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Intergration (by parts/substitution - Can I do it?

The integral of x^2 2 sec^2 x tan x dx . I have done one lesson of integration techniques and I am stuck on this, wondering if you need substitution method. I am yet to cover this! Many Thanks.

Reply 1

Justvisited

Yes, a product of a polynomial and a trigonometric function should be amenable to IBP.

You should also be able to use the fact that the derivative of tan x is sec^2 x.

Given standard formulae, you shouldn't need to use substitution.

Reply 2

tande33

Original post by Justvisited

Hmmm ok I have not been taught how tan x being derivative of sec ^2x will help me here, is it obvious?

Reply 3

RDKGames

Study Forum Helper

Original post by tande33

Hmmm ok I have not been taught how tan x being derivative of sec ^2x will help me here, is it obvious?

Under the integral sign you got

sec^2(x) tan(x) dx

If we transform to the u variable, then we have

sec^2(x) tan(x) dx/du du

and if we look at sec^2(x)tan(x), then we have

d(tan x)/dx tan(x) dx/du du

Notice that d(tanx)/dx * dx/du can be cancelled down to a constant if we impose that u=tanx.

Thus the stuff under the integral sign reduces to

tan(x) du

I.e.

u du

So this is why it is useful to notice that sec^2 is derivative of tan. Generally, if you are integrating f'(x)f(x) you can sub in u=f(x) to see that it will help.

(edited 3 years ago)

Reply 4

tande33

Ah ok thanks, I think I get that. This is likely something I will cover in class right?

Reply 5

RDKGames

Study Forum Helper

Original post by tande33

Ah ok thanks, I think I get that. This is likely something I will cover in class right?

Substitution? Of course. Alongside IBP it is a fundamental integration technique, and not at all obvious in some cases.

Reply 6

tande33

Original post by RDKGames

Substitution? Of course. Alongside IBP it is a fundamental integration technique, and not at all obvious in some cases.

great thanks again!

Reply 7

Justvisited

No need for substitution - more to the point to know the chain rule, and hence working backwards to see that integrating f'(x) f(x) gives you 1/2 (f(x))^2

Reply 8

tande33

Original post by Justvisited

No need for substitution - more to the point to know the chain rule, and hence working backwards to see that integrating f'(x) f(x) gives you 1/2 (f(x))^2

hmmm ok, I don't quite understand how that helps though apart from that I can find v?

Reply 9

RDKGames

Study Forum Helper

Original post by tande33

hmmm ok, I don't quite understand how that helps though apart from that I can find v?

sec^2(x) tan(x) is of the form f'(x)f(x) so it will integrate into 1/2[f(x)]^2 i.e. 1/2 tan^2(x).

I think this chain rule is what often flies over students heads to begin with ... Easy to forget that integration is just going the other way.