Difficult Integral

[AvgStudent]

[math]\int\frac{1}{1+\sin(x)}\space dx=?[/math]Hint given: Use trig sub.
Any idea what u should be?

 

[nasi112]

Half tangent formula will solve this integral easily.

\(\displaystyle \sin x = \frac{2\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}\)

 

[lex]

Yes, [imath]u=\tan \tfrac{x}{2}[/imath]

 

[AvgStudent]

Thanks for the hints, I'll try and report back

 

[AvgStudent]

[math] u=\tan\left(\frac{x}{2}\right),du=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx\\ \sin(x)=\frac{2u}{u^2+1},\cos(x)=\frac{1-u^2}{u^2+1}\implies dx=\frac{2}{u^2+1}du [/math]Substitution:
[math]\int\frac{1}{1+\sin(x)}\space dx= \int \frac{2du}{(u^2+1)(\frac{2u}{u^2+1}+1)} =\int \frac{2du}{u^2+2u+1}[/math]Stuck here, I feel it should be easy.

 

[topsquark]

[math] u=\tan\left(\frac{x}{2}\right),du=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)dx\\ \sin(x)=\frac{2u}{u^2+1},\cos(x)=\frac{1-u^2}{u^2+1}\implies dx=\frac{2}{u^2+1}du [/math]Substitution:
[math]\int\frac{1}{1+\sin(x)}\space dx= \int \frac{2du}{(u^2+1)(\frac{2u}{u^2+1}+1)} =\int \frac{2du}{u^2+2u+1}[/math]Stuck here, I feel it should be easy.

It is... get some coffee! Does the denominator of the integrand factor? What further substitution could you make?

-Dan

 

[AvgStudent]

It is... get some coffee! Does the denominator of the integrand factor? What further substitution could you make?

-Dan

Coffee needed, indeed!
[math]\int \frac{2du}{u^2+2u+1}=\int \frac{2du}{(u+1)^2}\\ s=u+1\implies ds=du\\ \int\frac{2}{s^2}ds=-\frac{2}{s}+C=\frac{-2}{u+1}+C=-\frac{2}{\tan\left(\frac{x}{2}\right)+1}+C[/math]Final answer?

 

[Dr.Peterson]

[math]\int\frac{1}{1+\sin(x)}\space dx=?[/math]Hint given: Use trig sub.
Any idea what u should be?

There's a much simpler way to integrate this; just multiply numerator and denominator by (1 - sin(x)).

But apparently that's not what you're expected to do.

I've struggled to prove (other than by graphing) that the answer I get this way is equal to the other (with C differing by 1), but it is.

 

[AvgStudent]

There's a much simpler way to integrate this; just multiply numerator and denominator by (1 - sin(x)).

But apparently that's not what you're expected to do.

I've struggled to prove (other than by graphing) that the answer I get this way is equal to the other (with C differing by 1), but it is.

I'll give your approach a go.

 

[lex]

There's a much simpler way to integrate this; just multiply numerator and denominator by (1 - sin(x)).

Nicely seen, as ever!

As you say, the two differ by 1:

 

[Steven G]

I agree with Dr Peterson. The first thing that came to my mind was to multiply the numerator and denominator by 1-sinx. Things work out nicely from there. Note that the 1st step in solving an integral is NOT always making a u-sub!

 

[nasi112]

There's a much simpler way to integrate this; just multiply numerator and denominator by (1 - sin(x)).

But apparently that's not what you're expected to do.

I've struggled to prove (other than by graphing) that the answer I get this way is equal to the other (with C differing by 1), but it is.

I agree with you Dr.Peterson that this is a simpler way. But trust me the idea of this type of integral is to use the half tangent formula. What if the integral has a negative sign instead of positive? What if there was \(\displaystyle \sin x + \cos x\) in the denominator? By coincidence in the OP case \(\displaystyle 1 - \sin x\) works pretty good.

If I were the OP, I would practise both methods.