What is the integral of 5sec^4(x) ?

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

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integral of 5sec^4(x)

\int 5 sec^{4} (x) d x

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

Solution steps

\int 5 sec^{4} (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^{4} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

= 5 \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

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

\int 1 d u = u

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

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

Substitute back

u = tan (x)

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

Add a constant to the solution

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

\frac{d}{d x} (r e^{r x})

\frac{d}{d x} (4 (5 - 7 x)^{5})

f (x) = \frac{3}{( x + 1 ) ^{2}}

e^{- 5 x^{4}}

\frac{d}{d x} (2^{arcsech (x)})

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