∫_( 0) ^( π/4) ((tan^2 x)/(1+sin x)) dx =?


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Question Number 187135 by cortano12 last updated on 14/Feb/23

$\underset{0}{\int^{π / 4}} \frac{\tan^{2} x}{1 + \sin x} d x = ?$
	Answered by horsebrand11 last updated on 14/Feb/23
	$I=∫_( 0) ^( π/4) ((tan^2 x)/(1+sin x)) dx Let tan ((x/2))=((1−t)/(1+t)) { ((tan x=((1−t^2 )/(2t)))),((dx=−(1+sin x)dt)) :} λ =∫ (((1−t^2 )/(2t)))^2 dt=∫(((t^4 −2t^2 +1)/(4t^2 )))dt λ= ∫ ((1/4)t^2 −(1/2)+(1/4)t^(−2) )dt λ=(1/(12))t^3 −(1/2)t−(1/(4t)) +c I= [ (1/(12))t^3 −(1/2)t−(1/(4t)) ]_((√2)−1) ^1 I=((1/(12))−(1/2)−(1/4))−(((((√2)−1)^3 )/(12))−((((√2)−1))/2)−(1/(4((√2)−1)))) I= (((√2)−1)/3)$
	$I = \underset{0}{\int^{π / 4}} \frac{\tan^{2} x}{1 + \sin x} d x$ $L e t \tan (\frac{x}{2}) = \frac{1 - t}{1 + t}$ ${\begin{cases} \tan x = \frac{1 - t^{2}}{2 t} \\ d x = - (1 + \sin x) d t \end{cases}$ $λ = \int {(\frac{1 - t^{2}}{2 t})}^{2} d t = \int (\frac{t^{4} - 2 t^{2} + 1}{4 t^{2}}) d t$ $λ = \int (\frac{1}{4} t^{2} - \frac{1}{2} + \frac{1}{4} t^{- 2}) d t$ $λ = \frac{1}{12} t^{3} - \frac{1}{2} t - \frac{1}{4 t} + c$ $I = {[\frac{1}{12} t^{3} - \frac{1}{2} t - \frac{1}{4 t}]}_{\sqrt{2} - 1}^{1}$ $I = (\frac{1}{12} - \frac{1}{2} - \frac{1}{4}) - (\frac{{(\sqrt{2} - 1)}^{3}}{12} - \frac{(\sqrt{2} - 1)}{2} - \frac{1}{4 (\sqrt{2} - 1)})$ $I = \frac{\sqrt{2} - 1}{3}$
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