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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{1}{8} tan^{2} (4 x) + C

Solution steps

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

Apply u-substitution

= \int \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u d u

Apply the Power Rule

= \frac{1}{4} \cdot \frac{u ^{2}}{2}

Substitute back

u = tan (4 x)

= \frac{1}{4} \cdot \frac{tan ^{2} ( 4 x )}{2}

Simplify

\frac{1}{4} \cdot \frac{tan ^{2} ( 4 x )}{2} : \frac{1}{8} tan^{2} (4 x)

= \frac{1}{8} tan^{2} (4 x)

Add a constant to the solution

= \frac{1}{8} tan^{2} (4 x) + C

\int \frac{1}{2 x ^{\frac{3}{2}}} d x

\int (4 sec (x) tan (x)) d x

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

\int sin (t - u) d u

\int \frac{1}{x x} d x

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