What is the integral of 6sin^3(x)cos^5(x) ?

The integral of 6sin^3(x)cos^5(x) is -cos^6(x)+3/4 cos^8(x)+C

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integral of 6sin^3(x)cos^5(x)

\int 6 sin^{3} (x) cos^{5} (x) d x

- cos^{6} (x) + \frac{3}{4} cos^{8} (x) + C

Solution steps

\int 6 sin^{3} (x) cos^{5} (x) d x

Take the constant out:

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

= 6 \cdot \int sin^{3} (x) cos^{5} (x) d x

Rewrite using trig identities

= 6 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{5} (x) d x

Apply u-substitution

: \int - u^{5} (1 - u^{2}) d u

= 6 \cdot \int - u^{5} (1 - u^{2}) d u

Expand

- u^{5} (1 - u^{2}) : - u^{5} + u^{7}

= 6 \cdot \int - u^{5} + u^{7} 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

= 6 (- \int u^{5} d u + \int u^{7} d u)

\int u^{5} d u = \frac{u ^{6}}{6}

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

= 6 (- \frac{u ^{6}}{6} + \frac{u ^{8}}{8})

Substitute back

u = cos (x)

= 6 (- \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8})

Simplify

6 (- \frac{cos ^{6} ( x )}{6} + \frac{cos ^{8} ( x )}{8}) : - cos^{6} (x) + \frac{3}{4} cos^{8} (x)

= - cos^{6} (x) + \frac{3}{4} cos^{8} (x)

Add a constant to the solution

= - cos^{6} (x) + \frac{3}{4} cos^{8} (x) + C

(x^{x})^{^{'}}

x \to 2 - lim (ln (4 - 2 x))

\int tan^{2} (t) sec^{3} (t) d t

n = 0 \sum \infty 5^{- n}

\frac{d}{d x} (ln (1 + 8 x))

What is the integral of 6sin^3(x)cos^5(x) ?
The integral of 6sin^3(x)cos^5(x) is -cos^6(x)+3/4 cos^8(x)+C