How do you integrate int sec^-1xsec1x by integration by parts method?

2 Answers
Mar 3, 2018

The answer is =x"arc"secx-ln(x+sqrt(x^2-1))+C=xarcsecxln(x+x21)+C

Explanation:

We need

(sec^-1x)'=("arc"secx)'=1/(xsqrt(x^2-1))

intsecxdx=ln(sqrt(x^2-1)+x)

Integration by parts is

intu'v=uv-intuv'

Here, we have

u'=1, =>, u=x

v="arc"secx, =>, v'=1/(xsqrt(x^2-1))

Therefore,

int"arc"secxdx=x"arc"secx-int(dx)/(sqrt(x^2-1))

Perform the second integral by substitution

Let x=secu, =>, dx=secutanudu

sqrt(x^2-1)=sqrt(sec^2u-1)=tanu

intdx/sqrt(x^2-1)=int(secutanudu)/(tanu)=intsecudu

=int(secu(secu+tanu)du)/(secu+tanu)

=int((sec^2u+secutanu)du)/(secu + tanu)

Let v=secu+tanu, =>, dv=(sec^2u+secutanu)du

So,

intdx/sqrt(x^2-1)=int(dv)/(v)=lnv

=ln(secu+tanu)

=ln(x+sqrt(x^2-1))

Finally,

int"arc"secxdx=x"arc"secx-ln(x+sqrt(x^2-1))+C

Mar 3, 2018

int\ sec^-1(x)\ dx=xsec^-1(x)-ln(|x|+sqrt(x^2-1))+C

Explanation:

Alternatively, we can use a little-known formula for working out integrals of inverse functions. The formula states:
int\ f^-1(x)\ dx=xf^-1(x)-F(f^-1(x))+C
where f^-1(x) is the inverse of f(x) and F(x) is the anti-derivative of f(x).

In our case, we get:
int\ sec^-1(x)\ dx=xsec^-1(x)-F(sec^-1(x))+C

Now all we need to work out is the anti-derivative F, which is the familiar secant integral:
int\ sec(x)\ dx=ln|sec(x)+tan(x)|+C

Plugging this back into the formula gives our final answer:
int\ sec^-1(x)\ dx=xsec^-1(x)-ln|sec(sec^-1(x))+tan(sec^-1(x))|+C

We need to be careful about simplifying tan(sec^-1(x)) to sqrt(x^2-1) because the identity is only valid if x is positive. We are lucky, however, because we can fix this by putting an absolute value on the other term inside the logarithm. This also removes the need for the first absolute value, since everything inside the logarithm will always be positive:
xsec^-1(x)-ln(|x|+sqrt(x^2-1))+C