What is the integral of 12cos^4(2x)sin^3(2x) ?

The integral of 12cos^4(2x)sin^3(2x) is 6(-1/5 cos^5(2x)+1/7 cos^7(2x))+C

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integral of 12cos^4(2x)sin^3(2x)

\int 12 cos^{4} (2 x) sin^{3} (2 x) d x

6 (- \frac{1}{5} cos^{5} (2 x) + \frac{1}{7} cos^{7} (2 x)) + C

Solution steps

\int 12 cos^{4} (2 x) sin^{3} (2 x) d x

Take the constant out:

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

= 12 \cdot \int cos^{4} (2 x) sin^{3} (2 x) d x

Apply u-substitution

: \int cos^{4} (u) sin^{3} (u) \frac{1}{2} d u

= 12 \cdot \int cos^{4} (u) sin^{3} (u) \frac{1}{2} d u

Take the constant out:

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

= 12 \cdot \frac{1}{2} \cdot \int cos^{4} (u) sin^{3} (u) d u

Rewrite using trig identities

= 12 \cdot \frac{1}{2} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{4} (u) d u

Apply u-substitution

: \int - v^{4} (1 - v^{2}) d v

= 12 \cdot \frac{1}{2} \cdot \int - v^{4} (1 - v^{2}) d v

Expand

- v^{4} (1 - v^{2}) : - v^{4} + v^{6}

= 12 \cdot \frac{1}{2} \cdot \int - v^{4} + v^{6} 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

= 12 \cdot \frac{1}{2} (- \int v^{4} d v + \int v^{6} d v)

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

\int v^{6} d v = \frac{v ^{7}}{7}

= 12 \cdot \frac{1}{2} (- \frac{v ^{5}}{5} + \frac{v ^{7}}{7})

Substitute back

= 12 \cdot \frac{1}{2} (- \frac{cos ^{5} ( 2 x )}{5} + \frac{cos ^{7} ( 2 x )}{7})

Simplify

12 \cdot \frac{1}{2} (- \frac{cos ^{5} ( 2 x )}{5} + \frac{cos ^{7} ( 2 x )}{7}) : 6 (- \frac{1}{5} cos^{5} (2 x) + \frac{1}{7} cos^{7} (2 x))

= 6 (- \frac{1}{5} cos^{5} (2 x) + \frac{1}{7} cos^{7} (2 x))

Add a constant to the solution

= 6 (- \frac{1}{5} cos^{5} (2 x) + \frac{1}{7} cos^{7} (2 x)) + C

x^{2} - 4 x - 4

9 x sin (x)

\frac{d}{d x} (\frac{2 sin ( x )}{1 + cos ( x )})

\int 2^{3 x + l o g_{2} (x)} d x

- 0.1 x^{2} + 0.8 x + 9

What is the integral of 12cos^4(2x)sin^3(2x) ?
The integral of 12cos^4(2x)sin^3(2x) is 6(-1/5 cos^5(2x)+1/7 cos^7(2x))+C