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Maths help integration

Yazomi

6A905C0E-8C55-4329-B3B4-6B3149996BA7.jpg.jpeg

Can someone explain where I went wrong pls

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Reply 1

mqb2766

Try differentiating your indefinite answer and compare with the question?
Looks like a tan() type substitution to me.

Reply 2

ㅋㅋㅋㅋㅋ

I think you forgot to integrate what’s inside the bracket?

Reply 3

ThatsAGoodOne349

Original post by mqb2766

Try differentiating your indefinite answer and compare with the question?
Looks like a tan() type substitution to me.

Yeah, It looks like an arctan(x) outcome somewhere along the line..
@Yazomi; don't remove the fraction and make it a negative power - that may be helpful to do

Reply 4

Yazomi

Original post by mqb2766

Try differentiating your indefinite answer and compare with the question?
Looks like a tan() type substitution to me.

Original post by ㅋㅋㅋㅋㅋ

I think you forgot to integrate what’s inside the bracket?

Hmm after tan substitution I got
169D2DE2-E6ED-499A-ABAF-CCCC83C42822.jpg.jpeg

And I think the answers a bit dodgy

Reply 5

davros

Original post by Yazomi

Hmm after tan substitution I got
169D2DE2-E6ED-499A-ABAF-CCCC83C42822.jpg.jpeg

And I think the answers a bit dodgy

I don't even know what you're doing there!

Your initial integral was in terms of x, so I would expect you to set

x = \tan \theta

and proceed from there. Where did u come from?

Reply 6

Yazomi

Original post by ThatsAGoodOne349

Yeah, It looks like an arctan(x) outcome somewhere along the line..
@Yazomi; don't remove the fraction and make it a negative power - that may be helpful to do

1A1F5F35-9E39-4DE8-9364-F2C63E458BE5.jpg.jpeg

I’m not sure if this is what you mean but I’ve got this instead?

Reply 7

Yazomi

Original post by davros

I don't even know what you're doing there!

Your initial integral was in terms of x, so I would expect you to set

x = \tan \theta

and proceed from there. Where did u come from?

95879BED-938D-47A0-B989-FD87AE1C5272.jpg.jpeg

Here’s the question is it helps
I was trying to do part be but I was trying to see if I could still find out the area without using the cos^2 theta given.

I think the part I messed up was trying to figure out what to put u=

Reply 8

davros

Original post by Yazomi

I’m not sure if this is what you mean but I’ve got this instead?

That is even worse than your original attempt!

I don't mean to be rude, but how long have you been studying integration? Are you getting support from a teacher or self-studying?

You've posted a number of integration questions recently where you seem to have invented rules for integration that just aren't valid, instead of following standard methods like recognition, substitution and IBP. This is leading you into all sorts of problems later on :smile:

Reply 9

RDKGames

Study Forum Helper

\displaystyle \int \dfrac{dx}{(1+x^2)^2}

Let

x=\tan \theta

, then

dx = \sec^2 \theta \ d \theta

The integral becomes;

\displaystyle \int \dfrac{dx}{(1+x^2)^2} = \int \dfrac{\sec^2 \theta \ d \theta}{(1+\tan^2 \theta)^2} = \int \dfrac{d \theta}{\sec^2 \theta} = ...

To integrate cos^2, you need double angle identity.

(edited 2 years ago)

Reply 10

Filthy Communist

Original post by Yazomi

Can someone explain where I went wrong pls

https://www.integral-calculator.com/ this shows you how to solve integrals step by step

Also man you gotta relearn integration from the ground up. I don't know how you're going from int((1+x^2)^-2)dx to -(1+x^2)^-1 + c but, in the nicest way possible, it shows that there are serious gaps in your understanding of how integration works. At this point you're inventing your own rules for integration, which aren't even remotely correct. Try TLMaths, ExamSolutions, Zeeshaan Zamurred, or some other maths teacher who makes videos on youtube, whoever you vibe best with.

EDIT: looking at a more recent post, my man this is worse than not understanding integration. Go back to GCSE maths, I'm not trying to be mean but if you don't understand fractions how are you going to be able to do integration?

(edited 2 years ago)

Reply 11

Yazomi

Oh dear welp I’ll start practising some more. I don’t think I’ll be prepared for the mock tmr 😂

Reply 12

mqb2766

Original post by Yazomi

Oh dear welp I’ll start practising some more. I don’t think I’ll be prepared for the mock tmr 😂

The cheatsheet
https://www.drfrostmaths.com/uploads/JFrost/files/C4CheatSheet.pdf
is decent, but an aid rather than substitute for practice.

Reply 13

Yazomi

Original post by mqb2766

The cheatsheet
https://www.drfrostmaths.com/uploads/JFrost/files/C4CheatSheet.pdf
is decent, but an aid rather than substitute for practice.

Ahhh thankssss it’s looking really helpful so far :smile:

Reply 14

mqb2766

Original post by Yazomi

Ahhh thankssss it’s looking really helpful so far :smile:

It's a bit basic, but something to build on.
When you think you've done the indefinite integrals, try differentiating them to check it gives the original function, if you've got time.

Reply 15

Filthy Communist

And also note you can instantly check definite integrals with your calculator, assuming you cave the 991-EX or CG-50

It does it numerically, so it will give a decimal answer. Subtract that answer from what you got and if you get 0 you did it right.

(edited 2 years ago)

Reply 16

Yazomi

Original post by Filthy Communist

Alright got ya thanks

Original post by mqb2766

It's a bit basic, but something to build on.
When you think you've done the indefinite integrals, try differentiating them to check it gives the original function, if you've got time.

Just got one more quick question
684ABABC-0CA7-49A3-A252-99754D573021.jpg.jpeg

For a question like this how do you know what to use as a substitute or it’s really just guess work?

Reply 17

mqb2766

Original post by Yazomi

Alright got ya thanks

Just got one more quick question
684ABABC-0CA7-49A3-A252-99754D573021.jpg.jpeg

For a question like this how do you know what to use as a substitute or it’s really just guess work?

1-x^2 sounds like the Pythagorean identity, as the limits also hint at (30-60-90 triangle). So something like cos^2 + sin^2 = 1 gives
cos^2 = 1 - sin^2

1+x^2 hints at:
1 + tan^2 = sec^2

(edited 2 years ago)

Reply 18

Yazomi

Original post by mqb2766

1-x^2 sounds like the Pythagorean identity, as the limits also hint at (30-60-90 triangle). So something like cos^2 = 1 - sin^2

1 x^2 hints at:
1 tan^2 = sec^2

What on earth
！(◎_◎; )
Ok I’ll leave that for later. I don’t think I’ll be able to take on or even spot these hints in the actual thing

(edited 2 years ago)

Reply 19

mqb2766

Original post by Yazomi

What on earth
！(◎_◎; )
Ok I’ll leave that for later. I don’t think I’ll be able to take on or even spot these hints in the actual thing

Thats fine, but these trig substitution questions come up a fair bit. There are only a few trig identities that frequently come up, so it's worth getting your head round them after tomorrow.