What is the integral of sqrt(tan(5x))(sec(5x))^2 ?

The integral of sqrt(tan(5x))(sec(5x))^2 is 2/15 tan^{3/2}(5x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(tan(5x))(sec(5x))^2

\int tan (5 x) (sec (5 x))^{2} d x

\frac{2}{15} tan^{\frac{3}{2}} (5 x) + C

Solution steps

\int tan (5 x) (sec (5 x))^{2} d x

Rewrite using trig identities

= \int \frac{sin ( 5 x )}{cos ( 5 x )} sec^{2} (5 x) d x

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

= \int sec^{2} (5 x) tan (5 x) d x

Apply u-substitution

: \int \frac{2 u ^{2}}{5} d u

= \int \frac{2 u ^{2}}{5} d u

Take the constant out:

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

= \frac{2}{5} \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{2}{5} \cdot \frac{u ^{3}}{3}

Substitute back

u = tan (5 x)

= \frac{2}{5} \cdot \frac{( tan ( 5 x ) ) ^{3}}{3}

Simplify

\frac{2}{5} \cdot \frac{( tan ( 5 x ) ) ^{3}}{3} : \frac{2}{15} tan^{\frac{3}{2}} (5 x)

= \frac{2}{15} tan^{\frac{3}{2}} (5 x)

Add a constant to the solution

= \frac{2}{15} tan^{\frac{3}{2}} (5 x) + C

\frac{d}{d x} (sin^{3} (4 x))

\int \frac{1}{x ( x + 0 )} d x

\frac{d}{d θ} (\frac{6}{2 - 3 sin ( θ )})

2 y^{^{''}} + y = 0

\int (\frac{ln ^{2} ( x )}{x ^{\frac{5}{3}}}) d x

What is the integral of sqrt(tan(5x))(sec(5x))^2 ?
The integral of sqrt(tan(5x))(sec(5x))^2 is 2/15 tan^{3/2}(5x)+C