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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 (\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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