What is the integral of sin^{10}(7x)cos^3(7x) ?

The integral of sin^{10}(7x)cos^3(7x) is 1/7 ((sin^{11}(7x))/(11)-(sin^{13}(7x))/(13))+C

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integral of sin^{10}(7x)cos^3(7x)

\int sin^{10} (7 x) cos^{3} (7 x) d x

\frac{1}{7} (\frac{sin ^{11} ( 7 x )}{11} - \frac{sin ^{13} ( 7 x )}{13}) + C

Solution steps

\int sin^{10} (7 x) cos^{3} (7 x) d x

Apply u-substitution

: \int sin^{10} (u) cos^{3} (u) \frac{1}{7} d u

= \int sin^{10} (u) cos^{3} (u) \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int sin^{10} (u) cos^{3} (u) d u

Rewrite using trig identities

= \frac{1}{7} \cdot \int (1 - sin^{2} (u)) cos (u) sin^{10} (u) d u

Apply u-substitution

: \int v^{10} (1 - v^{2}) d v

= \frac{1}{7} \cdot \int v^{10} (1 - v^{2}) d v

Expand

v^{10} (1 - v^{2}) : v^{10} - v^{12}

= \frac{1}{7} \cdot \int v^{10} - v^{12} d v

Apply the Sum Rule:

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

= \frac{1}{7} (\int v^{10} d v - \int v^{12} d v)

\int v^{10} d v = \frac{v ^{11}}{11}

\int v^{12} d v = \frac{v ^{13}}{13}

= \frac{1}{7} (\frac{v ^{11}}{11} - \frac{v ^{13}}{13})

Substitute back

= \frac{1}{7} (\frac{sin ^{11} ( 7 x )}{11} - \frac{sin ^{13} ( 7 x )}{13})

Add a constant to the solution

= \frac{1}{7} (\frac{sin ^{11} ( 7 x )}{11} - \frac{sin ^{13} ( 7 x )}{13}) + C

x \to 0 + lim (3^{\frac{1}{x}})

\int 2 x 3 1 + x^{2} d x

2 t y^{'} + (4 + 12 t^{3} - 120 t^{2}) y = 0

\frac{6 s + 3}{s ^{2}}

\int \frac{x}{( x ^{2} - 7 x + 12 )} d x

What is the integral of sin^{10}(7x)cos^3(7x) ?
The integral of sin^{10}(7x)cos^3(7x) is 1/7 ((sin^{11}(7x))/(11)-(sin^{13}(7x))/(13))+C