Come si integra $\int [{(sec (x))}^{5}] d x$ $int [(Sec (x)) ^ 5] dx$ ?

Risposta:

$\int {sec}^{5} x d x = \frac{2 tan x {sec}^{3} x + 3 tan x sec x + 3 ln | sec x + tan x |}{8} + C$ $int sec^5x dx = (2tanxsec^3x+ 3tanxsecx + 3ln abs(secx+tanx))/8 +C$

Spiegazione:

Scrivi l'integrando come: ${sec}^{5} (x) = {sec}^{2} (x) {sec}^{3} (x)$ $sec^5(x) = sec^2(x) sec^3(x)$ e si integrano per parti considerando che:

$\frac{d}{d x} (tan x) = {sec}^{2} (x)$ $d/dx (tanx) = sec^2(x)$ ,

Sun:

$\int {sec}^{5} x d x = \int {sec}^{2} (x) {sec}^{3} (x) d x$ $int sec^5x dx = int sec^2(x) sec^3(x)dx$

$\int {sec}^{5} x d x = \int {sec}^{3} (x) d (tan x)$ $int sec^5x dx = int sec^3(x)d(tanx)$

$\int {sec}^{5} x d x = tan x {sec}^{3} x - \int tan x d ({sec}^{3} (x))$ $int sec^5x dx = tanxsec^3x - int tanx d(sec^3(x))$

e come:

$\frac{d}{d x} ({sec}^{3} (x)) = 3 {sec}^{2} (x) \frac{d}{d x} sec (x) = 3 {sec}^{3} (x) tan x$ $d/dx (sec^3(x)) = 3sec^2(x) d/dx sec(x) = 3sec^3(x) tanx$

si ha:

$\int {sec}^{5} x d x = tan x {sec}^{3} x - 3 \int {tan}^{2} x {sec}^{3} x d x$ $int sec^5x dx = tanxsec^3x - 3int tan^2x sec^3x dx$

usa ora l'identità trigonometrica:

${tan}^{2} θ = \frac{{sin}^{2} θ}{{cos}^{2} θ} = \frac{1 - {cos}^{2} θ}{{cos}^{2} θ} = {sec}^{2} θ - 1$ $tan^2 theta = sin^2 theta/cos^2 theta = (1-cos^2 theta)/cos^2theta = sec^2theta -1$

avere:

$\int {sec}^{5} x d x = tan x {sec}^{3} x - 3 \int ({sec}^{2} x - 1) {sec}^{3} x d x$ $int sec^5x dx = tanxsec^3x - 3int (sec^2x -1) sec^3x dx$

e usando la linearità dell'integrale:

$\int {sec}^{5} x d x = tan x {sec}^{3} x + 3 \int {sec}^{3} x d x - 3 \int {sec}^{5} x d x$ $int sec^5x dx = tanxsec^3x + 3int sec^3x dx -3 int sec^5x dx$

L'integrale ora appare su entrambi i lati dell'equazione e possiamo risolverlo ottenendo una formula di riduzione:

$\int {sec}^{5} x d x = \frac{1}{4} (tan x {sec}^{3} x + 3 \int {sec}^{3} x d x)$ $int sec^5x dx = 1/4(tanxsec^3x + 3int sec^3x dx)$

Risolvi ora l'integrale risultante con la stessa procedura:

$\int {sec}^{3} x d x = \int sec x d (tan x)$ $int sec^3x dx = int secx d(tanx)$

$\int {sec}^{3} x d x = tan x sec x - \int tan x d (sec x)$ $int sec^3x dx = tanxsecx - int tanx d(secx)$

$\int {sec}^{3} x d x = tan x sec x - \int {tan}^{2} x sec x d x$ $int sec^3x dx = tanxsecx - int tan^2x secx dx$

$\int {sec}^{3} x d x = tan x sec x - \int ({sec}^{2} x - 1) sec x d x$ $int sec^3x dx = tanxsecx - int (sec^2x-1) secx dx$

$\int {sec}^{3} x d x = tan x sec x + \int sec x d x - \int {sec}^{3} x d x$ $int sec^3x dx = tanxsecx + int secx dx - int sec^3x dx$

$\int {sec}^{3} x d x = \frac{1}{2} (tan x sec x + \int sec x d x)$ $int sec^3x dx = 1/2(tanxsecx + int secx dx)$

Per risolvere la nota integrale risultante che:

$\frac{d}{d x} (tan x + sec x) = {sec}^{2} x + sec x tan x = sec x (tan x + sec x)$ $d/dx (tanx + secx) = sec^2x +secx tanx = secx(tanx+secx)$

così dividendo e moltiplicando l'integrando per $(sec x + tan x)$ $(secx+tanx)$ :

$\int sec x d x = \int \frac{sec x (sec x + tan x)}{sec x + tan x} d x$ $int secx dx = int (secx(secx+tanx))/(secx+tanx) dx$

$\int sec x d x = \int \frac{d (sec x + tan x)}{sec x + tan x}$ $int secx dx = int (d(secx+tanx))/(secx+tanx)$

$\int sec x d x = ln | sec x + tan x | + C$ $int secx dx = ln abs(secx+tanx) +C$

Mettere tutto insieme:

$\int {sec}^{5} x d x = \frac{2 tan x {sec}^{3} x + 3 tan x sec x + 3 ln | sec x + tan x |}{8} + C$ $int sec^5x dx = (2tanxsec^3x+ 3tanxsecx + 3ln abs(secx+tanx))/8 +C$

Come si integra int [(Sec (x)) ^ 5] dx ∫[(sec(x))5]dxint [(Sec (x)) ^ 5] dx ?

Risposta:

Spiegazione:

Lascia un commento Annulla risposta

Come si integra $\int [{(sec (x))}^{5}] d x$ $int [(Sec (x)) ^ 5] dx$ ?