What is the integral of 2+2cos^4(t) ?

The integral of 2+2cos^4(t) is 11/4 t-3/16 sin(4t)+2cos^3(t)sin(t)+C

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integral of 2+2cos^4(t)

integral

2 + 2 cos^{4} (t)

\frac{11}{4} t - \frac{3}{16} sin (4 t) + 2 cos^{3} (t) sin (t) + C

Solution steps

\int 2 + 2 cos^{4} (t) d t

Apply the Sum Rule:

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

= \int 2 d t + \int 2 cos^{4} (t) d t

\int 2 d t = 2 t

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

= 2 t + 2 (cos^{3} (t) sin (t) + \frac{3}{8} (t - \frac{1}{4} sin (4 t)))

Simplify

2 t + 2 (cos^{3} (t) sin (t) + \frac{3}{8} (t - \frac{1}{4} sin (4 t))) : \frac{11}{4} t - \frac{3}{16} sin (4 t) + 2 cos^{3} (t) sin (t)

= \frac{11}{4} t - \frac{3}{16} sin (4 t) + 2 cos^{3} (t) sin (t)

Add a constant to the solution

= \frac{11}{4} t - \frac{3}{16} sin (4 t) + 2 cos^{3} (t) sin (t) + C

\frac{( lo g _{10} ( x ) ) \cdot 1}{x}

\frac{x}{x ^{2} - 15}

(\frac{1 + 1}{x ^{2}}) e^{\frac{x - 1}{x}}

\frac{4 x}{( x ^{3} - x ^{2} - x + 1 )}

x \cdot (x - 1)^{3}

What is the integral of 2+2cos^4(t) ?
The integral of 2+2cos^4(t) is 11/4 t-3/16 sin(4t)+2cos^3(t)sin(t)+C