dx-csc-x-sec-x-

Question Number 132192 by benjo_mathlover last updated on 12/Feb/21

\int \frac{dx}{\csc x + \sec x}

Answered by Dwaipayan Shikari last updated on 12/Feb/21

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

= \frac{1}{2} \int s i n x + c o s x - \frac{1}{s i n x + c o s x} d x

= \frac{1}{2} (s i n x - c o s x) - \int \frac{d t}{1 - t^{2} + 2 t} t = t a n \frac{x}{2}

= \frac{1}{2} (s i n x - c o s x) + \int \frac{1}{{(t - 1)}^{2} - 2} d t

= \frac{1}{2} (s i n x - c o s x) + \frac{1}{2 \sqrt{2}} l o g (\frac{t - 1 - \sqrt{2}}{t - 1 + \sqrt{2}}) + C

= \frac{1}{2} (s i n x - c o s x) + \frac{1}{2 \sqrt{2}} l o g (\frac{t a n \frac{x}{2} - 1 - \sqrt{2}}{t a n \frac{x}{2} - 1 + \sqrt{2}}) + C

Answered by MJS_new last updated on 13/Feb/21

weird but possible

\int \frac{d x}{\csc x + \sec x} = \int \frac{\sin x \cos x}{\sin x + \cos x} d x =

= \frac{\sqrt{2}}{4} \int \frac{\sin 2 x}{\sin \frac{4 x + π}{4}} d x =

[t = \tan^{2} \frac{4 x + π}{8} \to d x = \frac{\cos^{2} \frac{4 x + π}{8}}{\tan \frac{4 x + π}{8}} d t]

= - \frac{\sqrt{2}}{8} \int \frac{t^{2} - 6 t + 1}{t {(t + 1)}^{2}} d t = \sqrt{2} \int \frac{d t}{{(t + 1)}^{2}} - \frac{\sqrt{2}}{8} \int \frac{d t}{t} =

= - \frac{\sqrt{2}}{t + 1} - \frac{\sqrt{2}}{8} \ln t =

= - \frac{\sqrt{2}}{4} \ln ∣ \tan \frac{4 x + π}{8} ∣ - \frac{\sqrt{2}}{2} \cos \frac{4 x + π}{4} + C

Answered by Ñï= last updated on 14/Feb/21

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

= \frac{1}{2} \int \frac{1 + 2 s i n x c o s x - 1}{s i n x + c o s x} d x

= \frac{1}{2} \int (s i n x + c o s x - \frac{1}{\sqrt{2}} c s c (x + \frac{π}{4})) d x

= \frac{1}{2} (s i n x - c o s x) - \frac{1}{2 \sqrt{2}} l n ∣ t a n (\frac{x}{2} + \frac{π}{8}) ∣ + C

Facebook Tweet Pin