Evaluate $I = \int {sec}^{3} (x) d x$ $I=\int\sec^3(x)dx$ ?

Question

Evaluate $I = \int {sec}^{3} (x) d x$ $I=\int\sec^3(x)dx$ ?

The problem suggests using integration by parts.
After splitting the integral into $\int {sec}^{2} (x) sec (x) d x$ $\int\sec^2(x)\sec(x)dx$ , I tried $u = sec (x)$ $u=\sec(x)$ and $d v = {sec}^{2} (x) d x$ $dv=\sec^2(x)dx$ ...

But I am stuck on the resulting $u v - \int v d u$ $uv-\intvdu$ form (heredx) is the equation I got).

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Answer 1 · 2018-04-09T11:51:27

$\int {sec}^{3} x d x = \frac{sec x tan x}{2} + \frac{1}{2} ln | sec x + tan x | + C$ $int sec^3x dx = (secxtanx)/2 + 1/2ln abs (secx+tanx)+C$

Explanation:

Note that:

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

$\frac{d}{d x} sec x = sec x tan x$ $d/dx secx = secx tanx$

so:

$\int {sec}^{3} x d x = \int sec x \cdot {sec}^{2} x d x = \int sec x d (tan x)$ $int sec^3x dx = int secx * sec^2x dx = int secx d(tanx)$

so, integrating by parts:

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

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

Use now the trigonometric identity:

${tan}^{2} x = {sec}^{2} x - 1$ $tan^2x = sec^2x -1$

to have:

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

and using the linearity of the integral:

$\int {sec}^{3} x d x = sec x tan x - \int {sec}^{3} x d x + \int sec x d x$ $int sec^3x dx = secxtanx - int sec^3xdx + int secxdx$

The integral to solve now appears on both sides of the equation, then:

$2 \int {sec}^{3} x d x = sec x tan x + \int sec x d x$ $2int sec^3x dx = secxtanx + int secxdx$

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

You can see here how to solve the resulting integral to have:

$\int {sec}^{3} x d x = \frac{sec x tan x}{2} + \frac{1}{2} ln | sec x + tan x | + C$ $int sec^3x dx = (secxtanx)/2 + 1/2ln abs (secx+tanx)+C$

Answer 2 · 2018-04-09T12:04:48

$I = \frac{1}{2} tan x sec x + \frac{1}{2} ln | tan x + sec x | + c$ $I=1/2tanxsecx+1/2ln|tanx+secx|+c$

Explanation:

Here,

$I = \int {sec}^{3} x d x$ $I=intsec^3xdx$

$= \int sec x {sec}^{2} x d x$ $=intsecxsec^2xdx$

$= \int \sqrt{1 + {tan}^{2} x} {sec}^{2} x d x$ $=intsqrt(1+tan^2x)sec^2xdx$

Let , $tan x = u \Rightarrow {sec}^{2} x d x = d u$ $tanx=u=>sec^2xdx=du$

$I = \int \sqrt{1 + u^{2}} d u$ $I=intsqrt(1+u^2)du$

$= \frac{u}{2} \sqrt{1 + u^{2}} + \frac{1}{2} ln ∣ ∣ u + \sqrt{u^{2} + 1} ∣ ∣ + c$ $=u/2sqrt(1+u^2)+1/2ln|u+sqrt(u^2+1)|+c$ ,where, $u = tan x$ $u=tanx$

$= \frac{1}{2} tan x sec x + \frac{1}{2} ln ∣ ∣ tan x + \sqrt{1 + {tan}^{2} x} ∣ ∣ + c$ $=1/2tanxsecx+1/2ln|tanx+sqrt(1+tan^2x)|+c$

$I = \frac{1}{2} tan x sec x + \frac{1}{2} ln | tan x + sec x | + c$ $I=1/2tanxsecx+1/2ln|tanx+secx|+c$

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