Question

help evaluating integral using u substitution

Answer 1

Try identifying a possible u and then find the associated du/dx (or du). Can you easily find both u and du in your integrand?

Answer 2

\[\int\limits \frac{ \cos(t) }{ (3\sin(t)+12)^2 } dt \]
im not sure what to use as u.

Answer 3

Where is the more mathematically complicated part of y our integrand?

Answer 4

the denominator

Answer 5

That's it!

Answer 6

\[u=3\sin(t)+12\]

Answer 7

Yes,a nd what's du?

Answer 8

What is the derivative of sint?

Answer 9

does it resemble the cos t dt of the integrand?

Answer 10

du /dt= 3cos(t)

Answer 11

So your du is the same as the numerator of the integrand except for a constant multiplier. That's a good sign, a very, very good sign. ;)

Answer 12

So, subst. u and du, we get what new integral?

Answer 13

So we have \[\int\limits\limits \frac{ \cos(t) }{ (3\sin(t)+12)^2 } dt \] where \[u= 3\sin(t)+12 \implies \frac{ du }{ dt } = 3\cos(t)\] keep going, I'll make it a bit more simpler, \[du = 3 \cos(t)dt\]

Answer 14

Very neat work. I appreciate it.

Answer 15

\[\int\limits \frac{ \cos(t) }{ (3\sin(t)+12)^2 }dt =\int\limits \frac{ \cos(t) }{ u^2 }dt=3\int\limits \frac{ du }{ u^2 }\]
or is it 1/3 ?

Answer 16

\[du = 3 \cos(t)dt\]

Answer 17

so du/3 = cos t dt.

What does this tell you? You're substituting du/3 for cos t dt.

Answer 18

Careful, should be 1/3 note \[du = 3 \cos(t)dt \implies \frac{ du }{ 3 } = \color{red}{cos(t) dt}\]\[\int\limits\limits\limits \frac{ \color{red} {\cos(t)} }{ (3\sin(t)+12)^2 } \color{red}{dt}\] compare the two

Answer 19

@astrophysics: Great graphics, accurate feedback!

Answer 20

Thanks @mathmale :)

Answer 21

\[\frac{ 1 }{ 3 } \int\limits \frac{ du }{ u^2} = \frac{ 1 }{ 3 } ( -\frac{ 1 }{ u })+c=\frac{ -1 }{ 9\sin(t)+36}\]

Answer 22

Yup, looks good!

Answer 23

thank you! @Astrophysics @mathmale

Answer 24

:-)