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

The integral of 6sec^4(t) is 6(tan(t)+(tan^3(t))/3)+C

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integral of 6sec^4(t)

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

6 (tan (t) + \frac{tan ^{3} ( t )}{3}) + C

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

= 6 \cdot \int (1 + tan^{2} (t)) sec^{2} (t) d t

Apply u-substitution

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

= 6 \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

= 6 (\int 1 d u + \int u^{2} d u)

\int 1 d u = u

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

= 6 (u + \frac{u ^{3}}{3})

Substitute back

u = tan (t)

= 6 (tan (t) + \frac{tan ^{3} ( t )}{3})

Add a constant to the solution

= 6 (tan (t) + \frac{tan ^{3} ( t )}{3}) + C

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