What is the integral of 1/(1-cos^4(x)) ?

The integral of 1/(1-cos^4(x)) is 1/(2sqrt(2))arctan((tan(x))/(sqrt(2)))+1/4 tan(x/2)-1/4 cot(x/2)+C

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integral of 1/(1-cos^4(x))

\int \frac{1}{1 - cos ^{4} ( x )} d x

\frac{1}{2 2} arctan (\frac{tan ( x )}{2}) + \frac{1}{4} tan (\frac{x}{2}) - \frac{1}{4} cot (\frac{x}{2}) + C

Solution steps

\int \frac{1}{1 - cos ^{4} ( x )} d x

Take the partial fraction of

\frac{1}{1 - cos ^{4} ( x )} : \frac{1}{2 ( cos ^{2} ( x ) + 1 )} + \frac{1}{4 ( cos ( x ) + 1 )} - \frac{1}{4 ( cos ( x ) - 1 )}

= \int \frac{1}{2 ( cos ^{2} ( x ) + 1 )} + \frac{1}{4 ( cos ( x ) + 1 )} - \frac{1}{4 ( cos ( x ) - 1 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{2 ( cos ^{2} ( x ) + 1 )} d x + \int \frac{1}{4 ( cos ( x ) + 1 )} d x - \int \frac{1}{4 ( cos ( x ) - 1 )} d x

\int \frac{1}{2 ( cos ^{2} ( x ) + 1 )} d x = \frac{1}{2 2} arctan (\frac{tan ( x )}{2})

\int \frac{1}{4 ( cos ( x ) + 1 )} d x = \frac{1}{4} tan (\frac{x}{2})

\int \frac{1}{4 ( cos ( x ) - 1 )} d x = \frac{1}{4} cot (\frac{x}{2})

= \frac{1}{2 2} arctan (\frac{tan ( x )}{2}) + \frac{1}{4} tan (\frac{x}{2}) - \frac{1}{4} cot (\frac{x}{2})

Add a constant to the solution

= \frac{1}{2 2} arctan (\frac{tan ( x )}{2}) + \frac{1}{4} tan (\frac{x}{2}) - \frac{1}{4} cot (\frac{x}{2}) + C

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What is the integral of 1/(1-cos^4(x)) ?
The integral of 1/(1-cos^4(x)) is 1/(2sqrt(2))arctan((tan(x))/(sqrt(2)))+1/4 tan(x/2)-1/4 cot(x/2)+C