Trigonometric Integration

[Yuuki]

how do i integrate the fifth power of a secant?
i broke it up into powers of two and three but that didn't seem to work

 

[MarkFL]

Re: trig integrate

I think your initial move is a good one:

\(\displaystyle \int\sec^5(x)\,dx=\int\sec^3(x)\cdot\sec^2(x)\,dx\)

Now, on the squared factor, apply a Pythagorean identity, and you should then have something you can work with.

 

[Yuuki]

Re: trig integrate

i still can't work it out.
how do i integrate sec^3[x]*tan^2[x]??

 

[shamieh]

Or you can use the \(\displaystyle tan^n x\) formula I think

 

[shamieh]

Integrate by parts

sec^3x * sec^2x

Great explanation here: RAndomly found your problem

 

[MarkFL]

Re: trig integrate

i still can't work it out.
how do i integrate sec^3[x]*tan^2[x]??

After having thought about it while I was away, I think integration by parts is a better method.

\(\displaystyle I=\int\sec^5(x)\,dx=\int\sec^3(x)\cdot\sec^2(x)\,dx\)

Let:

\(\displaystyle u=sec^3(x)\,\therefore\,du=3\sec^3(x)\tan(x)\,dx\)

\(\displaystyle dv=\sec^2(x)\,dx\,\therefore\,v=\tan(x)\)

And so we have:

\(\displaystyle I=\sec^3(x)\tan(x)-3\int \sec^3(x)\tan^2(x)\,dx\)

Now use a Pythagorean identity on $\tan^2(x)$.

 

[HallsofIvy]

Although it is probably harder, my first reaction would be to write that secant as "1 over cosine": [tex]\int \frac{dx}{cos^5(x)}[/tex]. And since that is an odd power, I can multiply numerator and denominator by cos(x) to get an even power in the denominator: [tex]\int \frac{cos(x)}{cos^6(x)}dx[/tex].

Now, use [tex]cos^2(x)= 1- sin^2(x)[/tex] (so essentially using that "Pythagorean theorem" as MarkFL suggested). [tex]\int \frac{cos(x)}{(1- sin^2(x))^6} dx[/tex].\

Let u= sin(x), du= cos(x) and that becomes [tex]\int \frac{1}{1- x^2)^6}dx= \int\frac{1}{(1- x)^6(1+ x)^6}dx[/tex] and we can use "partial fractions".

 

[Prove It]

Although it is probably harder, my first reaction would be to write that secant as "1 over cosine": [tex]\int \frac{dx}{cos^5(x)}[/tex]. And since that is an odd power, I can multiply numerator and denominator by cos(x) to get an even power in the denominator: [tex]\int \frac{cos(x)}{cos^6(x)}dx[/tex].

Now, use [tex]cos^2(x)= 1- sin^2(x)[/tex] (so essentially using that "Pythagorean theorem" as MarkFL suggested). [tex]\int \frac{cos(x)}{(1- sin^2(x))^6} dx[/tex].\

Let u= sin(x), du= cos(x) and that becomes [tex]\int \frac{1}{1- x^2)^6}dx= \int\frac{1}{(1- x)^6(1+ x)^6}dx[/tex] and we can use "partial fractions".

Surely you mean

[tex]\displaystyle \begin{align*} \frac{1}{\cos^5{(x)}} &= \frac{\cos{(x)}}{\cos^6{(x)}} \\ &= \frac{\cos{(x)}}{\left[ \cos^2{(x)} \right] ^3} \\ &= \frac{\cos{(x)}}{\left[ 1 - \sin^2{(x)} \right] ^3} \\ &= \frac{\cos{(x)}}{\left[ 1 - \sin{(x)} \right] ^3 \left[ 1 + \sin{(x)} \right] ^3 } \end{align*}[/tex]

On another note, this is the approach I would take too.