Simple Calc. Where's my error? (1 Viewer)

[ISAM77]

Please do not provide an alternate solution. Use the same method of working to get the right answer of pi/2

 

[killer queen]

try turning sin(2 theta) into 2sin(theta)cos(theta), then you get the integral of 2sin^2(theta) from 0 to pi/2? which should come out to the answer I would imagine

also, I note you used a math ext 2 technique in this question, but posted it in the math ext 1 forum - was this intentional?

 

[ISAM77]

try turning sin(2 theta) into 2sin(theta)cos(theta), then you get the integral of 2sin^2(theta) from 0 to pi/2? which should come out to the answer I would imagine

also, I note you used a math ext 2 technique in this question, but posted it in the math ext 1 forum - was this intentional?

No, I didn't even notice. Yeah, it goes smoothly with 2sin(theta)cos(theta). Thx!

I am being stubborn, I want to see it solved in the way I tried to go about it

 

[killer queen]

I genuinely don't know if it works after the integration by parts; in your working, you seem to have lost the bounds, but there's a tan in the first part, which you wouldn't be able to sub pi/2 into

sorry I can't help further!

 

[Trebla]

When you integrated by parts again you get the original integral I again, not 2I

Also, you need to be careful when dealing with integrals that have undefined values in the domain, you should be treading carefully and make your components converge to a limit, not diverge to infinity.

Generally speaking, such integrals are outside the syllabus.

 

[ISAM77]

When you integrated by parts again you get the original integral I again, not 2I

Also, you need to be careful when dealing with integrals that have undefined values in the domain, you should be treading carefully and make your components converge to a limit, not diverge to infinity.

Generally speaking, such integrals are outside the syllabus.

With my friend, Chat GPT I have discovered that indeed the error appears when the boundary term appears with a tan(theta) being evaluated with a pi/2 and 0. Basically, I can't use integration by parts because of this, is what I understood. Which agrees with what you're saying.

Thankyou!