What is the integral of 5sec^6(x) ?

The integral of 5sec^6(x) is 5tan(x)+10/3 tan^3(x)+tan^5(x)+C

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integral of 5sec^6(x)

\int 5 sec^{6} (x) d x

5 tan (x) + \frac{10}{3} tan^{3} (x) + tan^{5} (x) + C

Solution steps

\int 5 sec^{6} (x) d x

Take the constant out:

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

= 5 \cdot \int sec^{6} (x) d x

Rewrite using trig identities

= 5 \cdot \int sec^{2} (x) (1 + tan^{2} (x))^{2} d x

Apply u-substitution

= 5 \cdot \int (1 + u^{2})^{2} d u

Expand

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

= 5 \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

= 5 (\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}

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

Substitute back

u = tan (x)

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

Simplify

5 (tan (x) + \frac{2 tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5}) : 5 tan (x) + \frac{10}{3} tan^{3} (x) + tan^{5} (x)

= 5 tan (x) + \frac{10}{3} tan^{3} (x) + tan^{5} (x)

Add a constant to the solution

= 5 tan (x) + \frac{10}{3} tan^{3} (x) + tan^{5} (x) + C

\int_{3}^{8} \frac{x}{x ^{2} - 1} d x

\int cos (x - 2 y) d x

\int_{0}^{\infty} x e^{- \frac{x}{3}} d x

y^{^{''}} + 12 y^{^{'}} + 35 y = 70 t^{2} + 48 t + 4 + 48 e^{t}

l a pl a ce t r an s f or m h (t) = - 5 t^{2}

What is the integral of 5sec^6(x) ?
The integral of 5sec^6(x) is 5tan(x)+10/3 tan^3(x)+tan^5(x)+C