What is the integral of tan^4(r/2) ?

The integral of tan^4(r/2) is 2/3 tan^3(r/2)+r-2tan(1/2 r)+C

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integral of tan^4(r/2)

\int tan^{4} (\frac{r}{2}) d r

\frac{2}{3} tan^{3} (\frac{r}{2}) + r - 2 tan (\frac{1}{2} r) + C

Solution steps

\int tan^{4} (\frac{r}{2}) d r

\frac{r}{2} = r \frac{1}{2}

= \int tan^{4} (r \frac{1}{2}) d r

Apply u-substitution

: \int tan^{4} (u) \cdot 2 d u

= \int tan^{4} (u) \cdot 2 d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int tan^{4} (u) d u

Apply Integral Reduction

= 2 (\frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u)

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

= 2 (\frac{tan ^{3} ( u )}{3} - (- u + tan (u)))

Substitute back

u = \frac{1}{2} r

= 2 (\frac{tan ^{3} ( \frac{1}{2} r )}{3} - (- \frac{1}{2} r + tan (\frac{1}{2} r)))

Simplify

2 (\frac{tan ^{3} ( \frac{1}{2} r )}{3} - (- \frac{1}{2} r + tan (\frac{1}{2} r))) : \frac{2}{3} tan^{3} (\frac{r}{2}) + r - 2 tan (\frac{1}{2} r)

= \frac{2}{3} tan^{3} (\frac{r}{2}) + r - 2 tan (\frac{1}{2} r)

Add a constant to the solution

= \frac{2}{3} tan^{3} (\frac{r}{2}) + r - 2 tan (\frac{1}{2} r) + C

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What is the integral of tan^4(r/2) ?
The integral of tan^4(r/2) is 2/3 tan^3(r/2)+r-2tan(1/2 r)+C