Respuesta :
Answer:
[tex]\frac{x}{8}-\frac{\sin(4x)}{32}+C[/tex]
Step-by-step explanation:
[Most of the work here comes from manipulating the trig to make the term (integrand) integrable.]
Recall that we can express the squared trig functions in terms of cos(2x). That is,
[tex]\cos(2x)=2\cos^2x-1 \\ \cos(2x)=1 - 2\sin^2x.[/tex]
And so inverting these,
[tex]\cos^2x=\frac{1}{2} (1+\cos2x) \\ \sin^2x=\frac{1}{2} (1-\cos2x)[/tex].
Multiply them together to obtain an equivalent expression for sin^2(x)cos^2(x) in terms of cos(2x).
[tex]\sin^2x \cdot \cos^2x =\frac{1}{2} (1-\cos2x) \cdot \frac{1}{2} (1+\cos2x) = \frac{1}{4}(1-\cos^2(2x)).[/tex]
Notice we have cos^2(2x) in the integrand now. We've made it worse! Let's try plugging back in to the first identity for cos^2(2x).
[tex]\cos(2x)=2\cos^2x-1 \Rightarrow \cos(4x)=2\cos^2(2x)-1 \Rightarrow \cos^2(2x) = \frac{1}{2}(1+\cos(4x))[/tex]
So then,
[tex]\sin^2x \cdot \cos^2x = \frac{1}{4}(1-\cos^2(2x)) = \frac{1}{4}(1-\frac{1}{2}(1+\cos(4x))) = \frac{1}{4}(1-\frac{1}{2}-\frac{1}{2}\cos(4x))=\frac{1}{8}(1-\cos(4x)).[/tex]
This is now integrable (phew),
[tex]\int \sin^2x\cos^2x \ dx = \int \frac{1}{8}(1-\cos(4x)) \ dx = \frac{1}{8} \int (1-\cos(4x)) \ dx = \frac{1}{8}(x-\frac{1}{4}\sin(4x))+C.[/tex]
