrt{1 + \tan w}} so {you}^{'} re back at the start

Commented by MJS last updated on 28/Mar/20

we must decompose

2 \int \frac{d t}{t^{4} - 2 t^{2} + 2} =

= \int \frac{a t + b}{t^{2} - \sqrt{2 + 2 \sqrt{2}} t + \sqrt{2}} d t + \int \frac{c t + d}{t^{2} + \sqrt{2 + 2 \sqrt{2} t + \sqrt{2}}} d t =

= - \frac{\sqrt{- 1 + \sqrt{2}}}{2} \int \frac{t - \sqrt{2 + 2 \sqrt{2}}}{t^{2} - \sqrt{2 + 2 \sqrt{2}} t + \sqrt{2}} d t + \frac{\sqrt{- 1 + \sqrt{2}}}{2} \int \frac{t + \sqrt{2 + 2 \sqrt{2}}}{t^{2} + \sqrt{2 + \sqrt{2}} t + \sqrt{2}} d t

now use formula

Commented by jagoll last updated on 28/Mar/20

waw \dots yes sir . i try solve it

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