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

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

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

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

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

Solution steps

\int 3 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

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

Rewrite using trig identities

= 3 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{5} (x) d x

Apply u-substitution

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

= 3 \cdot \int - u^{5} (1 - u^{2}) d u

Expand

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

= 3 \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

= 3 (- \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}

= 3 (- \frac{u ^{6}}{6} + \frac{u ^{8}}{8})

Substitute back

u = cos (x)

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

Add a constant to the solution

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

\frac{8 x ^{2} + 6 x + 2}{x}

y = (arctan (4 x))^{2}

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

\frac{d}{d x} (x e^{x y})

\frac{d}{d x} (8 x^{- 3})

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