What is the integral of x^2cos^3(x^3)sin^2(x^3) ?

The integral of x^2cos^3(x^3)sin^2(x^3) is 1/3 ((sin^3(x^3))/3-(sin^5(x^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 x^2cos^3(x^3)sin^2(x^3)

\int x^{2} cos^{3} (x^{3}) sin^{2} (x^{3}) d x

\frac{1}{3} (\frac{sin ^{3} ( x ^{3} )}{3} - \frac{sin ^{5} ( x ^{3} )}{5}) + C

Solution steps

\int x^{2} cos^{3} (x^{3}) sin^{2} (x^{3}) d x

Apply u-substitution

: \int \frac{u cos ^{3} ( u ^{\frac{3}{2}} ) sin ^{2} ( u ^{\frac{3}{2}} )}{2 x} d u

= \int \frac{cos ^{3} ( u ) sin ^{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 cos^{3} (u) sin^{2} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int v^{2} (1 - v^{2}) d v

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

Expand

v^{2} (1 - v^{2}) : v^{2} - v^{4}

= \frac{1}{3} \cdot \int v^{2} - v^{4} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int v^{2} d v - \int v^{4} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{3} (\frac{v ^{3}}{3} - \frac{v ^{5}}{5})

Substitute back

= \frac{1}{3} (\frac{sin ^{3} ( x ^{3} )}{3} - \frac{sin ^{5} ( x ^{3} )}{5})

Add a constant to the solution

= \frac{1}{3} (\frac{sin ^{3} ( x ^{3} )}{3} - \frac{sin ^{5} ( x ^{3} )}{5}) + C

\int_{1}^{e} \frac{( ln ( x ) ) ^{2}}{x} d x

x \to 0 lim (\frac{1}{x ( x + 1 )})

\int_{0}^{9} \frac{6}{( x - 1 ) ^{\frac{1}{3}}} d x

10 x

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

What is the integral of x^2cos^3(x^3)sin^2(x^3) ?
The integral of x^2cos^3(x^3)sin^2(x^3) is 1/3 ((sin^3(x^3))/3-(sin^5(x^3))/5)+C