What is the integral of tan^3(x/2)sec^2(x/2) ?

The integral of tan^3(x/2)sec^2(x/2) is 1/2 tan^4(x/2)+C

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integral of tan^3(x/2)sec^2(x/2)

\int tan^{3} (\frac{x}{2}) sec^{2} (\frac{x}{2}) d x

\frac{1}{2} tan^{4} (\frac{x}{2}) + C

Solution steps

\int tan^{3} (\frac{x}{2}) sec^{2} (\frac{x}{2}) d x

Apply u-substitution

: \int 2 u^{3} d u

= \int 2 u^{3} d u

Take the constant out:

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

= 2 \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= 2 \cdot \frac{u ^{4}}{4}

Substitute back

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

= 2 \cdot \frac{tan ^{4} ( \frac{x}{2} )}{4}

Simplify

2 \cdot \frac{tan ^{4} ( \frac{x}{2} )}{4} : \frac{1}{2} tan^{4} (\frac{x}{2})

= \frac{1}{2} tan^{4} (\frac{x}{2})

Add a constant to the solution

= \frac{1}{2} tan^{4} (\frac{x}{2}) + C

\frac{d}{d x} (tan (x + y))

\frac{d}{d x} (\frac{( 5 - x ) ^{- 2}}{2})

\frac{d}{d x} (0.2 x^{2})

\frac{d ^{2}}{d x ^{2}} (14 x^{3} - 36 x^{2})

\frac{d Q}{d t} = \frac{12}{5} - 3 Q

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