What is the integral of 7sec^6(t) ?

The integral of 7sec^6(t) is 7(tan(t)+(2tan^3(t))/3+(tan^5(t))/5)+C

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integral of 7sec^6(t)

\int 7 sec^{6} (t) d t

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

Solution steps

\int 7 sec^{6} (t) d t

Take the constant out:

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

= 7 \cdot \int sec^{6} (t) d t

Rewrite using trig identities

= 7 \cdot \int sec^{2} (t) (1 + tan^{2} (t))^{2} d t

Apply u-substitution

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

= 7 \cdot \int (1 + u^{2})^{2} d u

Expand

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

= 7 \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

= 7 (\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}

= 7 (u + \frac{2 u ^{3}}{3} + \frac{u ^{5}}{5})

Substitute back

u = tan (t)

= 7 (tan (t) + \frac{2 tan ^{3} ( t )}{3} + \frac{tan ^{5} ( t )}{5})

Add a constant to the solution

= 7 (tan (t) + \frac{2 tan ^{3} ( t )}{3} + \frac{tan ^{5} ( t )}{5}) + C

\frac{d y}{d x}, x = y e^{y}

n = 1 \sum \infty 3 (- \frac{2}{3})^{n}

\frac{d y}{d x} = \frac{2 + y ^{2}}{2 x y}, y (1) = 1

\int x x^{2} + 3 d x

\int \frac{x}{- x ^{2} + 5} d x

What is the integral of 7sec^6(t) ?
The integral of 7sec^6(t) is 7(tan(t)+(2tan^3(t))/3+(tan^5(t))/5)+C