**Question**

**Introduction**

This
question is taken from the Techniques of Integration chapter of Thomas’
Calculus, 12th edition. It looks pretty
nasty in that the arcsecant is just one of those
things on the fringes of teachers’ and students’ minds.

**Strategy**

We apply the “Integration by Parts” Formula

**etail**)” heuristic. The expression

*t*is

**a**lgebraic whereas the arcsecant expression is of the “

**i**nverse” type. Since “

**a**” comes before “

**i**” in “d(

**etail**)”, we choose the

**a**lgebraic expression

*t*to serve as our

^{dv}/

_{dx}. We realise that we will need the derivative of the arcsecant

for sec

^{-1}

*x*being a cute angle ... I mean, an acute angle.

**Solution**

**Remarks**

A
slight modification of this approach is to first re-express the arcsecant
as arcsec

*t*= arccos(^{1}/*). One needs to work out the derivative of this arccosine expression when doing the integral.*_{t}
H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

H11. Solve part of the problem

**Suitable Levels**

*****GCE ‘A’ Levels H2 Mathematics (challenge)

* IB Mathematics HL (challenge)

* Advanced Placement (AP) Calculus
BC (challenge)

* University / College calculus

* other syllabuses that involve integration
and inverse trigonometric functions

* any precocious learner who loves a challenge
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