What is the integral of sin^3(5x)cos^4(5x) ?

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

\int sin^{3} (5 x) cos^{4} (5 x) d x

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

Solution steps

\int sin^{3} (5 x) cos^{4} (5 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{4} (u) \frac{1}{5} d u

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

Rewrite using trig identities

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

Expand

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

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

Substitute back

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

Add a constant to the solution

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

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

\int \frac{x}{1 + 2 x ^{4}} d x

\int \frac{( e ^{x} + 1 )}{e ^{x}} d x

\int (x^{1.2} + 7 x^{2.5}) d x

\int (x + 1) (3 - 3 x^{2} - 6 x)^{3} d x

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