What is the integral of tan^2(w/2)+sec^2(w/2) ?

The integral of tan^2(w/2)+sec^2(w/2) is -w+4tan(w/2)+C

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integral of tan^2(w/2)+sec^2(w/2)

\int tan^{2} (\frac{w}{2}) + sec^{2} (\frac{w}{2}) d w

- w + 4 tan (\frac{w}{2}) + C

Solution steps

\int tan^{2} (\frac{w}{2}) + sec^{2} (\frac{w}{2}) d w

Apply the Sum Rule:

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

= \int tan^{2} (\frac{w}{2}) d w + \int sec^{2} (\frac{w}{2}) d w

\int tan^{2} (\frac{w}{2}) d w = - w + 2 tan (\frac{w}{2})

\int sec^{2} (\frac{w}{2}) d w = 2 tan (\frac{w}{2})

= - w + 2 tan (\frac{w}{2}) + 2 tan (\frac{w}{2})

Simplify

= - w + 4 tan (\frac{w}{2})

Add a constant to the solution

= - w + 4 tan (\frac{w}{2}) + C

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What is the integral of tan^2(w/2)+sec^2(w/2) ?
The integral of tan^2(w/2)+sec^2(w/2) is -w+4tan(w/2)+C