Integration-how to integrate 1/(sin^3x+cos^3x). Denominator is sin cube x plus cos cube x. ?
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I = ∫ [ 1 / ( sin³ x + cos³ x ) ] dx
= ∫ { 1 / [ ( sin x + cos x )( sin² x - sin x cos x + cos² x ) ] } dx
= ∫ { 1 / [ ( sin x + cos x )( 1 - sin x cos x ) ] } dx
= (1/2) ∫ { [ (sin x + cos x)² + (sin x - cos x)² ] / [ (sin x + cos x)( 1 - sin x cos x ) ] } dx
= (1/2)·∫[(sin x+cos x)/(1-sin x cos x)]dx+(1/2)·∫{(1-sin 2x)/[(sin x + cos x)(1-sin x cos x)]}dx
= (1/2)·J + (1/2)·K, say ..................................................... (1)
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J = ∫ [ ( sin x + cos x ) / ( 1 - sin x cos x ) ] dx
= 2 ∫ [ ( sin x + cos x ) / ( 2 - 2 sin x cos x ) ] dx
= 2 ∫ { ( cos + sin x ) / [ 1 + (sin x - cos x)² ] } dx
= 2 ∫ [ 1 / ( 1 + t² ) ] dt, ............... t = sin x - cos x, dt = (cos x + sin x) dx
= 2· tanֿ¹ (t)
= 2· tanֿ¹ (sin x - cos x) ................................................... (2)
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K = ∫ { ( 1 - sin 2x ) / [ (sin x + cos x)( 1 - sin x cos x ) ] } dx
= 2 ∫ { [ ( 2 - sin 2x ) - 1 ] / [ (sin x + cos x)( 2 - sin 2x ) ] } dx
= 2 ∫ [ 1 / (sin x + cos x) ] dx - ∫ { { 1 / [ (sin x + cos x)( 1 - sin x cos x ) ] } dx
= 2 ∫ { 1 / [ √2. cos (x - π/4) ] } dx - ∫ [ 1 / ( sin³ x + cos³ x ) ] dx
= √2. ∫ sec (x - π/4) dx
= √2. ln | sec (x - π/4) + tan (x - π/4) | - I .................... from (1) ................. (3)
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From (1), (2), (3),
I = (1/2) { 2· tanֿ¹ (sin x - cos x) + √2. ln | sec (x - π/4) + tan (x - π/4) | - I }
Then, taking ' (-1/2)·I ' from RHS to LHS,
(3/2)·I = tanֿ¹ (sin x - cos x) + √2. ln | sec (x - π/4) + tan (x - π/4) |
that is,
I = (2/3)· tanֿ¹ ( sin x - cos x ) + (2√2 / 3)· ln | sec (x - π/4) + tan (x - π/4) | + C ... Ans.
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Happy To Help !
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Let sin^3x+cos^3x=k
Find out dk/dx which is equal to say y
dk/dx=y
Hence dk/y=dx.
I'll use f as sign of integral ok?
So in short We want to find f(dx/k)
=f(dk/y.k)........................Since dk/y=dx.
We can take out y as it is constant w.r.t k
Hence f(dk/k)/y
=(logk)/y
Substitute value of k and y!