How do you integrate #intxsec^-1(x)dx#?

Answer 1

The answer is #=x^2/2arcsecx-1/2sqrt(x^2-1)+C#

Perform the integration by parts

#intuv'=uv-intu'v#

Here,

#u=arcsecx#, #=>#, #u'=(dx)/(xsqrt(x^2-1))#
#v'=x#, #=>#, #v=x^2/2#

Therefore,

#intxarcsecxdx=x^2/2arcsecx-1/2int(xdx)/(sqrt(x^2-1))#
Let #u=x^2-1#, #=>#, #du=2xdx#

Therefore,

#1/2int(xdx)/(sqrt(x^2-1))=1/4int(du)/(sqrtu)#
#=1/2sqrtu#
#=1/2sqrt(x^2-1)#

And finally,

#intxarcsecxdx=x^2/2arcsecx-1/2sqrt(x^2-1)+C#
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Answer 2

To integrate ( \int x \sec^{-1}(x) , dx ), we use integration by parts. Let ( u = \sec^{-1}(x) ) and ( dv = x , dx ). Then, ( du = \frac{1}{|x|\sqrt{x^2 - 1}} , dx ) and ( v = \frac{1}{2}x^2 ).

Applying integration by parts formula ( \int u , dv = uv - \int v , du ), we get:

( \int x \sec^{-1}(x) , dx = \frac{1}{2}x^2 \sec^{-1}(x) - \frac{1}{2} \int \frac{x^2}{|x|\sqrt{x^2 - 1}} , dx ).

The remaining integral can be simplified using trigonometric substitution. Let ( x = \sec(\theta) ), then ( dx = \sec(\theta) \tan(\theta) , d\theta ). This simplifies the integral to:

( \int \frac{\sec^2(\theta)}{\sec(\theta)|\tan(\theta)|} , \sec(\theta) \tan(\theta) , d\theta ).

Simplify this expression and integrate it. Finally, substitute back ( \theta ) with ( \sec^{-1}(x) ) to get the final result.

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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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