What is the integral of 64cos^4(8x) ?

The integral of 64cos^4(8x) is 8(cos^3(8x)sin(8x)+3x-3/32 sin(32x))+C

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integral of 64cos^4(8x)

\int 64 cos^{4} (8 x) d x

8 (cos^{3} (8 x) sin (8 x) + 3 x - \frac{3}{32} sin (32 x)) + C

Solution steps

\int 64 cos^{4} (8 x) d x

Take the constant out:

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

= 64 \cdot \int cos^{4} (8 x) d x

Apply u-substitution

: \int cos^{4} (u) \frac{1}{8} d u

= 64 \cdot \int cos^{4} (u) \frac{1}{8} d u

Take the constant out:

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

= 64 \cdot \frac{1}{8} \cdot \int cos^{4} (u) d u

Simplify

cos^{4} (u) : cos^{3} (u) cos (u)

= 64 \cdot \frac{1}{8} \cdot \int cos^{3} (u) cos (u) d u

Apply Integration By Parts

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

= 64 \cdot \frac{1}{8} (cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u)

\int - 3 cos^{2} (u) sin^{2} (u) d u = - \frac{3}{8} (u - \frac{1}{4} sin (4 u))

= 64 \cdot \frac{1}{8} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = 8 x

= 64 \cdot \frac{1}{8} (cos^{3} (8 x) sin (8 x) - (- \frac{3}{8} (8 x - \frac{1}{4} sin (4 \cdot 8 x))))

Simplify

64 \cdot \frac{1}{8} (cos^{3} (8 x) sin (8 x) - (- \frac{3}{8} (8 x - \frac{1}{4} sin (4 \cdot 8 x)))) : 8 (cos^{3} (8 x) sin (8 x) + 3 x - \frac{3}{32} sin (32 x))

= 8 (cos^{3} (8 x) sin (8 x) + 3 x - \frac{3}{32} sin (32 x))

Add a constant to the solution

= 8 (cos^{3} (8 x) sin (8 x) + 3 x - \frac{3}{32} sin (32 x)) + C

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What is the integral of 64cos^4(8x) ?
The integral of 64cos^4(8x) is 8(cos^3(8x)sin(8x)+3x-3/32 sin(32x))+C