find-dx-cos-4-x-sin-4-x-

Question Number 36428 by prof Abdo imad last updated on 02/Jun/18

f i n d \int \frac{d x}{{c o s}^{4} x + {s i n}^{4} x}

Commented by abdo.msup.com last updated on 02/Jun/18

w e h a v e {c o s}^{4} x + {s i n}^{4} x = {({c o s}^{2} x + {s i n}^{2} x)}^{2}

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

= 1 - \frac{1}{2} {s i n}^{2} (2 x) = 1 - \frac{1}{2} (\frac{1 - c o s (4 x)}{2})

= 1 - \frac{1}{4} + \frac{1}{4} c o s (4 x) = \frac{3}{4} + \frac{1}{4} c o s (4 x)

I = \int \frac{4 d x}{3 + c o s (4 x)}

=_{4 x = t} \int \frac{d t}{3 + c o s t} a n d c h a n g e m e n t

t a n (\frac{t}{2}) = u g i v e

I = \int \frac{1}{3 + \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \int \frac{2 d u}{3 (1 + u^{2)} + 1 - u^{2}}

= \int \frac{2 d u}{4 + 2 u^{2}} = \int \frac{d u}{2 + u^{2}}

=_{u = x \sqrt{2}} \int \frac{\sqrt{2} d x}{2 (1 + x^{2})} = \frac{\sqrt{2}}{2} a r c t a n x + c

= \frac{\sqrt{2}}{2} a r c t a n (\frac{u}{\sqrt{2}}) + c

= \frac{\sqrt{2}}{2} a r c t a n {\frac{1}{\sqrt{2}} t a n (\frac{t}{2})} + c

I = \frac{\sqrt{2}}{2} a r c t a n {\frac{1}{\sqrt{2}} t a n (2 x)} + c .

Answered by MJS last updated on 02/Jun/18

\int \frac{d x}{\cos^{4} x + \sin^{4} x} =

[\sin x = \frac{\tan x}{\sec x}; \cos x = \frac{1}{\sec x}; \sec^{2} x = 1 + \tan^{2} x]

= \int \sec^{2} x \frac{1 + \tan^{2} x}{1 + \tan^{4} x} d x =

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

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

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

[\begin{matrix} \int \frac{d t}{t^{2} \pm p t + q} = \int \frac{d t}{{(t \pm \frac{p}{2})}^{2} + q - \frac{p^{2}}{4}} = \\ [u = \frac{2 t \pm p}{2 \sqrt{q - \frac{p^{2}}{4}}} \to d t = \sqrt{q - \frac{p^{2}}{4}} d u] \\ = \frac{2}{\sqrt{4 q - p^{2}}} \int \frac{d u}{u^{2} + 1} = \frac{2}{\sqrt{4 q - p^{2}}} \arctan u = \\ = \frac{2}{\sqrt{4 q - p^{2}}} \arctan (\frac{2 t \pm p}{\sqrt{4 q - p^{2}}}) \end{matrix}]

= \frac{\sqrt{2}}{2} (\arctan (\sqrt{2} \tan x - 1) + \arctan (\sqrt{2} \tan x + 1)) + C

Answered by MJS last updated on 02/Jun/18

better :

\int \frac{d x}{\cos^{4} x + \sin^{4} x} = 4 \int \frac{d x}{\cos 4 x + 3} =

[t = 4 x \to d x = \frac{d t}{4}]

= \int \frac{d t}{\cos t + 3} = \int \frac{\sec^{2} \frac{t}{2}}{2 (\tan^{2} \frac{t}{2} + 2)} d t =

[u = \frac{\sqrt{2}}{2} \tan \frac{t}{2} \to d t = \frac{2 \sqrt{2}}{\sec^{2} \frac{t}{2}} d u]

= \frac{\sqrt{2}}{2} \int \frac{d u}{u^{2} + 1} = \frac{\sqrt{2}}{2} \arctan u = \frac{\sqrt{2}}{2} \arctan (\frac{\sqrt{2}}{2} \tan \frac{t}{2}) =

= \frac{\sqrt{2}}{2} \arctan (\frac{\sqrt{2}}{2} \tan 2 x) + C

Commented by abdo.msup.com last updated on 02/Jun/18

t h a n k y o u s i r M j s y o u a r e r e a l l y

t a l e n t e d i n m a t h s . .

Commented by rahul 19 last updated on 03/Jun/18

Nice!