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

Grade:12

## 1 Answers

Vinod Ramakrishnan Eswaran
41 Points
5 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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