At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

\[\int\limits_{}^{}3\sin^2(t)\cos^4(t)dt\]

I don't know what to do after :( .

The 4th step should be a dt.

Is there no other way?

I mean I did get a little further.

\[\frac{ -3 }{ 8 }\int\limits\limits_{}^{}(\cos^2(2t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt\]

i would start with
\[\int\cos^4(x)dx-\int\cos^6(x)dx\] and then look in the back of the book

\[\frac{ -3 }{ 8 }\int\limits\limits\limits_{}^{}((1-\sin^2(t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt\]

Where did that come from?

Yeah, I know. I turn them into the half angles, which I did.

That stupid cos^2(2t) is getting in the way.

again*

I know. I have to do it though.

then i really recommend looking up the formula for \(\int\cos^n(x)dx\)

So I get:\[\frac{ -3 }{ 8 }\int\limits_{}^{}(1-\sin^2(2t))\cos(2t)+(1-\sin^2(2t)+\cos(2t)+1)dt\]

Hmm good point.

Sec.

Can I ask a dumb question?

sure so i can seem dumber if i cannot answer it

yes (whew)

ignore the annoying 3, put it in at the end

Hehe. Thanks. This should be easy now.

Ohh that's MASSIVELY convenient.

it is going to be a pain, but much less of a pain than what you were doing

Why 4 and 6?

I have 3 and 2.

\[\sin^2(x)\cos^4(x)=(1-\cos^2(x))\cos^4(x)=\cos^4(x)-\cos^6(x)\]

so 4 and 6

Right.

Thanks!

have fun!

yw