What is the integral of 7cos^5(x)sin^4(x) ?

The integral of 7cos^5(x)sin^4(x) is 7/5 sin^5(x)-2sin^7(x)+7/9 sin^9(x)+C

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integral of 7cos^5(x)sin^4(x)

\int 7 cos^{5} (x) sin^{4} (x) d x

\frac{7}{5} sin^{5} (x) - 2 sin^{7} (x) + \frac{7}{9} sin^{9} (x) + C

Solution steps

\int 7 cos^{5} (x) sin^{4} (x) d x

Take the constant out:

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

= 7 \cdot \int cos^{5} (x) sin^{4} (x) d x

Rewrite using trig identities

= 7 \cdot \int (1 - sin^{2} (x))^{2} cos (x) sin^{4} (x) d x

Apply u-substitution

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

= 7 \cdot \int u^{4} (1 - u^{2})^{2} d u

Expand

u^{4} (1 - u^{2})^{2} : u^{4} - 2 u^{6} + u^{8}

= 7 \cdot \int u^{4} - 2 u^{6} + u^{8} 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

= 7 (\int u^{4} d u - \int 2 u^{6} d u + \int u^{8} d u)

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

\int 2 u^{6} d u = \frac{2 u ^{7}}{7}

\int u^{8} d u = \frac{u ^{9}}{9}

= 7 (\frac{u ^{5}}{5} - \frac{2 u ^{7}}{7} + \frac{u ^{9}}{9})

Substitute back

u = sin (x)

= 7 (\frac{sin ^{5} ( x )}{5} - \frac{2 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{9})

Simplify

7 (\frac{sin ^{5} ( x )}{5} - \frac{2 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{9}) : \frac{7}{5} sin^{5} (x) - 2 sin^{7} (x) + \frac{7}{9} sin^{9} (x)

= \frac{7}{5} sin^{5} (x) - 2 sin^{7} (x) + \frac{7}{9} sin^{9} (x)

Add a constant to the solution

= \frac{7}{5} sin^{5} (x) - 2 sin^{7} (x) + \frac{7}{9} sin^{9} (x) + C

\int x^{2} - 3 d x

x \to a lim (3)

\int 8 x^{- 5} y^{7} - 10 x^{- 6} y^{8} d x

y = ln (ln (5 x))

\frac{d}{d x} (\frac{15}{x})

What is the integral of 7cos^5(x)sin^4(x) ?
The integral of 7cos^5(x)sin^4(x) is 7/5 sin^5(x)-2sin^7(x)+7/9 sin^9(x)+C