What is the integral of 4sec^6(x) ?

The integral of 4sec^6(x) is 4(tan(x)+(2tan^3(x))/3+(tan^5(x))/5)+C

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integral of 4sec^6(x)

\int 4 sec^{6} (x) d x

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

Solution steps

\int 4 sec^{6} (x) d x

Take the constant out:

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

= 4 \cdot \int sec^{6} (x) d x

Rewrite using trig identities

= 4 \cdot \int sec^{2} (x) (1 + tan^{2} (x))^{2} d x

Apply u-substitution

: \int (1 + u^{2})^{2} d u

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

Expand

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

= 4 \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

= 4 (\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}

= 4 (u + \frac{2 u ^{3}}{3} + \frac{u ^{5}}{5})

Substitute back

u = tan (x)

= 4 (tan (x) + \frac{2 tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5})

Add a constant to the solution

= 4 (tan (x) + \frac{2 tan ^{3} ( x )}{3} + \frac{tan ^{5} ( x )}{5}) + C

\int \frac{1}{s + 2} d s

(- 2, - 5), (2, - 8)

y = (1 + 5 x)^{11}, (0, 1)

\int_{4}^{8} \frac{1}{x ^{2} - 1} d x

- \frac{2}{( 1 - x ) ^{3}}

What is the integral of 4sec^6(x) ?
The integral of 4sec^6(x) is 4(tan(x)+(2tan^3(x))/3+(tan^5(x))/5)+C