Question

help with some integrals

Answer 1

\[\int\limits_{0}^{3\pi/2}\left| 3\sin(x) \right|\]

Answer 2

in 3rdQuadrant sinx is negative, but asyou are taking modulus it should be positive, so calculate that part separately

Answer 3

I got 3, but says it isn't right

Answer 4

|3sinx| = 3sinx when x is between 0 and pi.

When we go beyond pi,
our sine function is negative,
so our absolute must be giving us -3sinx to end up with a positive,
|3sinx| = -3sinx when x is between pi and 3pi/2.

Answer 5

\[\large\rm \int\limits_0^{3\pi/2}|3\sin x|dx\quad=\quad \int\limits_0^{\pi} 3\sin x dx-\int\limits_{\pi}^{3\pi/2}3\sin x dx\]

Answer 6

oh ok I see, thanks

Answer 7

:)

Answer 8

the ans is 9 i believe

Answer 9

what about for this one?
\[\int\limits_{}^{}\frac{ 2\sin(t) }{ 1-\sin ^2(t) }\]

Answer 10

Recall your Pythagorean Identity: \(\large\rm 1-sin^2x=cos^2x\)
That should help get things started.

Simple u-substitution from there.

Answer 11

still need some help with it

Answer 12

\[\large\rm \int\limits \frac{2\sin t}{1-\sin^2t}dt\quad=\quad 2\int\limits \frac{\sin t~dt}{\cos^2t}\]Ok with these steps? :)

Answer 13

Oh maybe we should write it like this,\[\large\rm =2\int\limits\frac{\sin t~dt}{(\cos t)^2}\]Might be easier to see your substitution that way.

Answer 14

\[\large\rm =2\int\limits\limits\frac{\sin t~dt}{(\color{royalblue}{\cos t})^2}\]So if we assign\[\large\rm \color{royalblue}{u=\cos t}\]What do we get for our du?

Answer 15

sin t

Answer 16

Derivative of cosine is -sine, right?
So we have a tiny bit of work to do,\[\large\rm du=-\sin t~dt\]

Answer 17

so wouldn't you move the negative over to the du

Answer 18

Hmm that seems good,\[\large\rm -du=\sin t~dt\]Ah nice! We were able to produce our numerator, ya?

Answer 19

yeah, so it's
\[2\int\limits_{}^{}\frac{ -1 }{u^2 }du\]

Answer 20

\[\large\rm =\quad-2\int\limits u^{-2}du\]Mmm k great.
Now power rule from this point.

Answer 21

power rule?

Answer 22

Yes, power rule for integration:\[\large\rm \int\limits x^n dx\quad=\quad\frac{1}{n+1}x^{n+1}+C\]

Answer 23

oh duh

Answer 24

I put 2arccos(t) in and it said it was wrong

Answer 25

Oh oh, ya you're being silly...
Lemme show you...

Answer 26

\[\Huge\rm \frac{1}{\cos x}\quad\ne\quad \cos^{-1}x\]

Answer 27

This -1 notation (which is the same as arc),
is reserved for the inverse function.

It's not the same as a reciprocal thing like this.

Answer 28

I feel stupid now

Answer 29

Yes, it works for most things,
not trig though :)
We reserve that -1 for inverse, not power.

Answer 30

I got it now, thanks