Question

Help me PLS...

Answer 1

I made a mistake somewhere.... lol

Answer 2

hmmm
\[\tan^2 \theta =\sec^\color{Red}{2}\theta -1 \]
correct ??

Answer 3

Oh yea.....

Answer 4

But then i get backed to where i started.... Eh

Answer 5

the identity wouldn't work for sec^3 theta
trickyyyy question

Answer 6

Wait... can this work?

Answer 7

are you going to post something ???

Answer 8

Yea sorry.

Answer 9

oh okay cool take ur time :=))

Answer 10

after using the identity you still get sec^3 but you have to do u sub there first and then sub in theta value

Answer 11

wait that looks good gimme a sec plz :D

Answer 12

Okay :3

Answer 13

Oh i think my professor solved \[\int\limits \sec^3 x dx\] using Integration by parts.

Answer 14

i just did this question last week and now i totally forgot
but i do remember we have to do u sub at the end that will give us same thing (Sec^3) and then we have to add (integral sec^3theta) both sides something like that hmm\[\int\limits_{0}^{\frac{\pi}{4}} \sec \theta + \int\limits_{0}^{\frac{\pi}{4}} \sec^3 \theta\]

Answer 15

Like this?

Answer 16

yeahhh _-_
and there is 81 so make sure you distribute that

Answer 17

Oh yea. thanks

Answer 18

I think i did something wrong.... This is so messy.

Answer 19

you forgot to distribute the numerator by 81 and there is sign error

Answer 20

Huh. Where? D:

Answer 21

\[81 \int\limits_{0}^{\frac{\pi}{4}} \sec^3 \theta ~~d \theta \]\[81 \int\limits_{0}^{\frac{\pi}{4}} sec \theta ~~~sec^2 \theta ~~d \theta \]\[81 \int\limits_{0}^{\frac{\pi}{4}} sec \theta ~~~\color{Red}{sec^2 \theta} ~~d \theta \]
\[u= \sec \theta ~~~~ du= \sec \theta ~\tan \theta ~d \theta\]
\[dv = \sec^2 \theta ~~~~ v= \tan \theta \]
\[81[ secx \tan x - \int\limits_{ }^{}\tan( \sec x \tan x) d \theta]\]\[81 [\sec x \tan x -(\int\limits_{ }^{} \sec^3 x -\sec x d \theta)]\]
i'm using x for theta( i'm tired typing 5 letter word lol

Answer 22

good job @Nnesha

Answer 23

put the limits in, please

Answer 24

\[81 [\sec x \tan x -(\int\limits_{ }^{} \sec^3 x -\sec x)]\]
\[81[\sec x \tan x + \int\limits_{ }^{} \sec x - \int\limits_{ }^{} \sec^3 x d \theta)]
\] distribute by 81
\[81 \sec x \tan x + 81 \ln \left| \sec x + \tan x \right| - 81 \int\limits_{ }^{} \sec^2 \theta d \theta\]
and ofcourse from 0 to pi/4
i'm too lazy .. and the headache is killing me

Answer 25

thanks Loser66 :=))

Answer 26

Is this calc? Seems like a lotttta work

Answer 27

that last sec is cubed right?

Answer 28

oh...................no!!

Answer 29

ohh yeahh sec^3 theta dtheta

Answer 30

Yea its Calc 2 @Seratul

Answer 31

@Loser66 ?

Answer 32

and i hope that's correct sid if not then... oh well

Answer 33

Haha no worries. Thanks for trying :P

Answer 34

you are correct until \(81\int_0^{\pi/4} sec^3 x dx\)

Answer 35

Now, just sec^3 x dx
let u = sec x, du = sec tan , dv = sec^2 dx, v = tan x
hence you have
\[81\int_0^{\pi/4}= sec xtanx|0^{\pi/4}-\int_0^{\pi/4}secx tan^2xdx\]

Answer 36

Sorry, I am just trying to process all this. My brain feel really dead right now... lol

Answer 37

multiple both sides by 81/82, you get what you want.

Answer 38

oh, too many experts here. hehehe. I should shut up.

Answer 39

but still have to use the triangle right ??

Answer 40

I don't think so. No need. I just use algebraic way, not geometric way to solve it.

Answer 41

yeah but there is sec theta + tan theta
we have to replace that with the values ??
sec x= sqrt{X^2+81}/9 and tan x= x/9

Answer 42

I am lost as to where you go \[\frac{ 2 }{ \sqrt{2} }\]

Answer 43

sec x tan x from 0 to pi/4= sec (pi/4) tan(pi/4)- sec (0) tan (0) = 2/sqrt2

Answer 44

I may make mistake at somewhere, but you got the idea!! hehehe... check it.

Answer 45

Lol okay... xD

Answer 46

right