MHB Integration Of trig. Functions sec^3

[paulmdrdo1]

what method should i use here(except integration by parts)?

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

 

[Prove It]

what method should i use here(except integration by parts)?

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

[math]\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*}[/math]

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

[math]\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*}[/math]

Now applying Partial Fractions

[math]\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*}[/math]

Let [math]\displaystyle \begin{align*} u = 1 \end{align*}[/math] and we find [math]\displaystyle \begin{align*} A = \frac{1}{2} \end{align*}[/math], and let [math]\displaystyle \begin{align*} u = -1 \end{align*}[/math] and we find [math]\displaystyle \begin{align*} B = \frac{1}{2} \end{align*}[/math], so

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

[math]\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*}[/math]

 

[MarkFL]

With a bit of manipulation, we can get this in a form we can integrate directly:

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

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

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

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

 

[alyafey22]

A nice method to evaluate it is by parts , try it !

 

[paulmdrdo1]

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*}

 

[MarkFL]

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.