could anybody guide me to the next step $$\int\frac{\sin{2x}}{\sqrt{\cos^4{x}+3}}dx=\int\frac{2\sin{x}\cos{x}}{\sqrt{\cos^4{x}+3}}dx=-\int\frac{\cos{x}}{\sqrt{\cos^4{x}+3}}d{\cos^2{x}}=...$$ and here I stuck. I don't know how to get rid of $cos{x}$ in the numerator. Any guidance is very welcome.
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Put $y=\cos^2(x)$ and the integral becomes $$-\int \frac{dy}{\sqrt{y^2+3}}$$
Then put $y=\sqrt{3}\sinh(t)$ and it becomes $$-\frac{1}{\sqrt{3}}\int dt.$$
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if I put $y=\cos^2{x}$ the integral becomes $-\int\frac{\sqrt{y}}{\sqrt{y^2+3}}dy$, isn't it? – dimaastronom Feb 24 '15 at 20:27
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thx, I got it. Didn't notice that we have a complicated function $\cos^2{x}$ – dimaastronom Feb 24 '15 at 20:39
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You made a mistake in the second step. You should have
$$-\int \frac{1}{\sqrt{\cos^4 x + 3}} d\cos^2 x$$
So letting $u = \cos^2 x$ results in
$$-\int \frac{1}{\sqrt{u^2 + 3}}\, du = -\ln|u + \sqrt{u^2 + 3}| + C = -\ln(\cos^2 x + \sqrt{\cos^4x + 3}) + C.$$
kobe
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Ok, in the second step $2\sin{x}$ goes into $d{\cos^2{x}}$ and where $\cos{x}$ gone? – dimaastronom Feb 24 '15 at 20:31
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1@dimaastronom note $d\cos^2 x = 2\cos x(\cos x)' dx = -2\sin x \cos x, dx$. So the cosine will not be present in the numerator in the second step. – kobe Feb 24 '15 at 20:33
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I have a different approach for this integral. It looks ripe for pattern matching, as $$\frac{\sin(2x)}{\sqrt{\cos^4(x)+3}}=\frac{2\sin(x)\cos(x)}{\sqrt{3}\sqrt{\left(\frac{\cos^2(x)}{\sqrt{3}}\right)^2+1}}$$ Now let $u = \frac{\cos^2(x)}{\sqrt{3}}$ with $du = \frac{-2}{\sqrt{3}}\sin(x)\cos(x)dx$. Then $$\int \frac{\sin(2x)}{\sqrt{\cos^4(x)+3}}dx = \int \frac{-1}{\sqrt{u^2+1}} du $$ where we can see that $\frac{-1}{\sqrt{u^2+1}} du$ is precisely the derivative of $-\text{arcsinh}(u)$. Hence, $$\int \frac{\sin(2x)}{\sqrt{\cos^4(x)+3}}dx = -\text{arcsinh}\left(\frac{\cos^2(x)}{\sqrt{3}}\right)+C$$
graydad
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interesting alternative. I like math for variety of ways you can do things. – dimaastronom Feb 24 '15 at 22:49
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@dimaastronom yup, math is pretty great :) This will still evaluate to the same answer as what kobe has above; just more compactly. – graydad Feb 24 '15 at 22:51