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

The integral of 2cos^2(t)sin^2(t) is 1/4 (t-1/4 sin(4t))+C

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integral of 2cos^2(t)sin^2(t)

\int 2 cos^{2} (t) sin^{2} (t) d t

\frac{1}{4} (t - \frac{1}{4} sin (4 t)) + C

Solution steps

\int 2 cos^{2} (t) sin^{2} (t) d t

Take the constant out:

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

= 2 \cdot \int cos^{2} (t) sin^{2} (t) d t

Rewrite using trig identities

= 2 \cdot \int \frac{1 - cos ( 4 t )}{8} d t

Take the constant out:

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

= 2 \cdot \frac{1}{8} \cdot \int 1 - 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

= 2 \cdot \frac{1}{8} (\int 1 d t - \int cos (4 t) d t)

\int 1 d t = t

\int cos (4 t) d t = \frac{1}{4} sin (4 t)

= 2 \cdot \frac{1}{8} (t - \frac{1}{4} sin (4 t))

Simplify

2 \cdot \frac{1}{8} (t - \frac{1}{4} sin (4 t)) : \frac{1}{4} (t - \frac{1}{4} sin (4 t))

= \frac{1}{4} (t - \frac{1}{4} sin (4 t))

Add a constant to the solution

= \frac{1}{4} (t - \frac{1}{4} sin (4 t)) + C

t an g e n t f (x) = x^{2} + 3 x + 5, (1, 9)

x \to 2 + lim (\frac{1}{∣ 2 - x ∣})

\frac{d}{d x} (\frac{1}{x ^{2} + a ^{2}})

\int_{0}^{\frac{π}{6}} sin^{4} (6 x) d x

in v erse l a pl a ce \frac{10}{( s ) ( s ^{2} + 8 s + 25 )}

What is the integral of 2cos^2(t)sin^2(t) ?
The integral of 2cos^2(t)sin^2(t) is 1/4 (t-1/4 sin(4t))+C