Question

Calculus 2 master needed; stuck on evaluating the integral please show steps
`\int {sec(x/2)tan^5(x/2)} \, dx` I am thinking that we want to split up the tan^5, making
`\int {sec(x/2)tan^2 tan^3(x/2)} \, dx` and then
`\int {sec(x/2)*sec^2x-1* tan^3(x/2)} \, dx` but I am not sure this is correct. Can anyone help? The thing I am unsure of is that tan^3 is still odd, so would we do the same thing again and factor so we will be left with just tanx?

Answer 1

= 2 (sec (x/2) - 2sec³(x/2) + sec⁵(x/2) ) + C

3 5

Explanation:

∫sec(x/2) tan⁵(x/2) dx

apply u substitute u = x/2

= ∫sec(u) tan⁵(u) * 2du

= 2 * ∫sec(u) tan⁴(u) tan(u) du

= 2 * ∫sec(u) (tan²(u))² tan(u) du

= 2 * ∫sec(u) (-1 + sec²(u))² tan(u) du

apply u substitute v = sec(u)

= 2 * ∫(-1 + v²)² dv

expand

= 2 * ∫1 - 2v² + v⁴ dv

sum

= 2 (v - 2v³ + v⁵ )

3 5

substitute it back

= 2 (sec (x/2) - 2sec³(x/2) + sec⁵(x/2) )

3 5

add constant to the solution.

= 2 (sec (x/2) - 2sec³(x/2) + sec⁵(x/2) ) + C

3 5

Answer 2

`\displaystyle \int \sec\left((x)/(2)\right)\tan^5\left((x)/(2)\right)dx=(2\sec^5\left((x)/(2)\right))/(5)-(4\sec^3\left((x)/(2)\right))/(3)+2\sec\left((x)/(2)\right)+C`

Explanation:

We want to find the integral:

`\displaystyle \int \sec\left((x)/(2)\right)\tan^5\left({(x)/(2)\right)dx`

First, perform the substitution y = x / 2. This yields:

`\displaystyle dy = (1)/(2) \, dx \Rightarrow dx = 2\, dy`

Hence, the integral becomes;

`\displaystyle =2\int \sec y\tan^5 y \, dy`

Next, as you had done, let's expand the tangent term but to the fourth:

`\displaystyle =2\int \sec y\tan^4 y\tan y\, dy`

Recall that:

`\tan^2(y)=\sec^2(y)-1`

Hence:

`\displaystyle =2\int \sec y(\sec^2 y-1)^2\tan y\, dy`

We can use substitution once more. This time, let u = sec(y). Hence:

`\displaystyle du = \sec y \tan y \, dy`

Rewrite:

`\displaystyle =2\int \left((\sec^2y-1)^2\right)\left(\sec y \tan y\right)dy`

Therefore:

`\displaystyle = 2\int (u^2 - 1)^2\, du`

Expand:

`\displaystyle =2\int u^4-2u^2+1\, du`

Reverse Power Rule:

`\displaystyle = 2\left((u^5)/(5) - (2u^3)/(3) + u\right) + C`

Simplify:

`\displaystyle = (2u^5)/(5) - (4u^3)/(3) + 2u + C`

Back-substitute:

`\displaystyle = (2\sec^5 y )/(5)-(4\sec^3 y)/(3)+2\sec y+C`

Back-substitute:

`\displaystyle =(2\sec^5\left((x)/(2)\right))/(5)-(4\sec^3\left((x)/(2)\right))/(3)+2\sec \left((x)/(2)\right)+C`

Therefore:

`\displaystyle \int \sec\left((x)/(2)\right)\tan^5\left((x)/(2)\right)dx=(2\sec^5\left((x)/(2)\right))/(5)-(4\sec^3\left((x)/(2)\right))/(3)+2\sec\left((x)/(2)\right)+C`