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Integral Calculus> Integration of cotx÷(cos^4x+sin^4x)^1÷2 P...

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

Akanksha , 6 Years ago

Grade 12

1 Answers

Vinod Ramakrishnan Eswaran

Last Activity: 6 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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