To mr mjs sir. integral lover ∫_0 ^(π/4) ((tan ((π/4)−x))/(cos^2 x (√(tan^3 x+tan^2 x+tan x)))) dx =?


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Question Number 123866 by benjo_mathlover last updated on 28/Nov/20

$T o m r m j s s i r .$ $i n t e g r a l l o v e r$ $\underset{0}{\int^{π / 4}} \frac{\tan (\frac{π}{4} - x)}{\cos^{2} x \sqrt{\tan^{3} x + \tan^{2} x + \tan x}} d x = ?$
	Answered by liberty last updated on 28/Nov/20
	$tan ((π/4)−x)=((1−tan x)/(1+tan x)) J = ∫_0 ^(π/4) (((1−tan x)/(1+tan x))/(cos^2 x (√(tan^3 x+tan^2 x+tan x)))) dx let tan x = h ⇒dx = (dh/(cos^2 x)) → { ((h=1)),((h=0)) :} J=∫_0 ^1 (((((1−h)/(1+h))))/( (√(h^3 +h^2 +h)))) dh = ∫_( 0) ^( 1) ((1−h)/((1+h)(√h) (√(h^2 +h+1)))) dh let h = r^2 , give J= 2∫((1−r^2 )/((1+r^2 )(√(r^4 +r^2 +1)))) dr J=2∫_( 0) ^( 1) ((−r^2 (1−r^(−2) ))/(r^2 (r^(−1) +r)(√(r^2 +1+r^(−2) )))) dr next let r+r^(−1) = t → { ((t=2)),((t=∞)) :} J=−2∫_( 0) ^( 2) (dt/(t(√(t^2 −1)))) = 2( arcsec t) ∣_2 ^∞ J= 2((π/2)−(π/3)) = (π/3).$
	$\tan (\frac{π}{4} - x) = \frac{1 - \tan x}{1 + \tan x}$ $J = \underset{0}{\int^{π / 4}} \frac{\frac{1 - \tan x}{1 + \tan x}}{\cos^{2} x \sqrt{\tan^{3} x + \tan^{2} x + \tan x}} d x$ $l e t \tan x = h \Rightarrow d x = \frac{d h}{\cos^{2} x} \to {\begin{cases} h = 1 \\ h = 0 \end{cases}$ $J = \int_{0}^{1} \frac{(\frac{1 - h}{1 + h})}{\sqrt{h^{3} + h^{2} + h}} d h = \int_{0}^{1} \frac{1 - h}{(1 + h) \sqrt{h} \sqrt{h^{2} + h + 1}} d h$ $l e t h = r^{2}, g i v e J = 2 \int \frac{1 - r^{2}}{(1 + r^{2}) \sqrt{r^{4} + r^{2} + 1}} d r$ $J = 2 \int_{0}^{1} \frac{- r^{2} (1 - r^{- 2})}{r^{2} (r^{- 1} + r) \sqrt{r^{2} + 1 + r^{- 2}}} d r$ $n e x t l e t r + r^{- 1} = t \to {\begin{cases} t = 2 \\ t = \infty \end{cases}$ $J = - 2 \int_{0}^{2} \frac{d t}{t \sqrt{t^{2} - 1}} = 2 (arcsec t) ∣_{2}^{\infty}$ $J = 2 (\frac{π}{2} - \frac{π}{3}) = \frac{π}{3} .$
	Answered by MJS_new last updated on 29/Nov/20

	$\tan (\frac{π}{4} - x) = \frac{1 - \tan x}{1 + \tan x}$ $\cos^{2} x = \frac{1}{1 + \tan^{2} x}$ $let me write τ for \tan x$ $now we have$ $- \underset{0}{\int^{π / 4}} \frac{(τ - 1) (τ^{2} + 1)}{(τ + 1) \sqrt{τ (τ^{2} + τ + 1)}} d x =$ $= [t = \frac{\sqrt{τ}}{\sqrt{τ^{2} + τ + 1}} \to d x = - \frac{2 {(τ^{2} + τ + 1)}^{3 / 2} \sqrt{τ}}{(τ^{2} - 1) (τ^{2} + 1)} d t]$ $= 2 \underset{0}{\int^{\sqrt{3} / 3}} \frac{τ^{2} + τ + 1}{{(τ^{2} + 1)}^{2}} d t =$ $[τ = \tan x = - \frac{t^{2} - 1 - \sqrt{- 3 t^{4} - 2 t^{2} + 1}}{2 t^{2}}]$ $= 2 \underset{0}{\int^{\sqrt{3} / 3}} \frac{d t}{t^{2} + 1} = 2 \arctan t = \frac{π}{3}$ $[2 \arctan t = \arccos \frac{1 - t^{2}}{1 + t^{2}} = \arccos \frac{1}{1 + \sin 2 x}]$
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