What is the integral of 7x^2tan(2x^3-5) ?

The integral of 7x^2tan(2x^3-5) is -7/6 ln|cos(2x^3-5)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 7x^2tan(2x^3-5)

\int 7 x^{2} tan (2 x^{3} - 5) d x

- \frac{7}{6} ln cos (2 x^{3} - 5) + C

Solution steps

\int 7 x^{2} tan (2 x^{3} - 5) d x

Take the constant out:

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

= 7 \cdot \int x^{2} tan (2 x^{3} - 5) d x

Apply u-substitution

: \int \frac{u tan ( 2 u ^{\frac{3}{2}} - 5 )}{2 x} d u

= 7 \cdot \int \frac{tan ( 2 u - 5 )}{3} d u

Take the constant out:

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

= 7 \cdot \frac{1}{3} \cdot \int tan (2 u - 5) d u

Rewrite using trig identities

= 7 \cdot \frac{1}{3} \cdot \int \frac{sin ( 2 u - 5 )}{cos ( 2 u - 5 )} d u

Apply u-substitution

: \int - \frac{1}{2 v} d v

= 7 \cdot \frac{1}{3} \cdot \int - \frac{1}{2 v} d v

Take the constant out:

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

= 7 \cdot \frac{1}{3} (- \frac{1}{2} \cdot \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 7 \cdot \frac{1}{3} (- \frac{1}{2} ln ∣ v ∣)

Substitute back

= 7 \cdot \frac{1}{3} (- \frac{1}{2} ln cos (2 x^{3} - 5))

Simplify

7 \cdot \frac{1}{3} (- \frac{1}{2} ln cos (2 x^{3} - 5)) : - \frac{7}{6} ln cos (2 x^{3} - 5)

= - \frac{7}{6} ln cos (2 x^{3} - 5)

Add a constant to the solution

= - \frac{7}{6} ln cos (2 x^{3} - 5) + C

\int \frac{x + 1}{( x - 5 ) ^{2}} d x

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

\int 8 x^{5} - 2 d x

\int z arctan (z) d z

\int \frac{9 t ^{2}}{2} d t

What is the integral of 7x^2tan(2x^3-5) ?
The integral of 7x^2tan(2x^3-5) is -7/6 ln|cos(2x^3-5)|+C