find-dx-acosx-bsinx-with-a-and-b-reals-

Question Number 59998 by Mr X pcx last updated on 16/May/19

f i n d \int \frac{d x}{a c o s x + b s i n x} w i t h a a n d b r e a l s

Commented by mr W last updated on 16/May/19

a \cos x + b \sin x = \sqrt{a^{2} + b^{2}} (\frac{a}{\sqrt{a^{2} + b^{2}}} \cos x + \frac{b}{\sqrt{a^{2} + b^{2}}} \sin x)

= \sqrt{a^{2} + b^{2}} (\sin θ \cos x + \cos θ \sin x)

= \sqrt{a^{2} + b^{2}} \sin (x + θ)

w i t h θ = \tan^{- 1} \frac{a}{b}

\int \frac{d x}{a c o s x + b s i n x}

= \frac{1}{\sqrt{a^{2} + b^{2}}} \int \frac{d x}{s i n (x + θ)}

= \frac{1}{2 \sqrt{a^{2} + b^{2}}} \times \ln \frac{1 - \cos (x + θ)}{1 + \cos (x + θ)} + C

Commented by maxmathsup by imad last updated on 17/May/19

t h a n k s y o u s i r .

Commented by maxmathsup by imad last updated on 17/May/19

l e t I = \int \frac{d x}{a c o s x + b s i n x} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

I = \int \frac{1}{a \frac{1 - t^{2}}{1 + t^{2}} + b \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int \frac{2 d t}{a - {a t}^{2} + 2 b t} = \int \frac{- 2 d t}{{a t}^{2} - 2 b t - a} = \int F (t) d t

r o o t s o f {a t}^{2} - 2 b t - a = 0 \to Δ^{^{'}} = b^{2} + a^{2} f o r a \neq 0 \Rightarrow Δ^{^{'}} > 0 \Rightarrow

t_{1} = \frac{b + \sqrt{a^{2} + b^{2}}}{a} a n d t_{2} = \frac{b - \sqrt{a^{2} + b^{2}}}{a} \Rightarrow F (t) = \frac{- 2}{a (t - t_{1}) (t - t_{2})}

= \frac{- 2}{a (t_{1} - t_{2})} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} = \frac{- 2}{2 \sqrt{a^{2} + b^{2}}} {\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}} \Rightarrow

I = \frac{1}{\sqrt{a^{2} + b^{2}}} l n ∣ \frac{t - t_{1}}{t - t_{2}} ∣ + C = \frac{1}{\sqrt{a^{2} + b^{2}}} l n ∣ \frac{t a n (\frac{x}{2}) - \frac{b + \sqrt{a^{2} + b^{2}}}{a}}{t a n (\frac{x}{2}) - \frac{b - \sqrt{a^{2} + b^{2}}}{a}} ∣ + C

= \frac{1}{\sqrt{a^{2} + b^{2}}} l n ∣ \frac{a t a n (\frac{x}{2}) - b - \sqrt{a^{2} + b^{2}}}{a t a n (\frac{x}{2}) - b + \sqrt{a^{2} + b^{2}}} ∣ + C .

Answered by kaivan.ahmadi last updated on 16/May/19