What is the integral of sin^5(5x)cos^2(5x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin^5(5x)cos^2(5x)

\int sin^{5} (5 x) cos^{2} (5 x) d x

\frac{1}{5} (- \frac{cos ^{3} ( 5 x )}{3} + \frac{2 cos ^{5} ( 5 x )}{5} - \frac{cos ^{7} ( 5 x )}{7}) + C

Solution steps

\int sin^{5} (5 x) cos^{2} (5 x) d x

Apply u-substitution

: \int sin^{5} (u) cos^{2} (u) \frac{1}{5} d u

= \int sin^{5} (u) cos^{2} (u) \frac{1}{5} d u

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

= \frac{1}{5} \cdot \int - v^{2} (1 - v^{2})^{2} d v

Expand

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

= \frac{1}{5} \cdot \int - v^{2} + 2 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}{5} (- \int v^{2} d v + \int 2 v^{4} d v - \int v^{6} d v)

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

\int 2 v^{4} d v = \frac{2 v ^{5}}{5}

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

= \frac{1}{5} (- \frac{v ^{3}}{3} + \frac{2 v ^{5}}{5} - \frac{v ^{7}}{7})

Substitute back

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

Add a constant to the solution

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

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

\frac{d}{d x} (\frac{2 x ^{3} + 5}{4 x ^{2} + 7})

\frac{d}{d x} ((f (x) (x^{3}))^{2})

\frac{d}{d x} (3 cot (2 x) sec (2 x))

\frac{d ^{2}}{d x ^{2}} (x^{4} - 2 x^{3})

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