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

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

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integral of 8cos^4(x)

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

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

Solution steps

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

Take the constant out:

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

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

Simplify

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

= 8 \cdot \int cos^{3} (x) cos (x) d x

Apply Integration By Parts

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

= 8 (cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x)

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

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

Simplify

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

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

Add a constant to the solution

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

\int 12 x^{2} + 2 d x

\int \frac{x ^{4}}{4 x ^{2} - 9} d x

x^{2} - 5 x + 3

x \to - 1 lim ((x - 31)^{\frac{2}{5}})

\frac{\partial}{\partial x} (x^{x^{2} + y^{2} + z^{2}})

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