What is the integral of 8sec^6(t) ?

The integral of 8sec^6(t) is 8(tan(t)+(2tan^3(t))/3+(tan^5(t))/5)+C

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integral of 8sec^6(t)

\int 8 sec^{6} (t) d t

8 (tan (t) + \frac{2 tan ^{3} ( t )}{3} + \frac{tan ^{5} ( t )}{5}) + C

Solution steps

\int 8 sec^{6} (t) d t

Take the constant out:

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

= 8 \cdot \int sec^{6} (t) d t

Rewrite using trig identities

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

Apply u-substitution

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

Expand

(1 + u^{2})^{2} : 1 + 2 u^{2} + u^{4}

= 8 \cdot \int 1 + 2 u^{2} + u^{4} 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 2 u^{2} d u + \int u^{4} d u)

\int 1 d u = u

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

\int u^{4} d u = \frac{u ^{5}}{5}

= 8 (u + \frac{2 u ^{3}}{3} + \frac{u ^{5}}{5})

Substitute back

u = tan (t)

= 8 (tan (t) + \frac{2 tan ^{3} ( t )}{3} + \frac{tan ^{5} ( t )}{5})

Add a constant to the solution

= 8 (tan (t) + \frac{2 tan ^{3} ( t )}{3} + \frac{tan ^{5} ( t )}{5}) + C

\int (r^{2} + r + 1) e^{r} d r

x \to 1 + lim (\frac{x ^{2} - 8 x + 15}{x - 1})

\frac{\partial}{\partial y} (x^{2} + 2 y^{2} - x^{2} y)

\frac{d}{d x} ((2 x - 9)^{4} (x^{2} + x + 1)^{5})

\frac{d}{d x} (3 ln (15 x))

What is the integral of 8sec^6(t) ?
The integral of 8sec^6(t) is 8(tan(t)+(2tan^3(t))/3+(tan^5(t))/5)+C