Find ∫(√(a^2 −tan^2 x)) dx (with a>0) (related to Q45187)


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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
	$I=∫(√(a^2 −tan^2 x)) dx let u=tan x du=(dx/(cos^2 x))=(1+tan^2 x)dx=(1+u^2 )dx I=∫((√(a^2 −u^2 ))/(1+u^2 ))du let u=a sin θ ⇒θ=sin^(−1) (u/a) du=a cos θ dθ I=a^2 ∫((cos^2 θ)/(1+a^2 sin^2 θ))dθ let v=tan θ dv=(dθ/(cos^2 θ))=(1+tan^2 θ)dθ=(v^2 +1)dθ I=a^2 ∫(dv/({(a^2 +1)v^2 +1}(v^2 +1))) (1/({(a^2 +1)v^2 +1}(v^2 +1)))=(A/((a^2 +1)v^2 +1))+(B/(v^2 +1)) A(v^2 +1)+B(a^2 +1)v^2 +B=1 {A+B(a^2 +1)}v^2 +(A+B)=1 ⇒A+B=1 ⇒A+B(a^2 +1)=0 ⇒B=−(1/a^2 ) ⇒A=((a^2 +1)/a^2 ) I=a^2 ∫[((a^2 +1)/(a^2 {(a^2 +1)v^2 +1}))−(1/(a^2 (v^2 +1)))]dv I=∫[((a^2 +1)/((a^2 +1)v^2 +1))−(1/(v^2 +1))]dv I=(a^2 +1)∫(1/((a^2 +1)v^2 +1))dv−∫(1/(v^2 +1))dv I=(a^2 +1)I_1 −I_2 I_1 =∫(1/((a^2 +1)v^2 +1))dv let p=(√(a^2 +1))v dp=(√(a^2 +1)) dv I_1 =(1/(√(a^2 +1)))∫(dp/(p^2 +1))=(1/(√(a^2 +1)))tan^(−1) p I_1 =(1/(√(a^2 +1))) tan^(−1) ((√(a^2 +1))v) I_1 =(1/(√(a^2 +1))) tan^(−1) ((√(a^2 +1)) tan θ) I_1 =(1/(√(a^2 +1))) tan^(−1) ((√(a^2 +1)) ((sin θ)/(√(1−sin^2 θ)))) I_1 =(1/(√(a^2 +1))) tan^(−1) ((√(a^2 +1)) ((u/a)/(√(1−(u^2 /a^2 ))))) I_1 =(1/(√(a^2 +1))) tan^(−1) ((√(a^2 +1)) (u/(√(a^2 −u^2 )))) ⇒I_1 =(1/(√(a^2 +1))) tan^(−1) ((((√(a^2 +1)) tan x)/(√(a^2 −tan^2 x)))) I_2 =∫(1/(v^2 +1))dv=tan^(−1) v=θ=sin^(−1) (u/a) ⇒I_2 =sin^(−1) (u/a)=sin^(−1) (((tan x)/a)) I=(a^2 +1)I_1 −I_2 ⇒I=((√(a^2 +1))) tan^(−1) ((((√(a^2 +1)) tan x)/(√(a^2 −tan^2 x))))−sin^(−1) (((tan x)/a))+C$
	$I = \int \sqrt{a^{2} - \tan^{2} x} d x$ $l e t u = \tan x$ $d u = \frac{d x}{\cos^{2} x} = (1 + \tan^{2} x) d x = (1 + u^{2}) d x$ $I = \int \frac{\sqrt{a^{2} - u^{2}}}{1 + u^{2}} d u$ $l e t u = a \sin θ \Rightarrow θ = \sin^{- 1} \frac{u}{a}$ $d u = a \cos θ d θ$ $I = a^{2} \int \frac{\cos^{2} θ}{1 + a^{2} \sin^{2} θ} d θ$ $l e t v = \tan θ$ $d v = \frac{d θ}{\cos^{2} θ} = (1 + \tan^{2} θ) d θ = (v^{2} + 1) d θ$ $I = a^{2} \int \frac{d v}{{(a^{2} + 1) v^{2} + 1} (v^{2} + 1)}$ $\frac{1}{{(a^{2} + 1) v^{2} + 1} (v^{2} + 1)} = \frac{A}{(a^{2} + 1) v^{2} + 1} + \frac{B}{v^{2} + 1}$ $A (v^{2} + 1) + B (a^{2} + 1) v^{2} + B = 1$ ${A + B (a^{2} + 1)} v^{2} + (A + B) = 1$ $\Rightarrow A + B = 1$ $\Rightarrow A + B (a^{2} + 1) = 0$ $\Rightarrow B = - \frac{1}{a^{2}}$ $\Rightarrow A = \frac{a^{2} + 1}{a^{2}}$ $I = a^{2} \int [\frac{a^{2} + 1}{a^{2} {(a^{2} + 1) v^{2} + 1}} - \frac{1}{a^{2} (v^{2} + 1)}] d v$ $I = \int [\frac{a^{2} + 1}{(a^{2} + 1) v^{2} + 1} - \frac{1}{v^{2} + 1}] d v$ $I = (a^{2} + 1) \int \frac{1}{(a^{2} + 1) v^{2} + 1} d v - \int \frac{1}{v^{2} + 1} d v$ $I = (a^{2} + 1) I_{1} - I_{2}$ $I_{1} = \int \frac{1}{(a^{2} + 1) v^{2} + 1} d v$ $l e t p = \sqrt{a^{2} + 1} v$ $d p = \sqrt{a^{2} + 1} d v$ $I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \int \frac{d p}{p^{2} + 1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} p$ $I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} (\sqrt{a^{2} + 1} v)$ $I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} (\sqrt{a^{2} + 1} \tan θ)$ $I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} (\sqrt{a^{2} + 1} \frac{\sin θ}{\sqrt{1 - \sin^{2} θ}})$ $I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} (\sqrt{a^{2} + 1} \frac{\frac{u}{a}}{\sqrt{1 - \frac{u^{2}}{a^{2}}}})$ $I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} (\sqrt{a^{2} + 1} \frac{u}{\sqrt{a^{2} - u^{2}}})$ $\Rightarrow I_{1} = \frac{1}{\sqrt{a^{2} + 1}} \tan^{- 1} (\frac{\sqrt{a^{2} + 1} \tan x}{\sqrt{a^{2} - \tan^{2} x}})$ $I_{2} = \int \frac{1}{v^{2} + 1} d v = \tan^{- 1} v = θ = \sin^{- 1} \frac{u}{a}$ $\Rightarrow I_{2} = \sin^{- 1} \frac{u}{a} = \sin^{- 1} (\frac{\tan x}{a})$ $I = (a^{2} + 1) I_{1} - I_{2}$ $\Rightarrow I = (\sqrt{a^{2} + 1}) \tan^{- 1} (\frac{\sqrt{a^{2} + 1} \tan x}{\sqrt{a^{2} - \tan^{2} x}}) - \sin^{- 1} (\frac{\tan x}{a}) + C$
	Commented bymalwaan last updated on 10/Oct/18

	$thank you$
	Commented byajfour 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

	$l e t \tan x = a \sin ψ$ $\Rightarrow d x = \frac{a \cos ψ d ψ}{1 + a^{2} \sin^{2} ψ}$ $I = \int \frac{a^{2} \cos^{2} ψ}{1 + a^{2} \sin^{2} ψ} d ψ$ $l e t \tan ψ = t \Rightarrow d ψ = \frac{d t}{1 + t^{2}}$ $I = \int \frac{a^{2} (\frac{1}{1 + t^{2}}) \frac{d t}{(1 + t^{2})}}{1 + \frac{a^{2}}{(1 + \frac{1}{t^{2}})}}$ $= \int \frac{a^{2} d t}{{(1 + t^{2})}^{2} + a^{2} t^{2} (1 + t^{2})}$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $= \int \frac{a^{2} d t}{(a^{2} + 1) t^{4} + (a^{2} + 2) t^{2} + 1}$ $= \frac{a^{2}}{\sqrt{a^{2} + 1}} \int \frac{(1 / t^{2})}{\sqrt{a^{2} + 1} t^{2} + \frac{1}{\sqrt{a^{2} + 1} t^{2}} + \frac{a^{2} + 2}{\sqrt{a^{2} + 1}}} d t$ $l e t t {(a^{2} + 1)}^{1 / 4} = b t = z$ $\Rightarrow b d t = d z$ $\Rightarrow I = \frac{b^{4} - 1}{b^{2}} \int \frac{\frac{b^{2}}{z^{2}} (\frac{d z}{b})}{{(z + \frac{1}{z})}^{2} + \frac{b^{4} + 1}{b^{2}} - 2}$ $= \frac{b^{4} - 1}{2 b} \int \frac{(1 + \frac{1}{z^{2}}) - (1 - \frac{1}{z^{2}})}{{(z + \frac{1}{z})}^{2} + {(\frac{b^{2} - 1}{b})}^{2}} d z$ $= \frac{b^{4} - 1}{2 b} \int \frac{(1 + \frac{1}{z^{2}}) d z}{{(z - \frac{1}{z})}^{2} + {(\frac{b^{2} + 1}{b})}^{2}}$ $- \frac{b^{4} - 1}{2 b} \int \frac{(1 - \frac{1}{z^{2}}) d z}{{(z + \frac{1}{z})}^{2} + {(\frac{b^{2} - 1}{b})}^{2}}$ $I = \frac{b^{4} - 1}{2 b} [\frac{b}{b^{2} + 1} \tan^{- 1} (\frac{z - \frac{1}{z}}{\frac{b^{2} + 1}{b}}) - \frac{b}{b^{2} - 1} \tan^{- 1} (\frac{z + \frac{1}{z}}{\frac{b^{2} - 1}{b}})] + c$ $A n d s i n c e b t = z, t = \tan ψ$ $a \sin ψ = \tan x$ $\Rightarrow \tan ψ = z / b$ $\frac{a z}{\sqrt{z^{2} + b^{2}}} = \tan x$ $∴ z^{2} = \frac{b^{2} \tan^{2} x}{a^{2} - \tan^{2} x}$ $\Rightarrow z = \frac{{(a^{2} + 1)}^{1 / 4} \tan x}{\sqrt{a^{2} - \tan^{2} x}}; b = {(a^{2} + 1)}^{1 / 4}$ $I = \frac{b^{4} - 1}{2 b} [\frac{b}{b^{2} + 1} \tan^{- 1} (\frac{z - \frac{1}{z}}{\frac{b^{2} + 1}{b}}) - \frac{b}{b^{2} - 1} \tan^{- 1} (\frac{z + \frac{1}{z}}{\frac{b^{2} - 1}{b}})] + c .$ $I = \frac{b^{2} - 1}{2} \tan^{- 1} \frac{b (z^{2} - 1)}{z (b^{2} + 1)}$ $- \frac{b^{2} + 1}{2} \tan^{- 1} \frac{b (z^{2} + 1)}{z (b^{2} - 1)} .$
	Commented byMrW3 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 bytanmay.chaudhury50@gmail.com last updated on 11/Oct/18

	$p l s c h e c k$
	Commented bytanmay.chaudhury50@gmail.com last updated on 11/Oct/18

	$m o s t w e l c o m e s i r . . .$
	Commented byMrW3 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 bytanmay.chaudhury50@gmail.com last updated on 11/Oct/18

	$s i r g i v e m e t i m e t o g o t h r o u g h t h e p r o b l e m o f$ $p h y s i c s b o t h o f y o u s i r s o l v e d . . .$
	Commented byMrW3 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 byMrW3 last updated on 11/Oct/18

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