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

The integral of sec^2(8x)tan^2(8x) is 1/24 tan^3(8x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{24} tan^{3} (8 x) + C

Solution steps

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

Apply u-substitution

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

= \int sec^{2} (u) tan^{2} (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 sec^{2} (u) tan^{2} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int v^{2} d v

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

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{8} \cdot \frac{v ^{3}}{3}

Substitute back

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

Simplify

\frac{1}{8} \cdot \frac{tan ^{3} ( 8 x )}{3} : \frac{1}{24} tan^{3} (8 x)

= \frac{1}{24} tan^{3} (8 x)

Add a constant to the solution

= \frac{1}{24} tan^{3} (8 x) + C

\int sin (20 x) d x

\int (3 x^{2} + 2 x) x^{3} + x^{2} d x

\int (3 cos (x)) d x

\int \frac{1}{y ( 1 - y )} d y

\int \frac{1}{9 x ^{2} - 6 x + 7} d x

What is the integral of sec^2(8x)tan^2(8x) ?
The integral of sec^2(8x)tan^2(8x) is 1/24 tan^3(8x)+C