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

The integral of sin^2(t)cos^2(t) is 1/8 (t-1/4 sin(4t))+C

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

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

\frac{1}{8} (t - \frac{1}{4} sin (4 t)) + C

Solution steps

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

Rewrite using trig identities

= \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

= \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

= \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)

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

Add a constant to the solution

= \frac{1}{8} (t - \frac{1}{4} sin (4 t)) + C

\frac{d}{d x} (x^{6 c o s (x)})

\frac{2}{1 + x ^{2}}

y^{^{'}} + y = e^{3 x}

\frac{d}{d x} (3^{x - 1})

\int \frac{1}{csc ^{4} ( 3 x ) sec ^{2} ( 3 x )} d x

What is the integral of sin^2(t)cos^2(t) ?
The integral of sin^2(t)cos^2(t) is 1/8 (t-1/4 sin(4t))+C