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

The integral of tan(x/2)sec^2(x/2) is tan^2(x/2)+C

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integral of tan(x/2)sec^2(x/2)

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

tan^{2} (\frac{x}{2}) + C

Solution steps

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

Apply u-substitution

: \int 2 u d u

= \int 2 u d u

Take the constant out:

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

= 2 \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= 2 \cdot \frac{u ^{2}}{2}

Substitute back

u = tan (\frac{x}{2})

= 2 \cdot \frac{tan ^{2} ( \frac{x}{2} )}{2}

Simplify

2 \cdot \frac{tan ^{2} ( \frac{x}{2} )}{2} : tan^{2} (\frac{x}{2})

= tan^{2} (\frac{x}{2})

Add a constant to the solution

= tan^{2} (\frac{x}{2}) + C

y^{^{'''}} + 4 y^{^{''}} = 3 x - 1

\int e^{(x + 3)} d x

\int x^{2} cos (9 x) d x

\frac{1}{5 s}

\frac{\partial}{\partial x} (x yz e^{x^{2} + y^{2} + z^{2}})

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