What is the integral of (sec(x/(24)))^2tan(x/(24)) ?

The integral of (sec(x/(24)))^2tan(x/(24)) is 12tan^2(x/(24))+C

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integral of (sec(x/(24)))^2tan(x/(24))

\int (sec (\frac{x}{24}))^{2} tan (\frac{x}{24}) d x

12 tan^{2} (\frac{x}{24}) + C

Solution steps

\int (sec (\frac{x}{24}))^{2} tan (\frac{x}{24}) d x

Simplify

= \int sec^{2} (\frac{x}{24}) tan (\frac{x}{24}) d x

Apply u-substitution

: \int sec^{2} (u) tan (u) \cdot 24 d u

= \int sec^{2} (u) tan (u) \cdot 24 d u

Take the constant out:

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

= 24 \cdot \int sec^{2} (u) tan (u) d u

Apply u-substitution

: \int v d v

= 24 \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= 24 \cdot \frac{v ^{2}}{2}

Substitute back

= 24 \cdot \frac{tan ^{2} ( \frac{x}{24} )}{2}

Simplify

24 \cdot \frac{tan ^{2} ( \frac{x}{24} )}{2} : 12 tan^{2} (\frac{x}{24})

= 12 tan^{2} (\frac{x}{24})

Add a constant to the solution

= 12 tan^{2} (\frac{x}{24}) + C

x \to - \infty lim (e)

(x^{2} + 49) \frac{d y}{d x} = 1, y (7) = 0

\int \frac{x + 8}{x ( x + 6 )} d x

h \to 0 lim (\frac{sinh ( x )}{h})

\frac{s + 2}{s ^{2} + 2 s + 1}

What is the integral of (sec(x/(24)))^2tan(x/(24)) ?
The integral of (sec(x/(24)))^2tan(x/(24)) is 12tan^2(x/(24))+C