Find-a-2-tan-2-x-dx-with-a-gt-0-related-to-Q45187-

Question Number 45213 by MrW3 last updated on 10/Oct/18

F i n d \int \sqrt{a^{2} - \tan^{2} x} d x (w i t h a > 0)

(r e l a t e d t o Q 45187)

Answered by MrW3 last updated on 10/Oct/18

I = \int \sqrt{a^{2} - \tan^{2} x} d x

l e t u = \tan x

d u = \frac{d x}{\cos^{2} x} = (1 + \tan^{2} x) d x = (1 + u^{2}) d x

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

l e t u = a \sin θ \Rightarrow θ = \sin^{- 1} \frac{u}{a}

d u = a \cos θ d θ

I = a^{2} \int \frac{\cos^{2} θ}{1 + a^{2} \sin^{2} θ} d θ

l e t v = \tan θ

d v = \frac{d θ}{\cos^{2} θ} = (1 + \tan^{2} θ) d θ = (v^{2} + 1) d θ

I = a^{2} \int \frac{d v}{{(a^{2} + 1) v^{2} + 1} (v^{2} + 1)}

\frac{1}{{(a^{2} + 1) v^{2} + 1} (v^{2} + 1)} = \frac{A}{(a^{2} + 1) v^{2} + 1} + \frac{B}{v^{2} + 1}

A (v^{2} + 1) + B (a^{2} + 1) v^{2} + B = 1

{A + B (a^{2} + 1)} v^{2} + (A + B) = 1

\Rightarrow A + B = 1

\Rightarrow A + B (a^{2} + 1) = 0

\Rightarrow B = - \frac{1}{a^{2}}

\Rightarrow A = \frac{a^{2} + 1}{a^{2}}

I = a^{2} \int [\frac{a^{2} + 1}{a^{2} {(a^{2} + 1) v^{2} + 1}} - \frac{1}{a^{2} (v^{2} + 1)}] d v

I = \int [\frac{a^{2} + 1}{(a^{2} + 1) v^{2} + 1} - \frac{1}{v^{2} + 1}] d v

I = (a^{2} + 1) \int \frac{1}{(a^{2} + 1) v^{2} + 1} d v - \int \frac{1}{v^{2} + 1} d v

I = (a^{2} + 1) I_{1} - I_{2}

I_{1} = \int \frac{1}{(a^{2} + 1) v^{2} + 1} d v

l e t p = \sqrt{a^{2} + 1} v

d p = \sqrt{a^{2} + 1} d v

I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \int \frac{d p}{p^{2} + 1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} p

I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} (\sqrt{a^{2} + 1} v)

I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} (\sqrt{a^{2} + 1} \tan θ)

I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} (\sqrt{a^{2} + 1} \frac{\sin θ}{\sqrt{1 - \sin^{2} θ}})

I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} (\sqrt{a^{2} + 1} \frac{\frac{u}{a}}{\sqrt{1 - \frac{u^{2}}{a^{2}}}})

I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} (\sqrt{a^{2} + 1} \frac{u}{\sqrt{a^{2} - u^{2}}})

\Rightarrow I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} (\frac{\sqrt{a^{2} + 1} \tan x}{\sqrt{a^{2} - \tan^{2} x}})

I_{2} = \int \frac{1}{v^{2} + 1} d v = \tan^{- 1} v = θ = \sin^{- 1} \frac{u}{a}

\Rightarrow I_{2} = \sin^{- 1} \frac{u}{a} = \sin^{- 1} (\frac{\tan x}{a})

I = (a^{2} + 1) I_{1} - I_{2}

\Rightarrow I = (\sqrt{a^{2} + 1}) \tan^{- 1} (\frac{\sqrt{a^{2} + 1} \tan x}{\sqrt{a^{2} - \tan^{2} x}}) - \sin^{- 1} (\frac{\tan x}{a}) + C

Commented by malwaan last updated on 10/Oct/18

thank you

Commented by ajfour last updated on 10/Oct/18

T h a n k s a l o t S i r!

Answered by ajfour last updated on 10/Oct/18

l e t \tan x = a \sin ψ

\Rightarrow d x = \frac{a \cos ψ d ψ}{1 + a^{2} \sin^{2} ψ}

I = \int \frac{a^{2} \cos^{2} ψ}{1 + a^{2} \sin^{2} ψ} d ψ

l e t \tan ψ = t \Rightarrow d ψ = \frac{d t}{1 + t^{2}}

I = \int \frac{a^{2} (\frac{1}{1 + t^{2}}) \frac{d t}{(1 + t^{2})}}{1 + \frac{a^{2}}{(1 + \frac{1}{t^{2}})}}

= \int \frac{a^{2} d t}{{(1 + t^{2})}^{2} + a^{2} t^{2} (1 + t^{2})}

\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots . .

= \int \frac{a^{2} d t}{(a^{2} + 1) t^{4} + (a^{2} + 2) t^{2} + 1}

= \frac{a^{2}}{\sqrt{a^{2} + 1}} \int \frac{(1 / t^{2})}{\sqrt{a^{2} + 1} t^{2} + \frac{1}{\sqrt{a^{2} + 1} t^{2}} + \frac{a^{2} + 2}{\sqrt{a^{2} + 1}}} d t

l e t t {(a^{2} + 1)}^{1 / 4} = b t = z

\Rightarrow b d t = d z

\Rightarrow I = \frac{b^{4} - 1}{b^{2}} \int \frac{\frac{b^{2}}{z^{2}} (\frac{d z}{b})}{{(z + \frac{1}{z})}^{2} + \frac{b^{4} + 1}{b^{2}} - 2}

= \frac{b^{4} - 1}{2 b} \int \frac{(1 + \frac{1}{z^{2}}) - (1 - \frac{1}{z^{2}})}{{(z + \frac{1}{z})}^{2} + {(\frac{b^{2} - 1}{b})}^{2}} d z

= \frac{b^{4} - 1}{2 b} \int \frac{(1 + \frac{1}{z^{2}}) d z}{{(z - \frac{1}{z})}^{2} + {(\frac{b^{2} + 1}{b})}^{2}}

- \frac{b^{4} - 1}{2 b} \int \frac{(1 - \frac{1}{z^{2}}) d z}{{(z + \frac{1}{z})}^{2} + {(\frac{b^{2} - 1}{b})}^{2}}

I = \frac{b^{4} - 1}{2 b} [\frac{b}{b^{2} + 1} \tan^{- 1} (\frac{z - \frac{1}{z}}{\frac{b^{2} + 1}{b}}) - \frac{b}{b^{2} - 1} \tan^{- 1} (\frac{z + \frac{1}{z}}{\frac{b^{2} - 1}{b}})] + c

A n d s i n c e b t = z, t = \tan ψ

a \sin ψ = \tan x

\Rightarrow \tan ψ = z / b

\frac{a z}{\sqrt{z^{2} + b^{2}}} = \tan x

∴ z^{2} = \frac{b^{2} \tan^{2} x}{a^{2} - \tan^{2} x}

\Rightarrow z = \frac{{(a^{2} + 1)}^{1 / 4} \tan x}{\sqrt{a^{2} - \tan^{2} x}}; b = {(a^{2} + 1)}^{1 / 4}

I = \frac{b^{4} - 1}{2 b} [\frac{b}{b^{2} + 1} \tan^{- 1} (\frac{z - \frac{1}{z}}{\frac{b^{2} + 1}{b}}) - \frac{b}{b^{2} - 1} \tan^{- 1} (\frac{z + \frac{1}{z}}{\frac{b^{2} - 1}{b}})] + c .

I = \frac{b^{2} - 1}{2} \tan^{- 1} \frac{b (z^{2} - 1)}{z (b^{2} + 1)}

- \frac{b^{2} + 1}{2} \tan^{- 1} \frac{b (z^{2} + 1)}{z (b^{2} - 1)} .

Commented by MrW3 last updated on 10/Oct/18

t h a n k s s i r!

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

\int \frac{a^{2} - {t a n}^{2} x}{\sqrt{a^{2} - {t a n}^{2} x}} d x

a^{2} \int \frac{d x}{\sqrt{a^{2} - {t a n}^{2} x}} - \int \frac{{s e c}^{2} x - 1}{\sqrt{a^{2} - {t a n}^{2} x}} d x

(a^{2} + 1) \int \frac{d x}{\sqrt{a^{2} - {t a n}^{2} x}} - \int \frac{{s e c}^{2} x d x}{\sqrt{a^{2} - {t a n}^{2} x}}

I_{1} - I_{2}

I_{1} = (a^{2} + 1) \int \frac{c o s x d x}{\sqrt{a^{2} {c o s}^{2} x - {s i n}^{2} x}}

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

(a^{2} + 1) \int \frac{d t}{\sqrt{a^{2} (1 - t^{2}) - t^{2}}}

(a^{2} + 1) \int \frac{d t}{\sqrt{a^{2} - t^{2} (a^{2} + 1)}}

(a^{2} + 1) \int \frac{d t}{\sqrt{(a^{2} + 1) (\frac{a^{2}}{a^{2} + 1}) - t^{2}}}

\sqrt{a^{2} + 1} \int \frac{d t}{\sqrt{{(\frac{a^{2}}{a^{2} + 1})}^{} - t^{2}}}

\sqrt{a^{2} + 1} \times {s i n}^{- 1} (\frac{t}{\sqrt{\frac{a^{2}}{a^{2} + 1}}})

\sqrt{a^{2} + 1} \times {s i n}^{- 1} (\frac{s i n x}{\sqrt{\frac{a^{2}}{a^{2} + 1}}}) + c_{1}

I_{2} = \int \frac{{s e c}^{2} x d x}{\sqrt{a^{2} - {t a n}^{2} x}} d x

k = t a n x d k = {s e c}^{2} x d x

\int \frac{d k}{\sqrt{a^{2} - k^{2}}}

{s i n}^{- 1} (\frac{k}{a}) + c_{2}

{s i n}^{- 1} (\frac{t a n x}{a}) + c_{2}

\sqrt{a^{2} + 1} \times {s i n}^{- 1} (\frac{s i n x}{\sqrt{\frac{a^{2}}{a^{2} + 1}}}) - {s i n}^{- 1} (\frac{t a n x}{a}) + c

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

p l s c h e c k

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

m o s t w e l c o m e s i r \dots

Commented by MrW3 last updated on 11/Oct/18

s i r, y o u r a n s w e r i s c o r r e c t, i t c a n b e

t r a n s f o r m e d t o t h e s a m e f o r m a s m i n e .

t h e r e i s o n e s t r a n g e t h i n g :

b o t h i n t h e r e s u l t f r o m m e a n d f r o m

a j f o u r s i r w e h a v e a l i m i t a t i o n t h a t

i s \tan x \neq a, b u t i n y o u r r e s u l t t h e r e

i s n o l i m i t a t i o n . b u t f r o m t h e p o i n t o f v i e w

o f p h y s i c t h e r e s h o u l d b e t h i s l i m i t a t i o n .

c a n y o u e x p l a i n i t ?

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

s i r g i v e m e t i m e t o g o t h r o u g h t h e p r o b l e m o f

p h y s i c s b o t h o f y o u s i r s o l v e d \dots

Commented by MrW3 last updated on 11/Oct/18

t h a n k y o u s i r .

{i t}^{'} s a b o u t t h e q u e s t i o n i f x c a n b e a l l

v a l u e s f r o m 0 t o \frac{π}{2} .

Commented by MrW3 last updated on 11/Oct/18

t h a n k y o u s i r!

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