What is the integral of 8sec^4(x) ?

The integral of 8sec^4(x) is 8(tan(x)+(tan^3(x))/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8sec^4(x)

\int 8 sec^{4} (x) d x

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

Solution steps

\int 8 sec^{4} (x) d x

Take the constant out:

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

= 8 \cdot \int sec^{4} (x) d x

Rewrite using trig identities

= 8 \cdot \int (1 + tan^{2} (x)) sec^{2} (x) d x

Apply u-substitution

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

= 8 \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

= 8 (\int 1 d u + \int u^{2} d u)

\int 1 d u = u

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

= 8 (u + \frac{u ^{3}}{3})

Substitute back

u = tan (x)

= 8 (tan (x) + \frac{tan ^{3} ( x )}{3})

Add a constant to the solution

= 8 (tan (x) + \frac{tan ^{3} ( x )}{3}) + C

\int 1 + sin (θ) d θ

f (x) = sin (x) - cos (x)

y = - 2 x^{2} + 32 x - 7

\frac{\partial}{\partial x} (x + y^{2} + x^{2} y)

\int 2^{t} d t

What is the integral of 8sec^4(x) ?
The integral of 8sec^4(x) is 8(tan(x)+(tan^3(x))/3)+C