What is the integral of tan^2(8x)sec^4(8x) ?

The integral of tan^2(8x)sec^4(8x) is 1/8 ((tan^3(8x))/3+(tan^5(8x))/5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of tan^2(8x)sec^4(8x)

\int tan^{2} (8 x) sec^{4} (8 x) d x

\frac{1}{8} (\frac{tan ^{3} ( 8 x )}{3} + \frac{tan ^{5} ( 8 x )}{5}) + C

Solution steps

\int tan^{2} (8 x) sec^{4} (8 x) d x

Apply u-substitution

: \int tan^{2} (u) sec^{4} (u) \frac{1}{8} d u

= \int tan^{2} (u) sec^{4} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int tan^{2} (u) sec^{4} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

= \frac{1}{8} \cdot \int v^{2} (1 + v^{2}) d v

Expand

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

= \frac{1}{8} \cdot \int v^{2} + v^{4} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int v^{2} d v + \int v^{4} d v)

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

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{8} (\frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{8} (\frac{tan ^{3} ( 8 x )}{3} + \frac{tan ^{5} ( 8 x )}{5})

Add a constant to the solution

= \frac{1}{8} (\frac{tan ^{3} ( 8 x )}{3} + \frac{tan ^{5} ( 8 x )}{5}) + C

\int y^{2} 4 y^{2} + 1 d y

\int_{0}^{1} x - x d x

\frac{d x}{d t} = sin (2 π (t + 3)), x (3) = 1

\frac{d}{d x} (cos (2 π (1 - x)))

y = 2 x^{2}, (6, 10)

What is the integral of tan^2(8x)sec^4(8x) ?
The integral of tan^2(8x)sec^4(8x) is 1/8 ((tan^3(8x))/3+(tan^5(8x))/5)+C