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My attempt:

$\int \frac {\sin (2x)} {\sin^4 (x) + \cos^4 (x)} dx$

$\int \frac {\sin (2x)} {(\sin^2 (x) + \cos^2 (x))^2 - 2\sin^2(x)\cos^2(x)}dx$

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

$\int \frac {\sin (2x)} {2 - {\sin^2(2x)}}dx$

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

$\int \frac {2\sin(x)\cos(x)} {2\cos^2(x)dx}$

$\int \tan(x)$

$\log(\cos(x)) + c$

But the answer in the textbook is $\arctan [2\sin^2(x) - 1]$

Have I done something wrong?

  • 0
    HINT: Set $\sin^2x=u$ So, $\cos^4x=(1-u)^2$ – 2017-02-23
  • 0
    You had an error near the end in which you assumed $1+\cos^2 2 x \propto \cos^2 x$. – 2017-02-23

2 Answers 2

2

There are a few mistakes in your work (typos included):

  1. In your $3$rd step, it will be $(1 - \frac {\sin^22x}{2})$ in the denominator and not $(1 - \frac {\sin^2 x}{2})$.
  2. From the $3$rd step onwards, there will be a $2$ in the numerator. You missed that.
  3. Very important to note: (don't know whether this is a typo)$$1 + \color{blue}{\cos^2(2x)} \not = 2\cos^2(x)$$ The actual identity is $$1 + \color{blue}{\cos(2x)} = 2\cos^2(x)$$

Now, this is where you went wrong.

So, if you correct those things, your integral boils down to: $$\int \frac {2\sin (2x)} {1 + \cos^2(2x)}dx$$ $$=-\int \frac {d\{\cos 2x\}} {1 + \cos^22x}$$ $$=-\arctan(\cos 2x) + c$$ The last line is from formula.

Hope this helps you.

1

Well, from step $3 \to $ step $4$, we have, $$I = \int \frac {\sin 2x}{1-0.5\sin^2 2x} \mathrm {d}x $$ $$ = \int \frac {2\sin 2x}{2-\sin^2 2x} \mathrm {d}x $$ $$= \int \frac {2\sin 2x}{1+\cos^2 2x} \mathrm {d}x$$

Now substitute $u = \cos 2x $ to simplify the integral and get the answer. Hope it helps.