What is the integral of \sqrt[7]{tan(3x)}sec^2(3x) ?

The integral of \sqrt[7]{tan(3x)}sec^2(3x) is 7/24 tan^{8/7}(3x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of \sqrt[7]{tan(3x)}sec^2(3x)

\int 7 tan (3 x) sec^{2} (3 x) d x

\frac{7}{24} tan^{\frac{8}{7}} (3 x) + C

Solution steps

\int 7 tan (3 x) sec^{2} (3 x) d x

Apply u-substitution

: \int \frac{7 u ^{7}}{3} d u

= \int \frac{7 u ^{7}}{3} d u

Take the constant out:

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

= \frac{7}{3} \cdot \int u^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= \frac{7}{3} \cdot \frac{u ^{8}}{8}

Substitute back

u = 7 tan (3 x)

= \frac{7}{3} \cdot \frac{( 7 tan ( 3 x ) ) ^{8}}{8}

Simplify

\frac{7}{3} \cdot \frac{( 7 tan ( 3 x ) ) ^{8}}{8} : \frac{7}{24} tan^{\frac{8}{7}} (3 x)

= \frac{7}{24} tan^{\frac{8}{7}} (3 x)

Add a constant to the solution

= \frac{7}{24} tan^{\frac{8}{7}} (3 x) + C

\frac{d}{d x} (x^{4} + 12 x^{2} + 42)

\frac{d}{d x} (\frac{6 x}{2 + x ^{2}})

x y^{^{'}} = 2 y + x^{3}

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

\frac{\partial}{\partial x} (k)

What is the integral of \sqrt[7]{tan(3x)}sec^2(3x) ?
The integral of \sqrt[7]{tan(3x)}sec^2(3x) is 7/24 tan^{8/7}(3x)+C