Answer: Solving Integral of sin2x/1+sin²x dx

[Cradle_of_Knowledge]

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"

 

[qbert]

rewrite the numerator using the trig identity sin(2x) = 2 sin(x)cos(x)
then look for a nice substitution.

 

[Cradle_of_Knowledge]

Could you explain your answer!

 

[James R]

$$\int \frac{\sin 2x}{1+sin^2 x}\,dx$$

Put $u = 1 + sin^2 x$. Then:

$$du = 2 \sin x \cos x\,dx = \sin 2x\,dx$$

So, your integral becomes:

$$\int \frac{1}{u}\,du = \ln u + c = \ln (1 + \sin^2 x) + c$$

 

[Cyrus]

We can do the following:

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

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

$$\frac{2sin(x)cos(x)}{1 + sin^2(x)}$$

Now we make a u-substitution:

$$u= 1+ sin^2 (x)$$

So

$$du = 2sin(x)cos(x)dx$$

and we can replace the integral as such:

$$\int \frac{1}{u}du$$

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:

$$ln(1+sin^2(x))+C$$

where C is the constant of integration.

I hope that helps you out.

Cheers,

Cyrus

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

 

[James R]

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

 

[quasar987]

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.

 

[grant555]

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.

 

[Cradle_of_Knowledge]

Can you solve this too
S (sinx-cos)² dx

Answer:- x+sin2x/2 +c

 

[Cyrus]

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.

 

[kant]

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...