hello-please-dx-a-2-cos-2-x-b-2-sin-2-x-

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 \dots

Answered by mathmax by abdo last updated on 22/Jan/21

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