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Question Number 7026 by Tawakalitu. last updated on 07/Aug/16

Commented by Yozzii last updated on 07/Aug/16
$Write ∫((sinx+cosx)/(sin^4 x+cos^4 x))dx=∫((sinx)/(sin^4 x+cos^4 x))dx+∫((cosx)/(sin^4 x+cos^4 x))dx. Evaluate each integral by the following steps (1) Substitute u=cosx in ∫((sinx)/(sin^4 x+cos^4 x))dx where sin^4 x+cos^4 x=2cos^4 x−2cos^2 x+1, {and substitute k=sinx in ∫((cosx)/(sin^4 x+cos^4 x))dx where sin^4 x+cos^4 x=2sin^4 x−2sin^2 x+1.} (2) Complete the square in the denominators of the resulting integrand. e.g 2u^4 −2u^2 +1=0.5((2u^2 −1)^2 +1). (3)Apply the identity z^2 +1=(z−i)(z+i) with z=2u^2 −1 for the first integrand for example. ⇒(2u^2 −1)^2 +1=(2u^2 −1−i)(2u^2 −1+i) or (2u^2 −1)^2 +1=((√2)u−(√(1+i)))((√2)u+(√(1+i)))(2u^2 −1+i) (4)Split (2/((2u^2 −1)^2 +1)) into partial fractions. Solve for a,b,c,d∈C in (2/((2u^2 −1)^2 +1))≡(a/((√2)u−(√(1+i))))+(b/((√2)u+(√(1+i))))+((cu+d)/(2u^2 +i−1)) (5)Then integrate and simplify as usual e.g: ∫((sinx)/(sin^4 x+cos^4 x))dx=−∫{(a/((√2)u−(√(1+i))))+(b/((√2)u+(√(1+i))))+((cu+d)/(2u^2 +i−1))}du=−F(u)+C=−F(cosx)+C where (d/du)F(u)=(a/((√2)u−(√(1+i))))+(b/((√2)u+(√(1+i))))+((cu+d)/(2u^2 +i−1)). −−−−−−−−−−−−−−−−−−−−−−−−−− For ∫((cosx)/(sin^4 x+cos^4 x))dx, this turns out to be ∫((cosx)/(sin^4 x+cos^4 x))dx=F(k)+T=F(sinx)+T ∴ ∫((sinx+cosx)/(sin^4 x+cos^4 x))dx=F(sinx)−F(cosx)+constant.$
$W r i t e \int \frac{s i n x + c o s x}{{s i n}^{4} x + {c o s}^{4} x} d x = \int \frac{s i n x}{{s i n}^{4} x + {c o s}^{4} x} d x + \int \frac{c o s x}{{s i n}^{4} x + {c o s}^{4} x} d x .$ $E v a l u a t e e a c h i n t e g r a l b y t h e f o l l o w i n g s t e p s$ $(1) S u b s t i t u t e u = c o s x i n \int \frac{s i n x}{{s i n}^{4} x + {c o s}^{4} x} d x$ $w h e r e {s i n}^{4} x + {c o s}^{4} x = 2 {c o s}^{4} x - 2 {c o s}^{2} x + 1,$ ${a n d s u b s t i t u t e k = s i n x i n \int \frac{c o s x}{{s i n}^{4} x + {c o s}^{4} x} d x$ $w h e r e {s i n}^{4} x + {c o s}^{4} x = 2 {s i n}^{4} x - 2 {s i n}^{2} x + 1 .}$ $(2) C o m p l e t e t h e s q u a r e i n t h e d e n o m i n a t o r s o f$ $t h e r e s u l t i n g i n t e g r a n d .$ $e . g 2 u^{4} - 2 u^{2} + 1 = 0.5 ({(2 u^{2} - 1)}^{2} + 1) .$ $(3) A p p l y t h e i d e n t i t y z^{2} + 1 = (z - i) (z + i)$ $w i t h z = 2 u^{2} - 1 f o r t h e f i r s t i n t e g r a n d$ $f o r e x a m p l e .$ $\Rightarrow {(2 u^{2} - 1)}^{2} + 1 = (2 u^{2} - 1 - i) (2 u^{2} - 1 + i)$ $o r {(2 u^{2} - 1)}^{2} + 1 = (\sqrt{2} u - \sqrt{1 + i}) (\sqrt{2} u + \sqrt{1 + i}) (2 u^{2} - 1 + i)$ $(4) S p l i t \frac{2}{{(2 u^{2} - 1)}^{2} + 1} i n t o p a r t i a l f r a c t i o n s .$ $S o l v e f o r a, b, c, d \in C i n$ $\frac{2}{{(2 u^{2} - 1)}^{2} + 1} \equiv \frac{a}{\sqrt{2} u - \sqrt{1 + i}} + \frac{b}{\sqrt{2} u + \sqrt{1 + i}} + \frac{c u + d}{2 u^{2} + i - 1}$ $(5) T h e n i n t e g r a t e a n d s i m p l i f y a s u s u a l$ $e . g : \int \frac{s i n x}{{s i n}^{4} x + {c o s}^{4} x} d x = - \int {\frac{a}{\sqrt{2} u - \sqrt{1 + i}} + \frac{b}{\sqrt{2} u + \sqrt{1 + i}} + \frac{c u + d}{2 u^{2} + i - 1}} d u = - F (u) + C = - F (c o s x) + C$ $w h e r e \frac{d}{d u} F (u) = \frac{a}{\sqrt{2} u - \sqrt{1 + i}} + \frac{b}{\sqrt{2} u + \sqrt{1 + i}} + \frac{c u + d}{2 u^{2} + i - 1} .$ $- - - - - - - - - - - - - - - - - - - - - - - - - -$ $F o r \int \frac{c o s x}{{s i n}^{4} x + {c o s}^{4} x} d x, t h i s t u r n s o u t t o b e$ $\int \frac{c o s x}{{s i n}^{4} x + {c o s}^{4} x} d x = F (k) + T = F (s i n x) + T$ $∴ \int \frac{s i n x + c o s x}{{s i n}^{4} x + {c o s}^{4} x} d x = F (s i n x) - F (c o s x) + c o n s t a n t .$
Commented by Tawakalitu. last updated on 07/Aug/16

$W o w . . . T h a n k s s o m u c h . . . i r e a l l y a p p r e c i a t e .$
Commented by Tawakalitu. last updated on 07/Aug/16

$I r e a l l y a p p r e c i a t e y o u r e f f o r t .$
Commented by Tawakalitu. last updated on 07/Aug/16

Commented by Tawakalitu. last updated on 07/Aug/16

$T h a n k y o u v e r y m u c h .$
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