What is the integral of 19sin^3(x)cos^2(x) ?

The integral of 19sin^3(x)cos^2(x) is 19(-(cos^3(x))/3+(cos^5(x))/5)+C

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integral of 19sin^3(x)cos^2(x)

\int 19 sin^{3} (x) cos^{2} (x) d x

19 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5}) + C

Solution steps

\int 19 sin^{3} (x) cos^{2} (x) d x

Take the constant out:

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

= 19 \cdot \int sin^{3} (x) cos^{2} (x) d x

Rewrite using trig identities

= 19 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{2} (x) d x

Apply u-substitution

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

= 19 \cdot \int - u^{2} (1 - u^{2}) d u

Expand

- u^{2} (1 - u^{2}) : - u^{2} + u^{4}

= 19 \cdot \int - u^{2} + u^{4} 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

= 19 (- \int u^{2} d u + \int u^{4} d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int u^{4} d u = \frac{u ^{5}}{5}

= 19 (- \frac{u ^{3}}{3} + \frac{u ^{5}}{5})

Substitute back

u = cos (x)

= 19 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5})

Add a constant to the solution

= 19 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5}) + C

f (x) = \frac{6}{x ^{2}}, at x = 1

y^{^{''}} - 3 y^{^{'}} - 10 y = 0

\frac{1}{( 1 + x ) ^{2}}, 1

\frac{d}{d x} (5 x)

\int 4 x^{2} + 1 d x

What is the integral of 19sin^3(x)cos^2(x) ?
The integral of 19sin^3(x)cos^2(x) is 19(-(cos^3(x))/3+(cos^5(x))/5)+C