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Integration !!!

Where am I going wrong? thank you
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Reply 1
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Notnek
21
Original post by otrivine
Where am I going wrong? thank you

When you substituted u, you forgot to write your integral in terms of u. What you've actually said is

−4∫sin⁡2tcos⁡t dt=4∫sin⁡tcos⁡t dt\displaystyle -4\int \sin^2 t \cos t \ dt = 4\int \sin t \cos t \ dt,

which is clearly not correct.

Try a substitution of u=sin⁡t\displaystyle u=\sin t instead of u=cos⁡t\displaystyle u=\cos t and post your working if you're still stuck.
Reply 2
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otrivine
OP
14
Original post by notnek
When you substituted u, you forgot to write your integral in terms of u. What you've actually said is

−4∫sin⁡2tcos⁡t dt=4∫sin⁡tcos⁡t dt\displaystyle -4\int \sin^2 t \cos t \ dt = 4\int \sin t \cos t \ dt,

which is clearly not correct.

Try a substitution of u=sin⁡t\displaystyle u=\sin t instead of u=cos⁡t\displaystyle u=\cos t and post your working if you're still stuck.



but its correct cause you use the identity of sin2x=2sinxcosx to get it into that form
Reply 3
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Notnek
21
Original post by otrivine
but its correct cause you use the identity of sin2x=2sinxcosx to get it into that form

You're not understanding what I'm getting at.

Notice that from the line −4∫sin⁡2tcos⁡t dt\displaystyle-4\int \sin^2 t \cos t \ dt

to the line

4∫12sin⁡2t dt\displaystyle 4\int\frac{1}{2}\sin 2t \ dt

you have used a substitution u=cos⁡tu=\cos t but your integral is still in terms of t.

Think about how you normally use a substitution. You normally end up with an integral in terms of u if you use a u substitution right?


This is how your working should read with a substitution u=cos⁡tu=\cos t:


−4∫sin⁡2tcos⁡t dt=−4∫sin⁡2tcos⁡t dtdu du=4∫sin⁡tcos⁡t du=...\displaystyle -4\int \sin^2 t \cos t \ dt = -4\int \sin^2 t \cos t \ \frac{dt}{du} \ du = 4\int \sin t\cos t \ du = ...


Notice that there is a du on the RHS which is different to your working. I recommend you don't continue the working from here and use a u=sin⁡tu=\sin t sub instead.

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