What is the antiderivative of sin^2(x)cos^2(x)?

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,

And so inverting these,

.

Multiply them together to obtain an equivalent expression for sin^2(x)cos^2(x) in terms of cos(2x).

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).

So then,

This is now integrable (phew),

the equation of the parallel line would be y = 2x + 9

step-by-step explanation:

to start we need to find the slope of the original line. we can do that by looking at the coefficient of x, which is 2

since the slope of the original line is 2 and parallel lines have the same slope, we know the new line will also have 2 slope. now we can use that along with the point given in point-slope form to get the find equation.

y - y1 = m(x - x1)

y - 5 = 2(x + 2)

y - 5 = 2x + 4

y = 2x + 9

the answer is -1 1/5, also a

hope i

-lavira : )