What is the integral of cos^4(2x)sin^3(2x) ?

The integral of cos^4(2x)sin^3(2x) is 1/2 (-(cos^5(2x))/5+(cos^7(2x))/7)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of cos^4(2x)sin^3(2x)

\int cos^{4} (2 x) sin^{3} (2 x) d x

\frac{1}{2} (- \frac{cos ^{5} ( 2 x )}{5} + \frac{cos ^{7} ( 2 x )}{7}) + C

Solution steps

\int cos^{4} (2 x) sin^{3} (2 x) d x

Apply u-substitution

: \int cos^{4} (u) sin^{3} (u) \frac{1}{2} d u

= \int cos^{4} (u) sin^{3} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int cos^{4} (u) sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{4} (u) d u

Apply u-substitution

: \int - v^{4} (1 - v^{2}) d v

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

Expand

- v^{4} (1 - v^{2}) : - v^{4} + v^{6}

= \frac{1}{2} \cdot \int - v^{4} + v^{6} 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}{2} (- \int v^{4} d v + \int v^{6} d v)

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

\int v^{6} d v = \frac{v ^{7}}{7}

= \frac{1}{2} (- \frac{v ^{5}}{5} + \frac{v ^{7}}{7})

Substitute back

= \frac{1}{2} (- \frac{cos ^{5} ( 2 x )}{5} + \frac{cos ^{7} ( 2 x )}{7})

Add a constant to the solution

= \frac{1}{2} (- \frac{cos ^{5} ( 2 x )}{5} + \frac{cos ^{7} ( 2 x )}{7}) + C

\int x^{21} e^{x^{22}} d x

ln^{2} (1 + x)

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

\int \frac{x ^{15}}{1 + x ^{16}} d x

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

What is the integral of cos^4(2x)sin^3(2x) ?
The integral of cos^4(2x)sin^3(2x) is 1/2 (-(cos^5(2x))/5+(cos^7(2x))/7)+C