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Integration of cotx÷(cos^4x+sin^4x)^1÷2 Please solve my doubt

Integration of cotx÷(cos^4x+sin^4x)^1÷2
Please solve my doubt

Grade:12

1 Answers

Vinod Ramakrishnan Eswaran
41 Points
2 years ago
I=\int (\cot x\: dx/(\cos ^{4}x+\sin^{4}x)^{1/2})
Dividing numerator and denominator with cos4x we get,
I=\int (\cot x*sec^{2}x\: dx/(1+\tan^{4}x)^{1/2})
But cot x=1/tan x
I=\int (sec^{2}x\: dx/\tan x(1+\tan^{4}x)^{1/2})
Let tan x=t
Hence sec2xdx=dt
I=\int ( dt/t(1+t^{4})^{1/2})
Multiplying numerator and denominator with 4t3 we get
I=\frac{1}{4}\int ( 4t^{3}\: dt/t^{4}(1+t^{4})^{1/2})
Let 1+t4=a
Hence 4t3dt=da
I=\frac{1}{4}\int ( \: da/(a-1)(a)^{1/2})
Now let a=z2
Hence da=2zdz
I=\frac{1}{4}\int \: 2\: z\: dz/(z^{2}-1)(z)
I=\frac{1}{4}\int \: 2\: dz/(z^{2}-1)
I=\frac{1}{2}\int \: dz/(z^{2}-1)
Using formula we get
I=\frac{1}{2}(\frac{1}{2*1}\log(\frac{z-1}{z+1}) )
Substituting back to orginal variable we get
I=\frac{1}{4}(\log(\frac{\sqrt{1+\tan^{4}x}-1}{\sqrt{1+\tan^{4}x}+1}) )

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