How do you integrate $\int {tan}^{5} (\frac{x}{4})$ $int tan^5(x/4)$ ?

Question

How do you integrate $\int {tan}^{5} (\frac{x}{4})$ $int tan^5(x/4)$ ?

1 Answer

Impact of this question

3412 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-07T00:00:12

Given $\int {tan}^{5} (\frac{x}{4}) d x$ $inttan^5(x/4)dx$

Begin with a substitution for $\frac{x}{4}$ $x/4$ :

$u = \frac{x}{4} \Rightarrow d u = \frac{1}{4} d x \Rightarrow 4 d u = d x$ $u=x/4 => du=1/4dx => 4du=dx$

We now have:

$4 \int {tan}^{5} (u) d u$ $4inttan^5(u)du$

Break up ${tan}^{5} (u)$ $tan^5(u)$ :

$4 \int {tan}^{2} (u) \cdot {tan}^{3} (u) d u$ $4inttan^2(u)*tan^3(u)du$

Use the trigonometric identity ${tan}^{2} (θ) = {sec}^{2} (θ) - 1$ $tan^2(theta)=sec^2(theta)-1$ :

$4 \int ({sec}^{2} (u) - 1) {tan}^{3} (u) d u$ $4int(sec^2(u)-1)tan^3(u)du$

Distribute ${tan}^{3}$ $tan^3$ :

$4 \int {sec}^{2} (u) {tan}^{3} (u) - {tan}^{3} (u) d u$ $4intsec^2(u)tan^3(u)-tan^3(u)du$

Split the integral:

$4 \int {sec}^{2} (u) {tan}^{3} (u) d u - 4 \int {tan}^{3} (u) d u$ $4intsec^2(u)tan^3(u)du-4inttan^3(u)du$

For the LH integral, we can perform a substitution:

$z = tan (u) \Rightarrow d z = {sec}^{2} (u) d u$ $z=tan(u) => dz=sec^2(u)du$

$4 \int {sec}^{2} (u) {tan}^{3} (u) d u \Rightarrow 4 \int z^{3} d z$ $4intsec^2(u)tan^3(u)du=>4intz^3dz$

This is basic integral. We will now move on to the RHS.

$4 \int {tan}^{3} (u) d u$ $4inttan^3(u)du$

Break up ${tan}^{3} (u)$ $tan^3(u)$ :

$4 \int {tan}^{2} (u) \cdot tan (u) d u$ $4inttan^2(u)*tan(u)du$

Apply trigonometric identity ${tan}^{2} (θ) = {sec}^{2} (θ) - 1$ $tan^2(theta)=sec^2(theta)-1$ :

$4 \int ({sec}^{2} (u) - 1) tan (u) d u$ $4int(sec^2(u)-1)tan(u)du$

Distribute $tan (u)$ $tan(u)$ :

$4 \int {sec}^{2} (u) tan (u) - tan (u) d u$ $4intsec^2(u)tan(u)-tan(u)du$

Split the integral:

$4 \int {sec}^{2} (u) tan (u) d u - 4 \int tan (u) d u$ $4intsec^2(u)tan(u)du-4inttan(u)du$

For the LH integral, a substitution:

$r = tan (u) \Rightarrow d r = {sec}^{2} (u) d u$ $r=tan(u) => dr=sec^2(u)du$

$4 \int {sec}^{2} (u) tan (u) d u = 4 \int r d r$ $4intsec^2(u)tan(u)du=4intrdr$

This is a basic integral, as is $4 \int tan (u) d u$ $4inttan(u)du$ .

We have:

$4 \int z^{3} d z - [4 \int r d r - 4 \int tan (u) d u]$ $4intz^3dz-[4intrdr-4inttan(u)du]$

We integrate, then substitute back in for all of the variables.

$4 (\frac{1}{4} z^{4}) - [4 (\frac{1}{2} r^{2}) - 4 ln | sec (u) |] + C$ $4(1/4z^4)-[4(1/2r^2)-4ln|sec(u)|]+C$

$\Rightarrow {tan}^{4} (u) - 2 {tan}^{u} + 4 ln | sec (u) | + C$ $=>tan^4(u)-2tan^(u)+4ln|sec(u)|+C$

$\Rightarrow {tan}^{4} (\frac{x}{4}) - 2 {tan}^{2} (\frac{x}{4}) + 4 ln ∣ ∣ sec (\frac{x}{4}) ∣ ∣ + C$ $=>tan^4(x/4)-2tan^2(x/4)+4ln|sec(x/4)|+C$

Science

Math

Humanities

... and beyond

How do you integrate $\int {tan}^{5} (\frac{x}{4})$ $int tan^5(x/4)$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate ∫tan5(x4)int tan^5(x/4)?

1 Answer

Related questions

Impact of this question

How do you integrate $\int {tan}^{5} (\frac{x}{4})$ $int tan^5(x/4)$ ?