Question

Evalute the integrals. Stuck on these two problems<< could someone explain?

Answer 1

O.O @Luigi0210 psst help meh here

Answer 2

Distribute the first one. Kainui will be of better help tho~

Answer 3

The first one becomes a lot simpler to look at if you just multiply it out. You should recognize immediately that you have 2 things you can integrate already and some extra piece that you might not.

But then remember, sin^2x+cos^2x=1 can be transformed by dividing both sides by cos^2x to get:
\[\tan^2x+1=\sec^2x\]

Answer 4

http://openstudy.com/study#/updates/52cf9501e4b039126010fa67 plz help

Answer 5

ok lemme try that out

Answer 6

lol Once you're done with the first one and it makes sense if you need to ask any questions, like maybe it's not that obvious, that's ok just ask!, then we can do the second one.

Answer 7

ok so to check if I'm starting correctly, I multiply it out which would give me \[\sec ^{2}x + 2secxtanx+\tan ^{2}x\]

Answer 8

Exactly.

Answer 9

If you're familiar with the derivative of tan(x) and sec(x) you should be fine. If not, it's time to review real quick.

Answer 10

Now you can just look at it as:
\[\LARGE \int sec^2xdx+\int 2secxtanxdx+\int tan^2xdx\]

Answer 11

Would \(t\)-substitution work?

Answer 12

@Luigi0210 It's subtle, but I think making the substitution tan^2x=sec^2x-1 before separating it apart makes it a little nicer to look at since the sec^2x terms will combine.

Answer 13

hold up one sec i think i i got it!

Answer 14

Awesome! =)

Answer 15

@Kainui I see what you mean now~

Answer 16

The second integral relies on this trigonometric identity:

\[\cos^2( \theta)=\frac{ 1+ \cos(2 \theta) }{ 2 }\]

That's all the help I'm going to give for the second integral until you try it out. You might not even need any more help than this if you're lucky and clever enough.

Answer 17

ok so for the first intera I got tanx, second, 2secx but what would be the last one, tan^2x?

Answer 18

Like kai said, replace the tan^2x with a trig identity c:

Answer 19

So try:
\[\LARGE \int (sec^2x-1)dx\]

Answer 20

ohh. got it. is the answer 2tanx+2secx-x+C?

Answer 21

That looks like it. =)

Answer 22

thank you! now I'll try the second one(:

Answer 23

Good luck~

Answer 24

i can use the double angle formula and change it so that that cos4x=1-2sin^2(2x) right??

Answer 25

@Kainui can help you with last one, I got no idea really on how to do it.

Answer 26

i'd have \[\int\limits_{0}^{\pi/4} \sqrt{2-2\sin ^{2}2x}\] look right?

Answer 27

and thanks @Luigi0210 :) haha I hate these problems cuz if u don't see what to do u just stare at it like wtf-_- lol

Answer 28

^Exactly

Answer 29

Sure, so there are two half angle formulas. Although you can use the one for sine like you are using, it's not the one I would suggest since it is still sort of confusing looking.

Here are both half angle formulas side by side:

\[-1+\cos(4x)=-2\sin^2(2x)\]\[1+\cos(4x)=2\cos^2(2x)\]

So here you can plug in the whole (1+cos4x) part into the bottom of the square root to get:

\[\sqrt{2}\int\limits_{0}^{\pi/4}\cos (2x)dx\]

Answer 30

ok i see! I've never been taught the half angle formulas so thank you for that(:

Answer 31

If you ever remember one half angle formula and not the other since one is basically adding and the other is subtracting and sine/cosine is if you remember:

\[\sin^2 \theta+ \cos^2 \theta =1\] and\[\frac{ 1 }{ 2}-\frac{ 1 }{ 2}\cos(2 \theta)=\sin^2(\theta)\]

You can plug in and find the other:
\[\frac{ 1 }{ 2}-\frac{ 1 }{ 2}\cos(2 \theta)+\cos^2(\theta)=1\]\[\cos^2(\theta)=\frac{ 1 }{ 2}+\frac{ 1 }{ 2}\cos(2 \theta)\]

Answer 32

got it! Thanks for that(:
and is the answer sq(2)/2?

Answer 33

I actually sort of don't memorize those though... I sort of have the double angle formulas and sinx*cosx=(1/2)sin(2x) as visualized.



So you can see it starts out at 0 and goes + and then 0 then negative, then 0. So it must be a sine function:

sinx*cosx=A*sin(f*x)

but what's the amplitude and frequency of this sine wave?

Well it looks like if you go around the circle it actually does two complete sine waves, +,-,+,- right? So f=2

What's the amplitude? Well if you think about it, right at 45 degrees both functions are sqrt(2)/2 and if you go any other direction from that, you'll have to decrease one of the functions. You could even use calculus to make sure that this is the maximum, but it's just intuition to help.

So both are sqrt(2)/2, multiply them together: (sqrt(2)/2)^2=2/4=1/2=A

sinxcosx=(1/2)sin(2x)

Maybe that's just weird or not worth your time. Not everyone likes math as much as me haha.

Answer 34

Also yes, you're right that's the answer! =)

Here's a good place to check your answers too, they have a show steps button that comes in handy:

http://www.wolframalpha.com/input/?i=integral+0+to+pi%2F4+sqrt%281%2Bcos%284x%29%29dx

Answer 35

Wow! You know you're really really good at teaching!! That is so clever. lol i just bookmarked this page so i could find this easily. Thank you so much! Can u be my private tutor lmao You rock!

Answer 36

If you want me to go into more depth on my intuition of the double angle formulas and sinx*cosx I can probably explain it better haha.

Also, when I reasoned out that you could find that 45 degrees is where the peaks are:

\[d/dx(sinx*cosx)=\cos^2x-\sin^2x=0\] The slope is zero on a peak.

\[\cos^2x=\sin^2x\]

which.... surprise... \[\cos x= \sin x\]

happens at 45. =P

Glad I could help!