Find the value of : 𝛀 = ∫_0 ^( (𝛑/2)) (( dx)/(sin^6 x + cos^6 x)) = ? −−−−−−−−−


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Question Number 207949 by mnjuly1970 last updated on 31/May/24

$F i n d t h e v a l u e o f :$ $Ω = \int_{0}^{\frac{π}{2}} \frac{d x}{{s i n}^{6} x + {c o s}^{6} x} = ?$ $- - - - - - - - -$
	Answered by Frix last updated on 31/May/24

	$\underset{0}{\int^{\frac{π}{2}}} \frac{d x}{\cos^{6} x + \sin^{6} x} = 2 \underset{0}{\int^{\frac{π}{4}}} \frac{d x}{\cos^{6} x + \sin^{6} x} =$ $= 16 \underset{0}{\int^{\frac{π}{4}}} \frac{d x}{5 + 3 \cos 4 x} \overset{t = \tan 2 x}{=} 4 \underset{0}{\int^{\infty}} \frac{d t}{t^{2} + 4} =$ $= 2 {[\tan^{- 1} \frac{t}{2}]}_{0}^{\infty} = π$
	Commented by mnjuly1970 last updated on 01/Jun/24

	$t h a n k s a l o t a l i . . .$
	Answered by Berbere last updated on 01/Jun/24
	$1=(sin^2 (x)+cos^2 (x))^3 =sin^6 (x)+cos^6 (x)+3sin^2 (x)cos^2 (x)(sin^2 (x)+cos^2 (x)) ⇒sin^6 (x)+cos^6 (x)=1−(3/4)(2sin(x)cos(x))^2 =1−((3sin^2 (2x))/4)=1−(3/8)(2sin^2 (2x)=1−(3/8)(1−cos(4x)) u=4x Ω=2∫_0 ^(2π) (du/(5+3cos(u))) cos(u)=((e^(iu) +e^(−iu) )/2),z=e^(iu) ⇒dz=izdu⇔du=(dz/(iz)) ⇒2∫_C (dz/(iz(5+(3/2)(((z^2 +1)/z)))))=(4/i)∫_C (dz/((3z^2 +10z+3)))=Ω 3z^2 +10z+3=0⇒z∈{−3,−(1/3)} Ω=2iπ.(4/i)Res((1/((3z^2 +10z+3))),z=−(1/3)}=8π.(1/(6z+10))=((8π)/8)=π$
	$1 = {({s i n}^{2} (x) + {c o s}^{2} (x))}^{3} = {s i n}^{6} (x) + {c o s}^{6} (x) + 3 {s i n}^{2} (x) {c o s}^{2} (x) ({s i n}^{2} (x) + {c o s}^{2} (x))$ $\Rightarrow {s i n}^{6} (x) + {c o s}^{6} (x) = 1 - \frac{3}{4} {(2 s i n (x) c o s (x))}^{2}$ $= 1 - \frac{3 {s i n}^{2} (2 x)}{4} = 1 - \frac{3}{8} (2 {s i n}^{2} (2 x) = 1 - \frac{3}{8} (1 - c o s (4 x))$ $u = 4 x$ $Ω = 2 \int_{0}^{2 π} \frac{d u}{5 + 3 c o s (u)}$ $c o s (u) = \frac{e^{i u} + e^{- i u}}{2}, z = e^{i u} \Rightarrow d z = i z d u \Leftrightarrow d u = \frac{d z}{i z}$ $\Rightarrow 2 \int_{C} \frac{d z}{i z (5 + \frac{3}{2} (\frac{z^{2} + 1}{z}))} = \frac{4}{i} \int_{C} \frac{d z}{(3 z^{2} + 10 z + 3)} = Ω$ $3 z^{2} + 10 z + 3 = 0 \Rightarrow z \in {- 3, - \frac{1}{3}}$ $Ω = 2 i π . \frac{4}{i} R e s (\frac{1}{(3 z^{2} + 10 z + 3)}, z = - \frac{1}{3}} = 8 π . \frac{1}{6 z + 10} = \frac{8 π}{8} = π$
	Commented by mnjuly1970 last updated on 01/Jun/24

	$g r a t e f u l s i r . . .$
	Commented by Berbere last updated on 01/Jun/24

	$W i t h e p l e a s u r$
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