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$$\int_{0}^{\pi/2}\frac{\cos^2x}{{\cos^2x + 4\sin^2x}}dx$$

Can I solve this question using p4 of definite integrals

James
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Mark Henry
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2 Answers2

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Hint: Recall that for any $f$ continuous on $[0,1]$ it is an elementary exercise to check

$$ \int\limits_0^{\pi/2} f(\sin x) d x = \int\limits_0^{\pi/2} f(\cos x) d x $$

James
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  • It's a good observation, and you are not under any obligation to solve problems that are posed without any evidence of effort, but I'd say this brief note is more of a Comment than an Answer. Of course a few more words about how to use the observation might well tip the balance. – hardmath Mar 01 '19 at 05:54
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    @hardmath I believe the current meta consensus is that hints are acceptable as answers, especially when there isn't much work shown in the problem. I personally think this is quite a good hint, as it not only offers help in solving the problem while leaving some work to the asker, but also helps in solving similar problems. It's much more useful IMO than the "other half" of the solution would be, and certainly helps the asker learn more than a full solution would. – Carl Schildkraut Mar 01 '19 at 06:23
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Define $I$ as the original definite integral .Note that $\sin x=\tan x/\sec x$ and $\cos x=1/\sec x$. Use $\sec^2x=\tan^2x+1$. Let $u=\tan x$ and $\mathrm du=\sec^2x \mathrm dx$.

$$\int \sec^2x\dfrac{1}{(\tan^2x+1)(4\tan^2x+1)}\mathrm dx=\int\dfrac{1}{(u^2+1)(4u^2+1)}\mathrm du$$

$$\int\dfrac{4/3}{4u^2+1}\mathrm du-\int\dfrac{1/3}{u^2+1}\mathrm du=\dfrac{2\arctan(2\tan x)-x}{3}\implies I=\dfrac{\pi}{6}$$

Paras Khosla
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