Integration question (1 Viewer)

[deepulse]

Integral of

x²tan³x.dx

I was thinking by parts, but i get stucks, help please!

 

[McLake]

u = tan^3x
du = 3tan^2xsec^2x

v = x^3/3
dv = x^2

I = 1/3*x^3*tan^3x - I(x^3*tan^2sec^2)

tan^2sec^2 = tan^2x - tan^4x

so I = 1/3*x^3*tan^3x - I(x^3*tan^2x) + I(x^3*(tan^2x)^2)

hmm ...

ANS is 1/3*x^3*tan^3

 

[Affinity]

maybe that's not the question? I plugged the thing into mathematica, turned out a quite weird.

maybe it meant x^2*tan(x^3) ?

 

[maniacguy]

I(f(x)) = integral of f(x) dx

I(x^2*tan^3(x))
= I(x^2*tan(x)*sec^2(x)) - I(x*tan(x))
= x^2*tan^2(x)/2 - I(x*tan^2(x)) - I(x*tan(x))

but I(x*tan^2(x))
= I(x*sec^2(x)) - I(x)
= x*tan(x) - I(tan(x)) - x^2/2
= x*tan(x) + ln(cos(x)) - x^2/2

As for I(x*tan(x)) - it boils down to integrating ln(cos(x)), which I can't see any way to do using HSC methods[1]. I'd put it in the box of questions too hard to solve in the HSC.

[1] or first year uni methods for that matter. There might be a couple of second-year techniques that could be applied, but you obviously aren't going to be expected to know them.

 

[Affinity]

the year 10 method:
flip to the backof the book.

 

[Harimau]

Originally posted by Affinity
the year 10 method:
flip to the backof the book.

Year 10? ive been doing that since year 7...

 

[deepulse]

It was in my friends trial 4u, and she asked if i could do it and i couldnt go figure, maybe they stuffed it up, but thats what was in the paper, it was cheltnum gurls btw