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

Answer 1

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

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#
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Answer 2

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

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#
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Answer 3

To integrate ∫sec^(-1)(x) dx using integration by parts method, you can set up the integral as follows:

Let u = sec^(-1)(x) and dv = dx. Then, differentiate u to find du and integrate dv to find v.

Differentiating u: du = (1 / (x√(x^2 - 1))) dx

Integrating dv: v = x

Now, apply the integration by parts formula:

∫u dv = uv - ∫v du

Substitute the values of u, dv, du, and v into the formula:

∫sec^(-1)(x) dx = x * sec^(-1)(x) - ∫x * (1 / (x√(x^2 - 1))) dx

Now, integrate the remaining integral, which is relatively simpler:

∫x * (1 / (x√(x^2 - 1))) dx

Letting u = x^2 - 1 and du = 2x dx, we have:

(1/2) * ∫(1 / √u) du

This integrates to:

(1/2) * 2 * √u + C

= √(x^2 - 1) + C

Substitute this back into the original equation:

∫sec^(-1)(x) dx = x * sec^(-1)(x) - √(x^2 - 1) + C

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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