What is the integral of 39sec^4(x) ?

The integral of 39sec^4(x) is 39(tan(x)+(tan^3(x))/3)+C

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integral of 39sec^4(x)

\int 39 sec^{4} (x) d x

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

Solution steps

\int 39 sec^{4} (x) d x

Take the constant out:

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

= 39 \cdot \int sec^{4} (x) d x

Rewrite using trig identities

= 39 \cdot \int (1 + tan^{2} (x)) sec^{2} (x) d x

Apply u-substitution

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

= 39 \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

= 39 (\int 1 d u + \int u^{2} d u)

\int 1 d u = u

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

= 39 (u + \frac{u ^{3}}{3})

Substitute back

u = tan (x)

= 39 (tan (x) + \frac{tan ^{3} ( x )}{3})

Add a constant to the solution

= 39 (tan (x) + \frac{tan ^{3} ( x )}{3}) + C

y^{^{'}} = (t + y)^{2} - 1, y (2) = 5

f (x) = \frac{1}{5 - x}, at x = 0

y = x^{3} - 14 x

\frac{d}{d x} ((x^{3} - 6 x) (2 - 4 x^{3}))

\int x (b - x)^{2} d x

What is the integral of 39sec^4(x) ?
The integral of 39sec^4(x) is 39(tan(x)+(tan^3(x))/3)+C