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

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

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

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

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

Solution steps

\int 3 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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= 3 \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

= 3 (- \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}

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

Substitute back

u = cos (x)

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

Simplify

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

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

Add a constant to the solution

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

\int 2 cos^{3} (t) d t

t y^{^{'}} - 3 y = t^{4}

\int \frac{x}{3 - x ^{2}} d x

\int \frac{x ^{2} - 2 x + 6}{( x + 1 ) ( x - 2 ) ^{2}} d x

x^{2} (120 - x)

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