`int(2-tan(theta/4))d theta=`

Use additivity of integral: `int(f(x)pm g(x))dx=int f(x)dx pm int g(x)dx.` `int2d theta-int tan(theta/4)d theta=`

Since the first integral is easy `int 2d theta=2theta+C` we will concentrate on the second integral. To solve it we will make substitution: `u=theta/4,` `du=(d theta)/4=>d theta=4du`

`int tan(theta/4)d theta=4int tan u du=-4ln|cos u|+C`

Return the substitution.

`-4ln|cos(theta/4)|+C`

Now we subtract the two integrals to obtain **the final result**.

`2theta-(-4ln|cos(theta/4)|)+C=2theta+4ln|cos(theta/4)|+C`

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