The Student Room Group
Reply 1
I would hedge a guess at the identities

cos2x = cos^2(x) - sin^2(x) & sin^2(x) + cos^2(x) = 1

You will end up with a cos^2(2x) which you can repeat to get a cos(4x)
Can you see that your expression is equal to:

sin^2(x)cos^2(x)

= [sin(x)cos(x)]^2

= [1/2 sin(2x)]^2

= 1/4 sin^2(2x)

= (1/4)(1/2)[cos(4x) - 1]

= 1/8 cos(4x) - 1/8

That should be easier to integrate, I hope.
Reply 3
tazarooni89
Can you see that your expression is equal to:

sin^2(x)cos^2(x)

= [sin(x)cos(x)]^2

= [1/2 sin(2x)]^2

= 1/4 sin^2(2x)

= (1/4)(1/2)[cos(4x) - 1]

= 1/8 cos(4x) - 1/8

That should be easier to integrate, I hope.


Elegant. I like it.
Reply 4
Ewan
I would hedge a guess at the identities

cos2x = cos^2(x) - sin^2(x) & sin^2(x) + cos^2(x) = 1

You will end up with a cos^2(2x) which you can repeat to get a cos(4x)


I was thinking about that, but not sure exactly what to do.

I'm also stuck on these :

Unparseable latex formula:

\displaystyle\int^1_-_1 \frac {x+1}{(2-x)^4} \, dx

(non calculator paper)

and

2tanxcos2xdx\int \frac {2tanx}{cos2x} \, dx
G O D I V A
I was thinking about that, but not sure exactly what to do.

I'm also stuck on these :

Unparseable latex formula:

\displaystyle\int^1_-_1 \frac {x+1}{(2-x)^4} \, dx

(non calculator paper)

and

2tanxcos2xdx\int \frac {2tanx}{cos2x} \, dx

first one substitution
second one identities. rewrite cos2x.
Reply 6
tazarooni89
Can you see that your expression is equal to:

sin^2(x)cos^2(x)

= [sin(x)cos(x)]^2

= [1/2 sin(2x)]^2

= 1/4 sin^2(2x)

= (1/4)(1/2)[cos(4x) - 1]

= 1/8 cos(4x) - 1/8

That should be easier to integrate, I hope.


how did you get to there?
G O D I V A
Unparseable latex formula:

\displaystyle\int^1_-_1 \frac {x+1}{(2-x)^4} \, dx

(non calculator paper)

x+1(2x)4=x2(2x)4+3(2x)4=(2x)3+3(2x)4\displaystyle \frac{x+1}{(2-x)^4} = \frac{x-2}{(2-x)^4} + \frac{3}{(2-x)^4} = -(2-x)^{-3} + 3(2-x)^{-4}.

G O D I V A
how did you get to there?

Double angle formula. sin 2x = 2 sin x cos x. You need to know these.
Reply 8
G O D I V A
how did you get to there?


http://en.wikipedia.org/wiki/List_of_trigonometric_identities

sin2x = 2sinxcosx
G O D I V A
how did you get to there?


I used the trigonometric identity:

sin(2x) = 2sin(x)cos(x)

i.e.

sin(x)cos(x) = 1/2 sin(2x)
Reply 10
Totally Tom
first one substitution
second one identities. rewrite cos2x.


left out the second part sorry. the thing said 'using substitution of t = tanx. I differentiated and got sec2x=1+tan2=1+t2sec^2x = 1+tan^2 = 1 + t^2 ??, but what do i do next?
Reply 11
cos2x=1tan2x1+tan2x\cos 2x = \frac{1 - \tan^2x}{1 + \tan^2x}


1cos2x=1+tan2x1tan2x\frac{1}{\cos 2x} = \frac{1 + \tan^2x}{1 - \tan^2x}


tanxcos2xdx=tanx1tan2x(tan2x+1)dx\int\frac{\tan x}{\cos 2x}dx = \int\frac{\tan x}{1 - \tan^2x}(\tan^2x + 1)dx
Reply 12
Thanks to all of you. I managed to get the right answers now.

Although I do have one question.

When integrating by parts, which do you differentiate and which do you integrate?
I mean like do you always differentiate the easier one, integrate the harder one? or something else?
G O D I V A
Thanks to all of you. I managed to get the right answers now.

Although I do have one question.

When integrating by parts, which do you differentiate and which do you integrate?
I mean like do you always differentiate the easier one, integrate the harder one? or something else?

As a rule of thumb, have a look for something you can integrate, and integrate that, and differentiate whatever's left. Differentiation is kinda easy, but integration is difficult because in most cases it can't be done.
Reply 14
udv=uvvdu\int u dv = uv - \int vdu

You want vdu to be easier to integrate than udv.

The rule of thumb is you let let u be the highest function on this list.

L - log function
I - inverse trig function
A - algebraic function (x^2 etc.)
T - trig function
E - exponential function

and you differentiate this to find du and integrate what's left (dv) to find v.
Reply 15
What if you can integrate both things or differentiate both things like:

xnlnxdx\int x^nlnx \, dx or
Unparseable latex formula:

\int sin^-^1x\, dx

or
Unparseable latex formula:

\int e^3^xcosx \, dx



then which one do you chose to do which?
G O D I V A
What if you can integrate both things or differentiate both things like:

xnlnxdx\int x^nlnx \, dx or
Unparseable latex formula:

\int sin^-^1x\, dx

or
Unparseable latex formula:

\int e^3^xcosx \, dx



then which one do you chose to do which?

Play around. Clearly, integrating ln x in the first integral is just gonna give you something more complicated to integrate. I don't see any particularly easy way of integrating arcsin x (otherwise you wouldn't be integrating it by parts, would you?!), so differentiate that and integrate 1. And e^3x cos x has to be done twice, and it doesn't matter whether you differentiate the exponential or the trig bit as long as you do it the same way round twice.
G O D I V A
What if you can integrate both things or differentiate both things like:

xnlnxdx\int x^nlnx \, dx or
Unparseable latex formula:

\int sin^-^1x\, dx

or
Unparseable latex formula:

\int e^3^xcosx \, dx



then which one do you chose to do which?


choose one function to be u and another one to be dv/dx

choose them such that v du/dx is easy to integrate.


So for the first one, it would be easier to differentiate ln(x) and integrate x^n, although it's possible to do it the other way round without too much difficulty.

For the second one... first use the substitution x = sin(t) and you get t cos(t) dt which you can integrate by parts by differentiating t and integrating cos(t)

For the third one... you're going to have to use integration by parts twice. It doesn't matter which one you differentiate and which one you integrate really, you'll end up with similar expressions ether way. Integrating by parts will get you your original integral back, and then you can just rearrange it.
Reply 18
cor wish i could do them...