What is the integral of cos^4(10x) ?

The integral of cos^4(10x) is 1/10 (cos^3(10x)sin(10x)+3/8 (10x-1/4 sin(40x)))+C

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integral of cos^4(10x)

\int cos^{4} (10 x) d x

\frac{1}{10} (cos^{3} (10 x) sin (10 x) + \frac{3}{8} (10 x - \frac{1}{4} sin (40 x))) + C

Solution steps

\int cos^{4} (10 x) d x

Apply u-substitution

= \int cos^{4} (u) \frac{1}{10} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int cos^{4} (u) d u

Simplify

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

= \frac{1}{10} \cdot \int cos^{3} (u) cos (u) d u

Apply Integration By Parts

= \frac{1}{10} (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))

= \frac{1}{10} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Substitute back

u = 10 x

= \frac{1}{10} (cos^{3} (10 x) sin (10 x) - (- \frac{3}{8} (10 x - \frac{1}{4} sin (4 \cdot 10 x))))

Simplify

= \frac{1}{10} (cos^{3} (10 x) sin (10 x) + \frac{3}{8} (10 x - \frac{1}{4} sin (40 x)))

Add a constant to the solution

= \frac{1}{10} (cos^{3} (10 x) sin (10 x) + \frac{3}{8} (10 x - \frac{1}{4} sin (40 x))) + C

What is the integral of cos^4(10x) ?
The integral of cos^4(10x) is 1/10 (cos^3(10x)sin(10x)+3/8 (10x-1/4 sin(40x)))+C