What is the integral of 6sin^3(x)cos^2(x) ?

The integral of 6sin^3(x)cos^2(x) is 6(-(cos^3(x))/3+(cos^5(x))/5)+C

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integral of 6sin^3(x)cos^2(x)

\int 6 sin^{3} (x) cos^{2} (x) d x

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

Solution steps

\int 6 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

= 6 \cdot \int sin^{3} (x) cos^{2} (x) d x

Rewrite using trig identities

= 6 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{2} (x) d x

Apply u-substitution

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

= 6 \cdot \int - u^{2} (1 - u^{2}) d u

Expand

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

= 6 \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

= 6 (- \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}

= 6 (- \frac{u ^{3}}{3} + \frac{u ^{5}}{5})

Substitute back

u = cos (x)

= 6 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5})

Add a constant to the solution

= 6 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5}) + C

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

f (x) = x^{2} - 1, (- 2, 3)

x \to 2 lim (\frac{3}{2 - x})

\int (3 t + t^{2}) sin (3 t) d t

\int ln (1 + t^{2}) d t

What is the integral of 6sin^3(x)cos^2(x) ?
The integral of 6sin^3(x)cos^2(x) is 6(-(cos^3(x))/3+(cos^5(x))/5)+C