What is the integral of 9sin^3(x)cos^7(x) ?

The integral of 9sin^3(x)cos^7(x) is 9(-(cos^8(x))/8+(cos^{10}(x))/(10))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9sin^3(x)cos^7(x)

\int 9 sin^{3} (x) cos^{7} (x) d x

9 (- \frac{cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10}) + C

Solution steps

\int 9 sin^{3} (x) cos^{7} (x) d x

Take the constant out:

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

= 9 \cdot \int sin^{3} (x) cos^{7} (x) d x

Rewrite using trig identities

= 9 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{7} (x) d x

Apply u-substitution

: \int - u^{7} (1 - u^{2}) d u

= 9 \cdot \int - u^{7} (1 - u^{2}) d u

Expand

- u^{7} (1 - u^{2}) : - u^{7} + u^{9}

= 9 \cdot \int - u^{7} + u^{9} d u

Apply the Sum Rule:

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

= 9 (- \int u^{7} d u + \int u^{9} d u)

\int u^{7} d u = \frac{u ^{8}}{8}

\int u^{9} d u = \frac{u ^{10}}{10}

= 9 (- \frac{u ^{8}}{8} + \frac{u ^{10}}{10})

Substitute back

u = cos (x)

= 9 (- \frac{cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10})

Add a constant to the solution

= 9 (- \frac{cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10}) + C

\int_{0}^{\infty} x d x

\frac{d}{d x} (a x e^{- 5 x})

\int 4 cot^{4} (x) d x

\int 4 sin (2 x) d x

\frac{d}{d x} ((x + π)^{2})

What is the integral of 9sin^3(x)cos^7(x) ?
The integral of 9sin^3(x)cos^7(x) is 9(-(cos^8(x))/8+(cos^{10}(x))/(10))+C