useful formula ======== ∀a∈R^+ :∀b ∈R: a sin x +b cos x =(√(a^2 +b^2 ))sin (x+arctan (b/a)) ∫(dx/(a sin x +b cos x))= =(1/(√(a^2 +b^2 )))∫(dx/(sin (x+arctan (b/a))))= [t=x+arctan (b/a) → dx=dt] (1/(√(a^2 +b^2 )))∫(dt/(sin t))=−(1/(√(a^2 +b^2 )))ln ((1/(sin t))+(1/(tan t))) = =−(1/(√(a^2 +b^2 )))ln ∣(((√(a^2 +b^2 ))−b sin x +a cos x)/(a sin x +b cos x))∣ +C


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Question Number 63738 by MJS last updated on 08/Jul/19

$useful formula$ $========$ $\forall a \in R^{+} : \forall b \in R : a \sin x + b \cos x = \sqrt{a^{2} + b^{2}} \sin (x + \arctan \frac{b}{a})$ $\int \frac{d x}{a \sin x + b \cos x} =$ $= \frac{1}{\sqrt{a^{2} + b^{2}}} \int \frac{d x}{\sin (x + \arctan \frac{b}{a})} =$ $[t = x + \arctan \frac{b}{a} \to d x = d t]$ $\frac{1}{\sqrt{a^{2} + b^{2}}} \int \frac{d t}{\sin t} = - \frac{1}{\sqrt{a^{2} + b^{2}}} \ln (\frac{1}{\sin t} + \frac{1}{\tan t}) =$ $= - \frac{1}{\sqrt{a^{2} + b^{2}}} \ln ∣ \frac{\sqrt{a^{2} + b^{2}} - b \sin x + a \cos x}{a \sin x + b \cos x} ∣ + C$
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