Question

How to evaluate the indefinite integral using u substitution?

Answer 1

\[\int\limits \frac{ \sin(4t-1) }{ 1-\sin^2(4t-1) }dx\]

Answer 2

would u be 4t-1?

Answer 3

it says " hint: express the sin in terms of cos and use substitution"

Answer 4

Your substitution MUST satisfy the following:

Both "u" and "du" MUST be present before you can do any integration.

Answer 5

sin^2 (4t-1) = 1 - cos^2 (4t-1)

Answer 6

If you choose u=4t+1, then du/dt = 4, so that du=4dt. Do you see a '4dt' in the given integral? If not, then u=4t+1 is inappropriate.

Answer 7

Hint: There is an identity involving the square of the sine function AND the square of the cosine function. Find and type in that identity here.

Answer 8

sin^2 (4t-1) = 1 - cos^2 (4t-1) is correct.

Use this to re-write the denominator.

Answer 9

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

Answer 10

You have merely substituted the square of the cosine of (4t-1) for the square of the sine of (4t-1). No. Troubleshoot this.

Answer 11

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

Answer 12

that's a lot better.

Try again: choose an appropriate u.

u=

du/dt =

Answer 13

u = sin(4t-1)
du/dt = 4cos(4t-1)

Answer 14

If this is correct, then what is du?

du=?

Answer 15

du=4cos(4t-1)dt

Answer 16

Do you have this du in the numerator of your original integrand?

Answer 17

no

Answer 18

Then please try out some other substitution, u.

u=

du=

Answer 19

u = cos(4t-1)
du/dt = -4sin(4t-1)
du = -4sin(4t-1)dt

Answer 20

Does the original integrand resemble =4sin(4t-1)dt?

Answer 21

Does the original integrand resemble -4sin(4t-1)dt?

Answer 22

sort of. would it be like this?
\[\int\limits \frac{ \sin(4t-1) }{ u^2 }dt=-4 \int\limits \frac{ du }{ u^2 }\]

Answer 23

Surely would!
What next?

Answer 24

\[-4 \int\limits \frac{ du }{ u^2 }=-4\int\limits u ^{-2}du=-4\frac{ -2 }{ u^3}=\frac{ 8 }{ \cos(4t-1) }+C\]

Answer 25

You have 3 = signs in there. The first 2 quantities, involving the first = sign, are correct. The last two quantities are incorrect. Double check your integration and substitution, please.

Answer 26

\[ -4 \int\limits u ^{-2}du = -4 \frac{ -1 }{ u }+C =\frac{ 4 }{ \cos(4t-1) }+C\]

Answer 27

that looks a lot better. any questions about this work?

Note that 1 / cos x = sec x. Would that have any impact on your final answer?

Answer 28

4sec(4t-1) ?
i put the answer in on my hw online and it says its wrong.

Answer 29

\[du = -4\sin(4t-1)dt\]

\[-\frac{1}{4}du = \sin(4t-1)dt\]

Answer 30

We have already covered the most important steps towards finding the integral.
You may have to do some follow-up checking of our work.
I would be happy to address specific questions from you, but not to go through the entire problem again.

Answer 31

ah got it, thank you @Zarkon

Answer 32

Zarkon: Thank you very much.
darkigloo: I'd suggest you backtrack to the point where you found and used "du" and incorporate Zarkon's contribution.

Answer 33

Thank you @mathmale

Answer 34

You're welcome! Thanks for your perseverance.