3
$\begingroup$

I am surprisingly having a bit of difficulty with an indefinite integral which is interesting since the integral I solved before is

$$ \int \frac{1+\cos2x}{\sin^2(2x)} dx$$

The integral I am currently working on is

$$ \int \frac{\sin^2(2x)}{1+\cos2x} dx$$

I first divided out giving: (this is the mistake I made early on, in this case you use the pythagorean identity for sin, and then cancel out $1+\cos2x$)

$$ \int \sin^2(2x)+ \sin^2(2x)\cdot \sec(2x) \,dx$$

Then I factored out the $\sin^2(2x)$ resulting in:

$$ \int \sin^2(2x)(1+ \sec(2x)) \,dx$$

Substituting for $u=2x$ and splitting the integral into two parts:

$$\frac{1}{2} \int \sin^2(u) du + \frac{1}{2} \int \sin^2(u)\sec(u) \,du$$ lets call this eq. 1.

Now, this is where I am having difficulty as 1.) dealing with even powers of sin and 2.) the $\sec(u)$ term is proving to be troublesome.

Another form of the above equation is:

$$ \frac{1}{2} \int \sin^2(u) \,du + \frac{1}{2} \int \sin(u)\tan(u) \,du$$

Some approaches I have tried are using different trigonometric identities e.g.

$$\sin^2(u) = \frac{1}{2} (1-\cos(2u))$$

however, this results in for eq. 1

$$ \frac{1}{4} \int 1-\cos(2u) \,du + \frac{1}{4} \int \frac{(1-\cos(2u)}{\cos(u)} du $$

Then I would have to use the cosine angle addition formula which quickly gets out of hand.

I understand there are different approaches to solving different indefinite integrals. The purpose of this problem is to only use substitutions.

Questions that I have are as follows, 1.) is it possible to continue along with the steps I have taken?, 2.) or must I do an entirely different substitution at the beginning. Sorry for the long post and thank you for your time.

$\endgroup$
  • $\begingroup$ why all the $2x$es in the integral if all you are is an indefinite integra? $\endgroup$ – abel Jan 12 '15 at 15:56
  • $\begingroup$ To everyone who posted. Thanks so much for your quick replies and insight. Mickep and abel for probably the most straightforward and easiest solution, amWhy for pointing at a mistake in the algebra, and idm for your unique solution. o.o, I'll need to spend more time looking at how you solved it. $\endgroup$ – user156926 Jan 12 '15 at 16:09
7
$\begingroup$

The function $\sin^2(2x)/(1+\cos(2x))$ can be simplified to $1-\cos 2x$: $$ \frac{\sin^2(2x)}{1+\cos(2x)}=\frac{1-\cos^2(2x)}{1+\cos(2x)}=\frac{(1-\cos(2x))(1+\cos(2x))}{1+\cos(2x)}=1-\cos(2x). $$

$\endgroup$
2
$\begingroup$

Note: While it is true that $\dfrac {a+b}c = \dfrac ac + \dfrac bc$,

$$\dfrac a{b+c}\neq \dfrac ab + \dfrac ac$$

In particular $$ \int \frac{\sin^2(2x)}{1+\cos2x} dx \neq \int \sin^2(2x)+ \sin^2(2x)\cdot \sec(2x) dx$$

$\endgroup$
  • $\begingroup$ Right! Thanks for pointing this out. I'll make the edits in my post accordingly. That's also another reason why I probably had such difficulty as well. $\endgroup$ – user156926 Jan 12 '15 at 16:06
  • 1
    $\begingroup$ I suspect so, since that step was very early in your work! $\endgroup$ – Namaste Jan 12 '15 at 16:09
0
$\begingroup$

$$\int \dfrac{\sin^2 t}{1+\cos t} dt = \int\dfrac{\sin^2 t (1 - \cos t)}{(1+\cos t)(1- \cos t)} dt = \int (1 - \cos t) dt = t + \sin t + C$$

change of variable $ t = 2x$ gives you $$\int \dfrac{\sin^2 2x}{1 + \cos 2x} dx = x + \dfrac{1}{2}\sin 2x + C $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy