C3 Integration question

Original post by Gondur

I cannot integrate this piece of maths in the question. Also why are both limits 5, will this not make the area 0?

Very confused here.

Help ASAP.

20140330_184048[1].jpg

If you're going to make a substitution, you need to make one that preserves sign, or else you need to note that the integrand is even and split up the range into [-1/2, 0] and [0, 1/2]. However, the better way to do it is to use an arctan substitution.

Reply 2

Original post by Gondur

I cannot integrate this piece of maths in the question. Also why are both limits 5, will this not make the area 0?

Very confused here.

Help ASAP.

20140330_184048[1].jpg

First things first, you can't take 1/8x outside of the integral, as it is a variable and needs to be integrated.

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Reply 3

Original post by Gondur

I cannot integrate this piece of maths in the question. Also why are both limits 5, will this not make the area 0?

Very confused here.

Help ASAP.

20140330_184048[1].jpg

Also, you seem to have misunderstood

du = \dfrac{du}{dx}dx

. You've pulled the

\dfrac{du}{dx}

expression out of the integral, which you can't do because it's not constant. That would be like saying

\int x^2 dx = x \int x dx

.

Reply 4

Gondur

OP

Original post by Smaug123

If you're going to make a substitution, you need to make one that preserves sign, or else you need to note that the integrand is even and split up the range into [-1/2, 0] and [0, 1/2]. However, the better way to do it is to use an arctan substitution.

What is an arctan substitution?

Reply 5

Original post by Gondur

I cannot integrate this piece of maths in the question. Also why are both limits 5, will this not make the area 0?

Very confused here.

Help ASAP.

20140330_184048[1].jpg

If I were you, I'd substute u=2x and then use any of the methods from this thread.

Reply 6

Gondur

OP

Original post by Smaug123

If I were you, I'd substute u=2x and then use any of the methods from this thread.

Haven't heard of any of those methods in that thread. I only know of integration by substitution and integration by inspection and these are the methods I tried with this question. It doesn't work if you let u=1+4x^2. Why are you suggesting to use u=2x? What is the logic behind this choice.

Reply 7

Principia

8

Original post by Gondur

Haven't heard of any of those methods in that thread. I only know of integration by substitution and integration by inspection and these are the methods I tried with this question. It doesn't work if you let u=1+4x^2. Why are you suggesting to use u=2x? What is the logic behind this choice.

It turns it into 1 / (1 + u^2) which is d arctanu / du

Reply 8

Original post by Principia

It turns it into 1 / (1 + u^2) which is d arctanu / du

You're trying to make this a lot more complex than C3 :/

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Reply 9

Original post by Gondur

Haven't heard of any of those methods in that thread. I only know of integration by substitution and integration by inspection and these are the methods I tried with this question. It doesn't work if you let u=1+4x^2. Why are you suggesting to use u=2x? What is the logic behind this choice.

Because I know my standard integrals - among them, the fact that

\int \dfrac{1}{1+x^2} dx = \tan^{-1}(x)

. That can itself be solved by substitution, as several of the methods listed in that thread state.

Reply 10

Nikki.N

Original post by Gondur

I cannot integrate this piece of maths in the question. Also why are both limits 5, will this not make the area 0?

Very confused here.

Help ASAP.

20140330_184048[1].jpg

Ah I've done these type of Q's before!
What you've done is fine, BUT you can't take the 1/8x out.
So put it back in, and you know u=1+4x^2 so what would x be? Make x the subject of the formula and sub it in for 1/8x

Reply 11

user2020user

16

Original post by Gondur

...

\displaystyle \begin{aligned} \int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{1}{1+4x^2} \ \mathrm{d}x & = \int_{-\frac{1}{2}}^{\frac{1}{2}} \dfrac{1}{1+(2x)^2} \ \mathrm{d}x \\ & \overset{2x=\tan \theta}= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \dfrac{\frac{1}{2} \sec^2 \theta}{1+\tan^2 \theta} \ \mathrm{d}\theta \\ & = \ ... \end{aligned}

(edited 10 years ago)

Reply 12

davros

16

Original post by Gondur

I cannot integrate this piece of maths in the question. Also why are both limits 5, will this not make the area 0?

Very confused here.

Help ASAP.

20140330_184048[1].jpg

If this is a C3 question, I'd be very surprised if you're expected to use a substitution anyway!

Take out a factor 1/4 so that the integrand looks like

\dfrac{1}{x^2 + (1/4)}

then look in your formula book for the result that tells you how to integrate

\dfrac{1}{x^2 + a^2}

Reply 13

Original post by davros

If this is a C3 question, I'd be very surprised if you're expected to use a substitution anyway!

Take out a factor 1/4 so that the integrand looks like

\dfrac{1}{x^2 + (1/4)}

then look in your formula book for the result that tells you how to integrate

\dfrac{1}{x^2 + a^2}

Just thought I'd say that my formula book tells you how to integrate it from the question, so unless it's a 'show that' question, you can just quote it. :smile:

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Reply 14

Davelittle

13

What exam board are you on?

My C3 exam doesn't include those types of integration, you need to use your formulae book :smile:

Reply 15

Principia

8

Original post by Dusky Mauve

You're trying to make this a lot more complex than C3 :/

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That is core 3 though ...

Reply 16

Original post by Principia

That is core 3 though ...

Well I've never seen arctan before and I'm sure I don't need it?

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Reply 17