What is the integral of x^2sec^2(x^3+3)tan^3(x^3+3) ?

The integral of x^2sec^2(x^3+3)tan^3(x^3+3) is 1/12 tan^4(x^3+3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^2sec^2(x^3+3)tan^3(x^3+3)

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

\frac{1}{12} tan^{4} (x^{3} + 3) + C

Solution steps

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

Apply u-substitution

: \int \frac{tan ^{3} ( u ) sec ^{2} ( u )}{3} d u

= \int \frac{tan ^{3} ( u ) sec ^{2} ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int tan^{3} (u) sec^{2} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 + tan^{2} (u)) tan^{3} (u) d u

Apply u-substitution

: \int v^{3} d v

= \frac{1}{3} \cdot \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

= \frac{1}{3} \cdot \frac{v ^{4}}{4}

Substitute back

= \frac{1}{3} \cdot \frac{tan ^{4} ( x ^{3} + 3 )}{4}

Simplify

\frac{1}{3} \cdot \frac{tan ^{4} ( x ^{3} + 3 )}{4} : \frac{1}{12} tan^{4} (x^{3} + 3)

= \frac{1}{12} tan^{4} (x^{3} + 3)

Add a constant to the solution

= \frac{1}{12} tan^{4} (x^{3} + 3) + C

\int \frac{1}{v ( 1 + v ) ^{5}} d v

\int \frac{x ^{4}}{2 e ^{x}} d x

\int - 2 x^{2} sin (2 x) d x

\int cot (31 x) d x

\int sin^{6} (d) x d x

What is the integral of x^2sec^2(x^3+3)tan^3(x^3+3) ?
The integral of x^2sec^2(x^3+3)tan^3(x^3+3) is 1/12 tan^4(x^3+3)+C