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

The integral of sin^3(1-2x)cos^3(1-2x) is -1/2 (-(cos^4(1-2x))/4+(cos^6(1-2x))/6)+C

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

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

- \frac{1}{2} (- \frac{cos ^{4} ( 1 - 2 x )}{4} + \frac{cos ^{6} ( 1 - 2 x )}{6}) + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2} sin^{3} (u) cos^{3} (u) d u

= \int - \frac{1}{2} sin^{3} (u) cos^{3} (u) d u

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int - v^{3} (1 - v^{2}) d v

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

Expand

- v^{3} (1 - v^{2}) : - v^{3} + v^{5}

= - \frac{1}{2} \cdot \int - v^{3} + v^{5} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \frac{1}{2} (- \int v^{3} d v + \int v^{5} d v)

\int v^{3} d v = \frac{v ^{4}}{4}

\int v^{5} d v = \frac{v ^{6}}{6}

= - \frac{1}{2} (- \frac{v ^{4}}{4} + \frac{v ^{6}}{6})

Substitute back

= - \frac{1}{2} (- \frac{cos ^{4} ( 1 - 2 x )}{4} + \frac{cos ^{6} ( 1 - 2 x )}{6})

Add a constant to the solution

= - \frac{1}{2} (- \frac{cos ^{4} ( 1 - 2 x )}{4} + \frac{cos ^{6} ( 1 - 2 x )}{6}) + C

\frac{d}{d x} (3 x - 4 x^{\frac{9}{10}})

\frac{d}{d x} (4 cos^{3} (x))

y = (9 x^{2} - 6 x + 2) (1 + 2 x), at x = 1

\frac{d}{d x} (a 1 + x)

x + 9, 0.25 x + 9

What is the integral of sin^3(1-2x)cos^3(1-2x) ?
The integral of sin^3(1-2x)cos^3(1-2x) is -1/2 (-(cos^4(1-2x))/4+(cos^6(1-2x))/6)+C