Question-85776

Question Number 85776 by M±th+et£s last updated on 24/Mar/20

Answered by MJS last updated on 24/Mar/20

\int \frac{a \sin^{2} x + 2 b \sin x \cos x + c \cos^{2} x}{\sin x + \cos x} d x =

[t = \tan \frac{x}{2} \to d x = \frac{2}{t^{2} + 1} d t]

= - 2 \int \frac{{c t}^{4} - 4 {b t}^{3} + 2 (2 a - c) t^{2} + 4 b t + c}{{(t^{2} + 1)}^{2} (t - 1 - \sqrt{2}) (t - 1 + \sqrt{2})} d t =

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

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

- \frac{\sqrt{2} (a - 2 b + c)}{4} \int \frac{d t}{t - 1 - \sqrt{2}} +

+ \frac{\sqrt{2} (a - 2 b + c)}{4} \int \frac{d t}{t - 1 + \sqrt{2}}

I think you can solve these

Commented by M±th+et£s last updated on 24/Mar/20

y e s s i r t h a n k y o u i t s e a s y t o s o l v e n o w

Commented by MJS last updated on 24/Mar/20

you can always use this substitution

(“ {Weierstrass}^{″}) for integrals with

trigonometric functions

t = \tan \frac{x}{2} \Leftrightarrow x = 2 \arctan t

\to d x = {2 \cos}^{2} \frac{x}{2} d t = (1 + \cos x) d t = \frac{2}{t^{2} + 1} d t

\sin x = \frac{2 t}{t^{2} + 1}

\cos x = - \frac{t^{2} - 1}{t^{2} + 1}

the main problem is to factorize the

denominator \dots

Commented by MJS last updated on 24/Mar/20

\dots also with \sinh x, \cosh x :

t = \tanh \frac{x}{2} \Leftrightarrow x = 2 arctanh t

\to d x = {2 \cosh}^{2} \frac{x}{2} d t = (1 + \cosh x) d t = - \frac{2}{t^{2} - 1} d t

\sinh x = - \frac{2 t}{t^{2} - 1}

\cosh x = - \frac{t^{2} + 1}{t^{2} - 1}

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