Answer:
[tex]\displaystyle \int \sec\left(\frac{x}{2}\right)\tan^5\left(\frac{x}{2}\right)dx=\frac{2\sec^5\left(\dfrac{x}{2}\right)}{5}-\frac{4\sec^3\left(\dfrac{x}{2}\right)}{3}+2\sec\left(\frac{x}{2}\right)+C[/tex]
Step-by-step explanation:
We want to find the integral:
[tex]\displaystyle \int \sec\left(\frac{x}{2}\right)\tan^5\left({\frac{x}{2}\right)dx[/tex]
First, perform the substitution y = x / 2. This yields:
[tex]\displaystyle dy = \frac{1}{2} \, dx \Rightarrow dx = 2\, dy[/tex]
Hence, the integral becomes;
[tex]\displaystyle =2\int \sec y\tan^5 y \, dy[/tex]
Next, as you had done, let's expand the tangent term but to the fourth:
[tex]\displaystyle =2\int \sec y\tan^4 y\tan y\, dy[/tex]
Recall that:
[tex]\tan^2(y)=\sec^2(y)-1[/tex]
Hence:
[tex]\displaystyle =2\int \sec y(\sec^2 y-1)^2\tan y\, dy[/tex]
We can use substitution once more. This time, let u = sec(y). Hence:
[tex]\displaystyle du = \sec y \tan y \, dy[/tex]
Rewrite:
[tex]\displaystyle =2\int \left((\sec^2y-1)^2\right)\left(\sec y \tan y\right)dy[/tex]
Therefore:
[tex]\displaystyle = 2\int (u^2 - 1)^2\, du[/tex]
Expand:
[tex]\displaystyle =2\int u^4-2u^2+1\, du[/tex]
Reverse Power Rule:
[tex]\displaystyle = 2\left(\frac{u^5}{5} - \frac{2u^3}{3} + u\right) + C[/tex]
Simplify:
[tex]\displaystyle = \frac{2u^5}{5} - \frac{4u^3}{3} + 2u + C[/tex]
Back-substitute:
[tex]\displaystyle = \frac{2\sec^5 y }{5}-\frac{4\sec^3 y}{3}+2\sec y+C[/tex]
Back-substitute:
[tex]\displaystyle =\frac{2\sec^5\left(\dfrac{x}{2}\right)}{5}-\frac{4\sec^3\left(\dfrac{x}{2}\right)}{3}+2\sec \left(\frac{x}{2}\right)+C[/tex]
Therefore:
[tex]\displaystyle \int \sec\left(\frac{x}{2}\right)\tan^5\left(\frac{x}{2}\right)dx=\frac{2\sec^5\left(\dfrac{x}{2}\right)}{5}-\frac{4\sec^3\left(\dfrac{x}{2}\right)}{3}+2\sec\left(\frac{x}{2}\right)+C[/tex]
