hello please. ∫(dx/(a^2 cos^2 x + b^2 sin^2 x))


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Question Number 130088 by stelor last updated on 22/Jan/21

$hello please .$ $\int \frac{d x}{a^{2} {c o s}^{2} x + b^{2} {s i n}^{2} x}$
	Answered by bobhans last updated on 22/Jan/21

	$I = \int \frac{\frac{d x}{\cos^{2} x}}{a^{2} + b^{2} \tan^{2} x} = \int \frac{\sec^{2} x d x}{a^{2} + b^{2} \tan^{2} x}$ $l e t \tan x = u \Rightarrow d u = \sec^{2} x d x$ $I = \int \frac{d u}{a^{2} + b^{2} u^{2}}; l e t b u = a \tan θ$ $I = \int \frac{\frac{a}{b} \sec^{2} θ d θ}{a^{2} \sec^{2} θ} = \frac{1}{a b} θ + c$ $I = \frac{1}{a b} \tan^{- 1} (\frac{b u}{a}) + c = \frac{1}{a b} \tan^{- 1} (\frac{b \tan x}{a}) + c$
	Commented by stelor last updated on 22/Jan/21

	$thank . . .$
	Answered by mathmax by abdo last updated on 22/Jan/21
	$I =∫ (dx/(a^2 (cos^2 x+(b^2 /a^2 )sin^2 x))) let α =(b^2 /a^2 ) (ab≠0and a≠b)) ⇒I=(1/a^2 )∫(dx/(cos^2 x+α^2 cos^2 x)) =(1/a^2 )∫ (dx/(((1+cos(2x))/2)+α^2 ×((1−cos(2x))/2))) =(2/a^2 )∫ (dx/(1+cos(2x)+α^2 −α^2 cos(2x))) =(2/a^2 )∫ (dx/(1+α^2 +(1−α^2 )cos(2x))) =_(2x=t) (1/a^2 )∫ (dt/((1+α^2 +(1−α^2 )cost))) =_(tan((t/2))=z) (1/a^2 )∫ ((2dz)/((1+z^2 ){1+α^2 +(1−α^2 )((1−z^2 )/(1+z^2 ))})) =(2/a^2 )∫ (dz/((1+α^2 )(1+z^2 )+(1−α^2 )(1−z^2 ))) =(2/a^2 )∫ (dz/(1+α^2 +(1+α^2 )z^2 +1−α^2 −(1−α^2 )z^2 )) =(2/a^2 )∫ (dz/(2+(2α^2 )z^2 )) =(1/a^2 )∫ (dz/(1+α^2 z^2 )) =(1/(αa^2 )) arctan(αz) +c =(1/(αa^2 ))arctan(αtan((t/2)))+C =(a^2 /(b^2 a^2 ))arctan((b^2 /a^2 )tan(x))+C =(1/b^2 )arctan((b^2 /a^2 )tan(x)) +C$
	$I = \int \frac{dx}{a^{2} (\cos^{2} x + \frac{b^{2}}{a^{2}} \sin^{2} x)} let α = \frac{b^{2}}{a^{2}} (ab \neq 0 and a \neq b)) \Rightarrow I = \frac{1}{a^{2}} \int \frac{dx}{\cos^{2} x + α^{2} \cos^{2} x}$ $= \frac{1}{a^{2}} \int \frac{dx}{\frac{1 + \cos (2 x)}{2} + α^{2} \times \frac{1 - \cos (2 x)}{2}} = \frac{2}{a^{2}} \int \frac{dx}{1 + \cos (2 x) + α^{2} - α^{2} \cos (2 x)}$ $= \frac{2}{a^{2}} \int \frac{dx}{1 + α^{2} + (1 - α^{2}) \cos (2 x)} =_{2 x = t} \frac{1}{a^{2}} \int \frac{dt}{(1 + α^{2} + (1 - α^{2}) cost)}$ $=_{\tan (\frac{t}{2}) = z} \frac{1}{a^{2}} \int \frac{2 dz}{(1 + z^{2}) {1 + α^{2} + (1 - α^{2}) \frac{1 - z^{2}}{1 + z^{2}}}}$ $= \frac{2}{a^{2}} \int \frac{dz}{(1 + α^{2}) (1 + z^{2}) + (1 - α^{2}) (1 - z^{2})}$ $= \frac{2}{a^{2}} \int \frac{dz}{1 + α^{2} + (1 + α^{2}) z^{2} + 1 - α^{2} - (1 - α^{2}) z^{2}}$ $= \frac{2}{a^{2}} \int \frac{dz}{2 + (2 α^{2}) z^{2}} = \frac{1}{a^{2}} \int \frac{dz}{1 + α^{2} z^{2}} = \frac{1}{α a^{2}} \arctan (α z) + c$ $= \frac{1}{α a^{2}} \arctan (α \tan (\frac{t}{2})) + C$ $= \frac{a^{2}}{b^{2} a^{2}} \arctan (\frac{b^{2}}{a^{2}} \tan (x)) + C$ $= \frac{1}{b^{2}} \arctan (\frac{b^{2}}{a^{2}} \tan (x)) + C$
	Commented by mathmax by abdo last updated on 22/Jan/21

	$sorry α = \frac{b}{a} \Rightarrow I = \frac{a}{{ba}^{2}} \arctan (\frac{b}{a} \tan (x)) + C$ $= \frac{1}{ab} \arctan (\frac{b}{a} tanx) + C$
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