What is the integral of 5sin^3(x)cos^7(x) ?

The integral of 5sin^3(x)cos^7(x) is 5(-(cos^8(x))/8+(cos^{10}(x))/(10))+C

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integral of 5sin^3(x)cos^7(x)

\int 5 sin^{3} (x) cos^{7} (x) d x

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

Solution steps

\int 5 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

= 5 \cdot \int sin^{3} (x) cos^{7} (x) d x

Rewrite using trig identities

= 5 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{7} (x) d x

Apply u-substitution

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

= 5 \cdot \int - u^{7} (1 - u^{2}) d u

Expand

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

= 5 \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

= 5 (- \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}

= 5 (- \frac{u ^{8}}{8} + \frac{u ^{10}}{10})

Substitute back

u = cos (x)

= 5 (- \frac{cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10})

Add a constant to the solution

= 5 (- \frac{cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10}) + C

\int_{1}^{2} ln x^{2} + 1 d x

\frac{d}{d x} (ln (4 x - 1))

\int_{4}^{5} \frac{1}{5 x - 1} d x

x y^{^{'}} + 3 y = \frac{4 e ^{2 x}}{x ^{2}}

x \to 1 lim (\frac{2 x}{x - 1})

What is the integral of 5sin^3(x)cos^7(x) ?
The integral of 5sin^3(x)cos^7(x) is 5(-(cos^8(x))/8+(cos^{10}(x))/(10))+C