What is the integral of-1sin^3(x)cos^2(x) ?

The integral of-1sin^3(x)cos^2(x) is 1/3 cos^3(x)-1/5 cos^5(x)+C

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integral of-1sin^3(x)cos^2(x)

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

\frac{1}{3} cos^{3} (x) - \frac{1}{5} cos^{5} (x) + C

Solution steps

\int - 1 \cdot 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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= - \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

= - (- \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}

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

Substitute back

u = cos (x)

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

Simplify

- (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5}) : \frac{1}{3} cos^{3} (x) - \frac{1}{5} cos^{5} (x)

= \frac{1}{3} cos^{3} (x) - \frac{1}{5} cos^{5} (x)

Add a constant to the solution

= \frac{1}{3} cos^{3} (x) - \frac{1}{5} cos^{5} (x) + C

\int x (tan (x))^{2} d x

x^{2} + 1, 3 - x, x = 0

\frac{d}{d x} ((\frac{tan ( x )}{x})^{2})

\int \frac{x - 6}{( x + 3 ) ^{2} + 36} d x

n = 1 \sum \infty n ln (2)

What is the integral of-1sin^3(x)cos^2(x) ?
The integral of-1sin^3(x)cos^2(x) is 1/3 cos^3(x)-1/5 cos^5(x)+C