Question

How do you integrate the following below?

Answer 1

\[\int\limits_{}^{}-cosx \cos ^{2}x\]

Answer 2

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

Answer 3

Use that property to apply substitution-method.

Answer 4

\[\cos ^2(x)\] can be rewritten as \[1 - \sin ^2 (x)\]

which gives you

\[\int\limits_{}^{}-cosx(1-\sin^2x)\]

Multiply it out

\[\int\limits_{}^{}-cosx + cosxsin^2x\]
Separate into two integrals

\[\int\limits_{}^{}-cosx + \int\limits_{}^{}cosxsin^2x\]
The first integral becomes \[-sinx\]

The second integral requires u-substitution

Let u = sinx and du = cosxdx

Then the integral becomes

\[\int\limits_{}^{} u^2du\]

And when we integrate that, we get

\[u^3/3\]

Plug sinx back in for u to get

\[\sin^3(x)/3\]

Add this second integral to the first one to get

\[\sin^3(x)/3 - sinx\]

Answer 5

okay maybe i messed up my work before I got to that... the original problem states...

Find the general solution of \[y'-y \tan x = \cos ^{2} x\]

and what I had was \[y(-\cos x)=\int\limits_{}^{}-\cos x \cos ^{2}x\]

is that wrong?

Answer 6

and then i needed to integrate the right side... but doing that i do not get the correct answer.

Answer 7

\[\large y'+(-\tan x)y=\cos^2x\]

\[\huge \mu=e^{\int\limits -\tan x dx}\]\[\huge \mu=e^{\ln|\cos x|} \qquad \rightarrow \qquad \mu=\cos x\]

I think u missed the negative sign at the start. The rest looks good though.

Answer 8

my fault... it was supposed to be \[y'+y \tan x = \cos ^{2} x\]

but then i would still get \[\mu = -\cos x\]

then when i multiple by mu i get\[y'(-\cos x) -\cos x \tan x = - \cos x \cos ^{2}x\]

Answer 9

which equals \[[(-\cos x) y']'=-\cos x \cos ^{2}\]

then i just have to integrate both side right?

Answer 10

\[\large \left[(-\cos x)y\right]'=-\cos^3 x\]Yes, correct :) Looks good so far.

Answer 11

You accidentally had an extra prime on the y, i was just fixin that :3

Answer 12

opps yes... and then when i integrate the right side... i get \[-\sin x + \frac{ 1 }{ 3 }\sin ^{3} x\]

Answer 13

is that correct?

Answer 14

then dividing both sides by -cos x i get\[\frac{ -\sin x }{ -\cos x }+\frac{ \sin ^{3}x}{-3\cos x }\]

Answer 15

but that is way different than the answer i am given

Answer 16

ohhh opps and there is also a C/-cos x in there

Answer 17

I think I'm getting \[\large \sin x-\frac{1}{3}\sin^3x+C\]Way different? :O oh my

Answer 18

1/2 sin 2x + C cos x

this is what the answer says

Answer 19

Hmmmmmm

Answer 20

The book clearly applied the Sine Double Angle formula.
So if we undo that, we're trying to get to an answer of,
\[\large y=\sin x \cos x+ C \cos x\]

Answer 21

OH OH OH i see the problem.

Answer 22

It was way back near the start, darn it!!!

Answer 23

\[\large \int\limits \tan x dx=-\ln |\cos x|\]

\[\huge e^{- \ln|\cos x|}\]We are NOT allowed to cancel the base e and the log since the negative is in the way.
We must bring the negative into the log first.

Answer 24

ohh i just assumed it was just -cos x, so then this is not the case? what do you do with the negative then

Answer 25

Remember your rule of logs? You can bring the constant multiplier INTO the log as an exponent.\[-\ln(\cos x) \qquad = \qquad \ln\left(\frac{1}{\cos x}\right) \qquad = \qquad \ln(\sec x)\]

I brought the negative one in as a power. I didn't want to write it as cos^-1 x since it's easi to confuse that with the inverse trig function :) lol

Answer 26

ohhh so mu = sec x then

Answer 27

Yah looks like it :)
That's where we'll end up getting all those extra cosines from.

Answer 28

i will attempt the calculation again.

Answer 29

so since it is now \[\int\limits_{?}^{?}\sec x \cos ^{2}x\
so i still use u=sin x

Answer 30

\[\large \left(y \sec x\right)'=\int\limits \sec x \cos^2x dx \quad = \quad \int\limits \cos x dx\]

Answer 31

Convert your secant into it's cosine equivalent, and you'll make a nice cancellation from there.

Answer 32

\[\large \int\limits \left(\frac{1}{\cos x}\right)\cos^2x dx\]

Answer 33

ohhh 1/cos x

Answer 34

and then once you divide by sec x, you convert it back to cos x and get cos x sin x + c cos x

Answer 35

Hmm yah that worked out nicely! :)

Answer 36

and then the sinx cosx = 1/2sin 2x

Answer 37

got it... awesome... thank you... only 15 more to go. ekks.