What is the integral of 8tan(x)sec^3(x) ?

The integral of 8tan(x)sec^3(x) is 8/3 sec^3(x)+C

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integral of 8tan(x)sec^3(x)

\int 8 tan (x) sec^{3} (x) d x

\frac{8}{3} sec^{3} (x) + C

Solution steps

\int 8 tan (x) sec^{3} (x) d x

Take the constant out:

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

= 8 \cdot \int tan (x) sec^{3} (x) d x

Apply u-substitution

: \int \frac{1}{3} d u

= 8 \cdot \int \frac{1}{3} d u

Integral of a constant:

\int a d x = a x

= 8 \cdot \frac{1}{3} u

Substitute back

u = sec^{3} (x)

= 8 \cdot \frac{1}{3} sec^{3} (x)

Simplify

8 \cdot \frac{1}{3} sec^{3} (x) : \frac{8}{3} sec^{3} (x)

= \frac{8}{3} sec^{3} (x)

Add a constant to the solution

= \frac{8}{3} sec^{3} (x) + C

y = (x^{2} + 2)^{2} (x^{4} + 4)^{4}

\int \frac{1}{y ^{3} - y} d y

h (x) = \frac{7}{x}, at x = 5

\frac{d}{d x} (- \frac{1}{2} e^{- \frac{x}{4}})

x \to 0 + lim (x lo g_{e} (x))

What is the integral of 8tan(x)sec^3(x) ?
The integral of 8tan(x)sec^3(x) is 8/3 sec^3(x)+C