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Integration Of trig. Functions sec^3

paulmdrdo

Active member
May 13, 2013
386
what method should i use here(except integration by parts)?

\begin{align*}\displaystyle \int sec^3\theta \,d\theta\end{align*}
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
what method should i use here(except integration by parts)?

\begin{align*}\displaystyle \int sec^3\theta \,d\theta\end{align*}
\(\displaystyle \displaystyle \begin{align*} \int{\sec^3{(\theta)}\,d\theta} &= \int{\frac{1}{\cos^3{(\theta)}}\,d\theta} \\ &= \int{ \frac{\cos{(\theta)}}{\cos^4{(\theta)}}\,d\theta} \\ &= \int{ \frac{\cos{(\theta)}}{\left[ \cos^2{(\theta)}\right] ^2 } \,d\theta } \\ &= \int{ \frac{\cos{(\theta)}}{ \left[ 1 - \sin^2{(\theta)} \right] ^2 } \,d\theta} \end{align*}\)

Now let \(\displaystyle \displaystyle \begin{align*} u = \sin{(\theta)} \implies du = \cos{(\theta)}\,d\theta \end{align*}\) and the integral becomes

\(\displaystyle \displaystyle \begin{align*} \int{ \frac{\cos{(\theta)}}{ \left[ 1 - \sin^2{(\theta)} \right] ^2 } \,d\theta} &= \int{ \frac{1}{ \left( 1 - u^2 \right) ^2 } \, du } \\ &= \int{ \frac{1}{ \left[ \left( 1 - u \right) \left( 1 + u \right) \right] ^2 } \,du} \\ &= \int{ \left[ \frac{1}{ \left( 1- u \right) \left( 1 + u \right) } \right] ^2 \,du} \end{align*}\)

Now applying Partial Fractions

\(\displaystyle \displaystyle \begin{align*} \frac{A}{1 - u} + \frac{B}{1 + u} &\equiv \frac{1}{(1 - u)(1 + u)} \\ \frac{A(1 + u) + B(1 - u)}{(1 - u)(1 + u)} &\equiv \frac{1}{(1 - u)(1 + u)} \\ A(1 + u) + B(1 -u) &\equiv 1 \end{align*}\)

Let \(\displaystyle \displaystyle \begin{align*} u = 1 \end{align*}\) and we find \(\displaystyle \displaystyle \begin{align*} A = \frac{1}{2} \end{align*}\), and let \(\displaystyle \displaystyle \begin{align*} u = -1 \end{align*}\) and we find \(\displaystyle \displaystyle \begin{align*} B = \frac{1}{2} \end{align*}\), so

\(\displaystyle \displaystyle \begin{align*} \frac{1}{(1 - u)(1 + u)} = \frac{1}{2(1 - u)} + \frac{1}{2(1 + u)} \end{align*}\), so continuing with the integral...

\(\displaystyle \displaystyle \begin{align*} \int{ \left[ \frac{1}{(1 - u)(1 + u)} \right] ^2 \, du} &= \int{ \left[ \frac{1}{2(1 - u)} + \frac{1}{2(1 + u)} \right] ^2 \, du } \\ &= \int{ \frac{1}{4(1 - u)^2} + \frac{1}{2(1 - u)(1 + u)} + \frac{1}{4(1 + u)^2}\, du } \\ &= \int{ \frac{1}{4( 1- u)^2} + \frac{1}{4(1 - u)} + \frac{1}{4(1 + u)} + \frac{1}{4( 1 + u)^2} \, du} \\ &= \frac{1}{4} \int{ (1 - u)^{-2} + \frac{1}{1 - u} + \frac{1}{1 + u} + (1 + u)^{-2} \,du} \\ &= \frac{1}{4} \left[ (1 - u)^{-1} - \ln{ | 1 - u | } + \ln{ |1 + u| } - (1 + u)^{-1} \right] + C \\ &= \frac{1}{4} \left[ \frac{1}{1 - \sin{(\theta)} } - \ln{ \left| 1 - \sin{(\theta)} \right| } + \ln{ \left| 1 + \sin{(\theta)} \right| } - \frac{1}{1 + \sin{(\theta)} } \right] + C \end{align*}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
With a bit of manipulation, we can get this in a form we can integrate directly:

\(\displaystyle \sec^3(\theta)=\frac{1}{2}\left(\sec^3(\theta)+ \sec^3(\theta) \right)=\frac{1}{2}\left(\sec^3(\theta)+ \sec(\theta)\left(\tan^2(\theta)+1 \right) \right)=\)

\(\displaystyle \frac{1}{2}\left(\sec^3(\theta)+ \sec(\theta)\tan^2(\theta)+\sec(\theta)\frac{\tan(\theta)+ \sec(\theta)}{\tan(\theta)+\sec(\theta)} \right)=\)

\(\displaystyle \frac{1}{2}\left(\left(\sec(\theta)\sec^2(\theta)+\sec(\theta)\tan^2(\theta) \right)+\left(\frac{\sec(\theta)\tan(\theta)+\sec^2(\theta)}{\sec(\theta)+\tan(\theta)} \right) \right)=\)

\(\displaystyle \frac{1}{2}\frac{d}{d\theta}\left(\sec(\theta)\tan(\theta)+\ln\left|\sec(\theta)+\tan(\theta) \right| \right)\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
A nice method to evaluate it is by parts , try it !
 

paulmdrdo

Active member
May 13, 2013
386
this is my solution using integration by parts. :D

\begin{align*}\displaystyle \int \sec^3\,(\theta)\,d\theta\\ let&u\,=\, \sec(\theta)\\ du&=\, \sec(\theta) \tan(\theta)\\\int dv&=\,\int \sec^2(\theta)\,d\theta\\v&=\,\tan^2(\theta)\\ \sec(\theta)& \tan(\theta)-\int \tan^2(\theta)\,\sec(\theta)\,d\theta\\ \sec(\theta)&\tan(\theta)-\int \sec(\theta)\,(\sec^2(\theta)-1)\,d\theta\\ \sec(\theta)&\tan(\theta)-\int \sec^3(\theta)\,d\theta+\int \sec(\theta)\,d\theta\\ \sec(\theta)&\tan(\theta)-\int \sec^3(\theta)\,d\theta+\ln |\sec(\theta)+\tan(\theta)|+C\\\int \sec^3(\theta)& d\theta=\,\sec(\theta) \tan(\theta)-\int \sec^3(\theta)\,d\theta+\ln |\sec(\theta)+\tan(\theta)|-\int \sec^3(\theta) d\theta\\2\int \sec^3(\theta)&=\,\sec(\theta) \tan(\theta)-\int \sec^3(\theta)\,d\theta+\ln |\sec(\theta)+\tan(\theta)|\\ \sec^3(\theta)&=\,\frac{1}{2}\sec(\theta)\, \tan(\theta)+\frac{1}{2}\ln |\sec(\theta)+\tan(\theta)|+C\end{align*}
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
While your method is solid, you have several typos which distract from an otherwise good post. Can you spot them?

Also, allow me to suggest that in your use of $\LaTeX$, you precede predefined functions (trigonometric, logarithmic, etc.) with a backslash so that their names are not italicized. I also suggest enclosing function parameters with parentheses so that it is clear just what the parameters are.