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3 years ago

Help Me Please !! THX!

Mathematics

1 answer:

solong [7]3 years ago

8 0

Answer: decreases

Step-by-step explanation:

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Wewaii [24]

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3 years ago

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HELP ASAP<br> Find the coordinates of B

Elza [17]

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the answer is B

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1 year ago

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3 years ago

Need help on this calculus

N76 [4]

$\displaystyle\int\sin^3t\cos^3t\,\mathrm dt$

One thing you could do is to expand either a factor of $\sin^2t$ or $\cos^2t$ , then expand the integrand. I'll do the first.

You have

$\sin^2t=1-\cos^2t$

which means the integral is equivalent to

$\displaystyle\int\sin t(1-\cos^2t)\cos^3t\,\mathrm dt$

Substitute $u=\cos t$ , so that $\mathrm du=-\sin t\,\mathrm dt$ . This makes it so that the integral above can be rewritten in terms of $u$ as

$\displaystyle-\int(1-u^2)u^3\,\mathrm du=\int(u^5-u^3)\,\mathrm du$

Now just use the power rule:

$\displaystyle\int(u^5-u^3)\,\mathrm du=\dfrac16u^6-\dfrac14u^4+C$

Back-substitute to get the antiderivative back in terms of $t$ :

$\dfrac16\cos^6t-\dfrac14\cos^4t+C$

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2 years ago

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Pls answer this i really need it

jeka57 [31]

Answer:

8.15

Step-by-step explanation:

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2 years ago