Problem about integretaion

Hi,

I don't know where is to place this thread. Please can you solve the following integeration problem.
(Note: sign of integeration is replaced by "S").

S sin2x/1+sin²x dx

the answer in the book is "ln(1+sin²x)+c"
 
185
3
rewrite the numerator using the trig identity sin(2x) = 2 sin(x)cos(x)
then look for a nice substitution.
 
Could you explain your answer!
 

James R

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[tex]\int \frac{\sin 2x}{1+sin^2 x}\,dx[/tex]

Put [itex]u = 1 + sin^2 x[/itex]. Then:

[tex]du = 2 \sin x \cos x\,dx = \sin 2x\,dx[/tex]

So, your integral becomes:

[tex]\int \frac{1}{u}\,du = \ln u + c = \ln (1 + \sin^2 x) + c[/tex]
 
2,903
13
We can do the following:

[tex] sin(2x) = 2 sin(x)cos(x) [/tex]

This can be placed in for sin(2x), giving us:

[tex] \frac{2sin(x)cos(x)}{1 + sin^2(x)} [/tex]

Now we make a u-substitution:

[tex] u= 1+ sin^2 (x) [/tex]

So

[tex] du = 2sin(x)cos(x)dx [/tex]

and we can replace the integral as such:

[tex] \int \frac{1}{u}du [/tex]

Now this is simply 1/u which equals ln (u)

so the anwser is ln(u), where u is equal to what we previously stated:

[tex] ln(1+sin^2(x))+C [/tex]

where C is the constant of integration.

I hope that helps you out.

Cheers,

Cyrus

Edit: Damn, james beat me to it!GRRRRRRRRRRRRRRRRR :rofl:
 
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James R

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Doh! Don't you hate it when that happens! :grumpy:
 

quasar987

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But it's never a wasted post. Seeing two ways of doing it, even though they're essentially the same, are always better then just one.
 
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quasar987 said:
But it's never a wasted post. Seeing two ways of doing it, even though they're essentially the same, are always better then just one.


I agree with you totally on this. I always found it comforting having more than one way to work a problem.
 
Can you solve this too
S (sinx-cos)² dx

Answer:- x+sin2x/2 +c
 
2,903
13
Expand it out, and solve the integral, its not hard once you expand it. You will have a sin squared term, plus a cos squared term, which add up to one, plus a -2sinxcosx term. And if you notice, you can use the trig identity from your first question.
 
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Cradle_of_Knowledge said:
Hi,

I don't know where is to place this thread. Please can you solve the following integeration problem.
(Note: sign of integeration is replaced by "S").

S sin2x/1+sin²x dx

the answer in the book is "ln(1+sin²x)+c"
1) sin2x=2sinxcosx

2) u=sinx => du=cosx dx

the intergral come:

2 * intergral{ u/(1+u^2) du}

make substitution as again.

this time let w= 1+u^2 => dw= 2u du

blah blah....
 

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