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Trigonometric Integrals [1]

shamieh

Active member
Sep 13, 2013
539
Stuck on this problem.

Evaluate
\(\displaystyle
\int \cos^{2}x \, \tan^{3}x \, dx\)

What I have so far:

used the trig identity sin/cos = tan
factored out a sin so I can have a even power.
changed \(\displaystyle \sin^{2}x\) to its identity = 1/2(1 - cos2x)
combined like terms and canceled out the cos

\(\displaystyle \int \cos^{2}x * \frac{\sin^{2}x}{\cos^{3}x} * \sin x \, dx\)

\(\displaystyle \frac{1}{2} \int \frac{(1 - \cos 2x)}{\cos x} * \sin x \, dx\)

Now I'm not sure what to do.

Should i do u = cosx? but then won't i get -2sin(2x)dx which isn't in the problem above
 
Last edited by a moderator:

Pranav

Well-known member
Nov 4, 2013
428
I suggest writing $\sin^2x$ as $1-\cos^2x$.
 

shamieh

Active member
Sep 13, 2013
539
I suggest writing $\sin^2x$ as $1-\cos^2x$.
That's what I saw someone else doing but I thought $\sin^2x$ = 1/2(1 - cos2x) ? So how can you write it as that??
 
Last edited:

Pranav

Well-known member
Nov 4, 2013
428
That's what I thought someone else doing but I thought $\sin^2x$ = 1/2(1 - cos2x) ? So how can you write it as that??
I am not sure what you ask. $\sin^2x=(1/2)(1-\cos(2x))$ is perfectly valid but it doesn't help in the given problem. We are using the substitution u=cos(x) so we would like to have most of the terms as $\cos(x)$.
 

shamieh

Active member
Sep 13, 2013
539
so would I have


$-\frac{1}{2} \int \frac{1}{u} - \frac{u^2}{u} * du$

which turns out to be

$-\frac{1}{2} [ln|u| - \frac{u^2}{2}] = -\frac{1}{2} ln|cosx| + \frac{cos^2x}{4} + c$ ?
 

soroban

Well-known member
Feb 2, 2012
409
Hello, shamieh!

[tex]\int \cos^2\!x \tan^3\!x \, dx[/tex]

[tex]\cos^2\!x\tan^3\!x \:=\:\cos^2\!x\frac{\sin^3\!x}{\cos^3\!x} \:=\;\frac{\sin^3\!x}{\cos x} \:=\:\frac{\sin^2\!x\cdot\sin x}{\cos x} [/tex]

. . [tex]=\;\frac{(1-\cos^2\!x)\sin x}{\cos x} \:=\: \left(\frac{1}{\cos x} - \cos x\right)\sin x [/tex]

. . [tex]=\;\frac{\sin x}{\cos x} - \sin x\cos x \:=\:\tan x - \sin x\cos x[/tex]


[tex]\int(\tan x - \sin x\cos x)dx \:=\:-\ln|\cos x| - \tfrac{1}{2}\sin^2x + C[/tex]
 

Pranav

Well-known member
Nov 4, 2013
428

shamieh

Active member
Sep 13, 2013
539
Hello, shamieh!


[tex]\cos^2\!x\tan^3\!x \:=\:\cos^2\!x\frac{\sin^3\!x}{\cos^3\!x} \:=\;\frac{\sin^3\!x}{\cos x} \:=\:\frac{\sin^2\!x\cdot\sin x}{\cos x} [/tex]

. . [tex]=\;\frac{(1-\cos^2\!x)\sin x}{\cos x} \:=\: \left(\frac{1}{\cos x} - \cos x\right)\sin x [/tex]

. . [tex]=\;\frac{\sin x}{\cos x} - \sin x\cos x \:=\:\tan x - \sin x\cos x[/tex]


[tex]\int(\tan x - \sin x\cos x)dx \:=\:-\ln|\cos x| - \tfrac{1}{2}\sin^2x + C[/tex]
I believe there is a error here. I don't understand hwo you have a sin^2x in the final answer

- - - Updated - - -

How do you get a factor of $1/2$?

I see what I was doing i was substituting the 1/2(1 - cosx) or whatever it was and bringing the 1/2 out as a constant which was WRONG.
(I was using the wrong identity.) Changed that to $ 1 - cos^2x$ to fix the problem


so you end up with just \(\displaystyle -[ln|u| - \frac{u^2}{2}] = -ln|cosx| + \frac{cos^2x}{2} + c\)
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
shamieh said:
I believe there is a error here. I don't understand how you have a sin^2x in the final answer
Well, it's equivalent to your result if you see it carefully enough. $\sin^2(x) = 1 - \cos^2(x)$

$$\ln|\cos x| - \tfrac{1}{2}\sin^2x + C = \ln|\cos x| - \tfrac{1}{2}\left[1 - \cos^2(x)\right] + C = \ln|\cos x| + \tfrac{1}{2}\cos^2x + \left [C - 1/2\right]$$

Since $C$ is arbitrary, replace $C - 1/2$ by $C$.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here is a more circuitous route:

\(\displaystyle \cos^2(x)\tan^3(x)=\frac{1}{2}\frac{2\sin^2(x)}{2 \cos^2(x)}2\sin(x)\cos(x)=\frac{1}{4}\frac{\cos(2x)-1}{\cos(2x)+1}(-2\sin(2x))\)

Now, use the substitution:

\(\displaystyle u=\cos(2x)\,\therefore\,du=-2\sin(2x)\,dx\)

And the integral becomes:

\(\displaystyle \frac{1}{4}\int\frac{u-1}{u+1}\,du=\frac{1}{4}\int 1-\frac{2}{u+1}\,du=\)

\(\displaystyle \frac{1}{4}\left(u-\ln\left((u+1)^2 \right) \right)+C\)

Back-substitute for $u$:

\(\displaystyle \frac{1}{4}\left(\cos(2x)-\ln\left((\cos(2x)+1)^2 \right) \right)+C\)

\(\displaystyle \frac{\cos(2x)}{4}-\ln\left(\sqrt{\cos(2x)+1)} \right)+C\)

Using double-angle identities for cosine, we have:

\(\displaystyle -\frac{1}{2}\sin^2(x)-\ln|\cos(x)|+C+\frac{1}{4}\)

Since an arbitrary constant plus a constant is still an arbitrary constant, the result may be written as:

\(\displaystyle -\frac{1}{2}\sin^2(x)-\ln|\cos(x)|+C\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Hello, shamieh!


[tex]\cos^2\!x\tan^3\!x \:=\:\cos^2\!x\frac{\sin^3\!x}{\cos^3\!x} \:=\;\frac{\sin^3\!x}{\cos x} \:=\:\frac{\sin^2\!x\cdot\sin x}{\cos x} [/tex]

. . [tex]=\;\frac{(1-\cos^2\!x)\sin x}{\cos x} \:=\: \left(\frac{1}{\cos x} - \cos x\right)\sin x [/tex]


The integral can be evaluated from here using the substitution [tex]\displaystyle \begin{align*} u = \cos{(x)} \end{align*}[/tex] (you will need to negate the integrand though as [tex]\displaystyle \begin{align*} du = -\sin{(x)}\,dx \end{align*}[/tex]).