What is the integral of sin^{18}(x)cos^3(x) ?

The integral of sin^{18}(x)cos^3(x) is (sin^{19}(x))/(19)-(sin^{21}(x))/(21)+C

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integral of sin^{18}(x)cos^3(x)

\int sin^{18} (x) cos^{3} (x) d x

\frac{sin ^{19} ( x )}{19} - \frac{sin ^{21} ( x )}{21} + C

Solution steps

\int sin^{18} (x) cos^{3} (x) d x

Rewrite using trig identities

= \int (1 - sin^{2} (x)) cos (x) sin^{18} (x) d x

Apply u-substitution

: \int u^{18} (1 - u^{2}) d u

= \int u^{18} (1 - u^{2}) d u

Expand

u^{18} (1 - u^{2}) : u^{18} - u^{20}

= \int u^{18} - u^{20} 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

= \int u^{18} d u - \int u^{20} d u

\int u^{18} d u = \frac{u ^{19}}{19}

\int u^{20} d u = \frac{u ^{21}}{21}

= \frac{u ^{19}}{19} - \frac{u ^{21}}{21}

Substitute back

u = sin (x)

= \frac{sin ^{19} ( x )}{19} - \frac{sin ^{21} ( x )}{21}

Add a constant to the solution

= \frac{sin ^{19} ( x )}{19} - \frac{sin ^{21} ( x )}{21} + C

g (x) = 4 x^{2} (x^{3} + 5 x)

\int (- 8 sec (x) tan (x) + 5 sec^{2} (x)) d x

\frac{d}{d x} ((2 - 7 x^{\frac{3}{2}})^{2})

y^{^{'}} + e^{x} y = 1

2 x^{^{''}} + 4 x^{^{'}} + 8 x = 0, x (0) = 0, x^{^{'}} (0) = 1

What is the integral of sin^{18}(x)cos^3(x) ?
The integral of sin^{18}(x)cos^3(x) is (sin^{19}(x))/(19)-(sin^{21}(x))/(21)+C