Integration by parts problem

[James10492]

Hi there, I decided today I would revise integration by parts and I have come across a problem in one of my exercises which I cannot solve.

∫(2x²)(sec²x)tanx dx

I don't how how to select the terms for u and dv- nothing I have tried works.

For example if you take secxtanx as dv (writing the original problem as ∫(2x²secx)(secxtanx),

2x²sec²x - ∫(2x²)(sec²x)tanx + 4 sec²x

leads to a circular problem. I have also tried using 2x²sec²x as the u term with tanx as the dv term (integrating to ln secx) but to no avail. Please could someone show me what I am missing?

 

[Deleted member 4993]

Hi there, I decided today I would revise integration by parts and I have come across a problem in one of my exercises which I cannot solve.

∫(2x²)(sec²x)tanx dx

I don't how how to select the terms for u and dv- nothing I have tried works.

For example if you take secxtanx as dv (writing the original problem as ∫(2x²secx)(secxtanx),

2x²sec²x - ∫(2x²)(sec²x)tanx + 4 sec²x

leads to a circular problem. I have also tried using 2x²sec²x as the u term with tanx as the dv term (integrating to ln secx) but to no avail. Please could someone show me what I am missing?

d/dx[tan(x)] = [sec(x)]^2

So "natural" substitution would be:

dv = (sec²x) * tanx dx .........→.................. v = (1/2) * [tan(x)]^2

Continue....................

 

[BigBeachBanana]

d/dx[tan(x)] = [sec(x)]^2

So "natural" substitution would be:

dv = (sec²x) * tanx dx .........→.................. v = (1/2) * [tan(x)]^2

Continue....................

Shouldn't [imath]v=\frac{\sec^2(x)}{2}?[/imath]

 

[Deleted member 4993]

Shouldn't [imath]v=\frac{\sec^2(x)}{2}?[/imath]

I think:

[tan(x)] * [sec(x)]^2 dx = [tan(x)] d[tan(x)] = (1/2)d[{tan(x)}^2] = d[{tan(x)}^2]...........→ .......................v = {(1/2) * tan(x)}^2

If we use 1 + tan^2(Θ) = sec^2 (Θ) ................................. Then we have the same expression with a difference of constant.

 

[BigBeachBanana]

I think:

[tan(x)] * [sec(x)]^2 dx = [tan(x)] d[tan(x)] = (1/2)d[{tan(x)}^2] = d[{tan(x)}^2]...........→ .......................v = {(1/2) * tan(x)}^2

If we use 1 + tan^2(Θ) = sec^2 (Θ) ................................. Then we have the same expression with a difference of constant.

Yes, I didn't consider that.