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I can't solve this integral, can anybody help me ?

$$\int\frac{\cos^2x}{\sin^6x}\mathrm dx$$

I've tried with $\frac{\cos2x+1}{2}=\cos^2x$ transformations but still cannot get the result.

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    Start by noting the $cot^2 (x) $ – 2017-02-27
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    HINT: use the identity $\cos^2x+\sin^2x=1$. – 2017-02-27

2 Answers 2

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$\displaystyle \dfrac{\cos^2x}{\sin^6 x}=\csc^4 x \cot^2 x= (\cot x+\cot^3 x)^2 \csc^2 x$

Now integrating and letting $\cot x=y$

$I=-\int (y^2+y^6+2y^4)\; dy$

Now you can proceed.

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    Could you resolve this integral without $cscx$ function ? – 2017-02-27
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    @Mikkey Actually the power of the sine in the denominator is even, so using the $csc^2$ is really the way to go. – 2017-02-27
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    Every possible way would lead to the same answer. Try multiplying $\sec^6 x$ up and down instead – 2017-02-27
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    The problem is that i'm not allowed to use $cscx$ and $sec$, only $sinx$ $cosx$ $tanx$ $cotx$ – 2017-02-27
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    @Mikkey Every step is still true in this answer. Just replace each function with sine and cosine and you are golden – 2017-02-27
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    @Mikkey Don't call it cosecant: observe that $\;\csc x=\frac1{\sin x}\;$ and etc... – 2017-02-27
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If you absolutely want to avoid trig, here you go
$$\int\frac{\cos^2x}{\sin^6x}dx$$ $$\int\frac{1-\sin^2x}{\sin^6x}dx$$ We now let $u = \sin^2(x) \implies u' = 2\sqrt{u-u^2}$
This shrinks our interval a bit since the square root actually gives the absolute value, so we need really have the above for $[0+\pi n,\pi/2+\pi n]$. $$\frac{1}{2}\int\frac{1-u}{u^6\sqrt{u-u^2}}du$$ You can now multiply out the denominator, make a substitution to move the square root to the top, and simplify. Good luck.

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    I still do not know what is $sec(x)$. I'm obligated to use only $sinx$ ,$cosx$, $tanx$ and $cotx$ – 2017-02-27
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    Would you mind explaining how you rewrote that numerator? – 2017-02-27