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Okay, I understood they split $\cos^4t$ in half but I don't know what was used to get $1/4$ out in front and the $4$ in front of $\sin^2 x\cos^2 x$.

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    Associativity of scalar multiplication and linearity of the integral. – 2017-02-05
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    The $3$ was brought out of the integral, then a $\frac{1}{4}$ was also added to the outside, requiring that the inside be multiplied by $4$. Think of it like this: on the outside you multiplied by a clever form of $1$, in this case $\frac{4}{4}$ – 2017-02-05

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Note that: $$\int_a^b c\cdot f(t) ~dt\equiv c \int_a^b f(t)~dt\tag{1}$$ Therefore: $$\int_0^{\pi} 3\sin^2 t \cos^4 t ~dt=3\int_0^{\pi} \sin^2 t \cos^4 t ~dt$$ You can now divide the outside of the integral by $4$ while multiplying the inside, it is a consequence of $(1)$, since you are essentially multiplying by $\frac{4}{4}$: $$3\int_0^{\pi} \sin^2 t \cos^4 t ~dt=\frac{3}{4}\int_0^{\pi} 4\sin^2 t \cos^4 t~dt$$ Now, note that: $$\cos^4 t\equiv\cos^2 t\cdot \cos^2 t$$ And that scalar multiplication is associative: $$(a\cdot b)\cdot c\equiv a\cdot (b\cdot c)$$ Therefore, we obtain the expression you need.

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    But why was 4 chosen? Is it because it would easily fit the identity? – 2017-02-05
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    @cout Are you asking why I chose to use $4$, or why they used $4$ in the identity you were given? If you are asking why I chose to use it, it is simply because it would fit the identity. – 2017-02-05