Question

Calculus 2 master needed; stuck on evaluating the integral please show steps 8x%2F2%29tan%5E5%28x%2F2%29%7D%20%5C%2C%20dx" id="TexFormula1" title="\int {sec(x/2)tan^5(x/2)} \, dx" alt="\int {sec(x/2)tan^5(x/2)} \, dx" align="absmiddle" class="latex-formula"> 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

Answer:

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

                              3                     5

Step-by-step 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

Answer:

$\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C$

Step-by-step explanation:

So we have the integral:

$\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx$

First, let's use substitution to get rid of the x/2. I'm going to use the variable y. So, let y be x/2. Thus:

$y=\frac{x}{2}\\dy=\frac{1}{2}dx\\2dy=dx$

Therefore, the integral is:

$=2\int \sec(y)\tan^5(y)dy$

Now, as you had done, let's expand the tangent term. However, let's do it to the fourth. Thus:

$=2\int \sec(y)\tan^4(y)\tan(y)dy$

Now, we can use a variation of the trigonometric identity:

$\tan^2(y)+1=\sec^2(y)$

So:

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

Substitute this into the integral. Note that tan^4(x) is the same as (tan^2(x))^2. Thus:

$=2\int \sec(y)(\tan^2(y))^2\tan(y)dy\\=2\int \sec(y)(\sec^2(y)-1)^2\tan(y)dy$

Now, we can use substitution. We will use it for sec(x). Recall what the derivative of secant is. Thus:

$u=\sec(y)\\du=\sec(y)\tan(y)dy$

Substitute:

$2\int (\sec^2(y)-1)^2(\sec(y)\tan(y))dy\\=2\int(u^2-1)^2 du$

Expand the binomial:

$=2\int u^4-2u^2+1du$

Spilt the integral:

$=2(\int u^4 du+\int-2u^2du+\int +1du)$

Factor out the constant multiple:

$=2(\int u^4du-2\int u^2du+\int(1)du)$

Reverse Power Rule:

$=2(\frac{u^{4+1}}{4+1}-2(\frac{u^{2+1}}{2+1})+(\frac{u^{0+1}}{0+1}}))$

Simplify:

$=2(\frac{u^5}{5}-\frac{2u^3}{3}+u)$

Distribute the 2:

$=\frac{2u^5}{5}-\frac{4u^3}{3}+2u$

Substitute back secant for u:

$=\frac{2\sec^5(y)}{5}-\frac{4\sec^3(y)}{3}+2\sec(y)$

And substitute back 1/2x for y. Therefore:

$=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)$

And, finally, C:

$=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C$

And we're done!