solve this integral :/ not sure how to do this!

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- anonymous

solve this integral :/ not sure how to do this!

- jamiebookeater

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- anonymous

\[\int\limits_{} \sin^5 t \cos^4 t dt\]

- Zale101

|dw:1442984817219:dw|

- Astrophysics

Some general rules with these trig functions |dw:1442984972414:dw|

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## More answers

- Zale101

Any ideas of what to do next?

- Zale101

That chart is very helpful! @astrophysics

- Astrophysics

Yeah, it's pretty awesome :P

- Astrophysics

http://puu.sh/kl22A/732d78a86f.png if anyone wants the link XD

- anonymous

sorry!! i am back!! :) i am reading the chart right now :) one sec!! hopefully i can have a clue what happens next:)

- anonymous

okay, i see now what you did @Zale101 so i strip one sin away and get|dw:1442985650253:dw|

- anonymous

not sure what to do next though :/

- Astrophysics

|dw:1442985664022:dw|

- anonymous

ohh so i would get this? |dw:1442985687508:dw|

- Zale101

|dw:1442985746191:dw|

- anonymous

okay!! i forget though, from here, do i sub something in? :/

- Zale101

U-Sub! :)

- anonymous

so we get integral sin t (sin^2 t)^2 u^4 dt ?

- Zale101

Nope. First of all, you cant have u while the rest of the terms are in t. Second, you have to use trig identities for (sin^2t)^2

- anonymous

ohhh how do i do that then?

- Zale101

\((sin^2t)^2=(1-cos^2t)^2\)

- anonymous

ohh the square thing?

- Zale101

\(sin^4t=(sin^2t)^2=(1âˆ’cos^2t)^2\)

- anonymous

confused which trig functions to insert in now?

- Zale101

Yes. If you wan to replace sin^2 with the trig identities 1-cos^2t. But sin is squared, so the identity has to be squared as well.

- anonymous

ohh but it is fine leaving like sin? what happens next?

- Zale101

Next, you do u-sub. But first show me your integral now after using trig identities.

- anonymous

\[\int\limits_{} (1-\cos^2 t)((1-\cos^2)^2 t)^2 cost^4 t dt\] this?

- Zale101

\[\int\limits_{}^{}\sin(t) (\sin^2(t))^2\cos^4(t)dt\]
\[\int\limits_{}^{}\sin(t) (1-cos^2(t))^2\cos^4(t)dt\]

- anonymous

ohh okay oopsies :P what happens now? :O

- Zale101

What you did was wrong because you can replace sint with 1-cos^2
trig identities say
\(sin^2t=1-cos^2\)

- anonymous

ohhhh:O

- Zale101

\(sin^2t+cos^2t=1\)

- Zale101

What is your next step now?

- anonymous

i integrate? :/

- Zale101

How?

- anonymous

i am not sure :(

- anonymous

i am sorry :( i am very bad at calc:(

- Zale101

We need u-sub here. What would be you u in this case?

- Zale101

your*

- anonymous

u=sin(t) ?

- Zale101

Would u=sint work?

- anonymous

i think so :/

- Zale101

I will try u=sint and see if it works.

- anonymous

ok!! sorry, i get very confused with these concepts :/

- Zale101

|dw:1442986993486:dw|

- anonymous

ohhh okay :/ what should we do then?

- Zale101

try u=cos(t)

- Astrophysics

A saying I remember, get the odd man out! You want the odd man out!

- anonymous

okay!! where do we sub that into ? :/

- Zale101

With integration, here is what you should do. Every time you make an educated guess on what might the u be, do the integration and see if it works.If not, then try something else. This is what i did with integral and it was helpful.

- Zale101

integral problems*

- anonymous

ohhh okie :) so we should use u=cos(t)
and so we do this? \[\int\limits_{}sint(1-u)^2 u^4 dt \]

- Zale101

You cant place in the u when you have not finished u-subbing

- anonymous

ohh no! I'm sorry, i don't know what to do here :(

- Astrophysics

When you do a u sub, you must also get the derivative, that's what will cancel out the sint \[u= cost \implies du= - sint dt\]

- Astrophysics

You don't want t and u's, just one variable

- anonymous

okay! what would it look like then?

- Zale101

Do u=cost and du=-sint dt and tell me what you will get

- anonymous

i get this?
is it correct?
integral with plug in u and du udu and i sub in and replace and get this? -1/5cos^5t+2/7cost^7t-1/9cos^9t+C?

- anonymous

thank you so much @Zale101 @Astrophysics i will try to figure this one out some more tomorrow! i need to sleep now because it is 2am here and i have to be up in four and a half hrs!! thanks so much for your help!!! :D sorry it was hard for me to understand!! have a nice night!!

- Zale101

Alright, good luck and take care!

- Astrophysics

Good night! :)

- Zale101

Just a type error i made
"What you did was wrong because you \(\color{red}{can't}\) replace sint with 1-cos^2"

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