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

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

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

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

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

Solution steps

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

Rewrite using trig identities

= 8 \cdot \int (- 1 + sec^{2} (x)) tan (x) sec (x) d x

Apply u-substitution

: \int - 1 + u^{2} d u

= 8 \cdot \int - 1 + u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 (- \int 1 d u + \int u^{2} d u)

\int 1 d u = u

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

= 8 (- u + \frac{u ^{3}}{3})

Substitute back

u = sec (x)

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

Add a constant to the solution

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

\frac{d}{d x} (3 x^{2} - 10)

x^{2} y^{^{''}} - 7 x y^{^{'}} + 15 y = 0, y (1) = 0, y^{^{'}} (1) = 2

\frac{d}{d x} (\frac{x}{x ^{4} + 4})

\frac{d}{d x} (\frac{x ^{4}}{5 - x ^{3}})

f (x) = ln (5 + ln (x))

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