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With $(a)$ I got that $-y^2 dx = \sec^2x\ dy$, but it makes no sense. Hence, no Idea how to handle $(b)$.

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    $\begingroup$ For $(i)$, since $0\le x\le\dfrac\pi2$, I think we can use the facts that $\cos x\le1$ and $\sin x +\cos x\ge1$. Divide them by $\cos x$ and combine. $\endgroup$
    – Tunk-Fey
    Jun 14, 2014 at 22:16

2 Answers 2

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Hint :

For $(a)$, using separation of variables, we obtain \begin{align} \frac{dy}{y^2}&=-\frac1{\sec^2x}\ dx\\ \int y^{-2}\ dy&=-\int \cos^2x\ dx\\ \int y^{-2}\ dy&=-\int\frac{1+\cos 2x}{2}\ dx. \end{align}

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    $\begingroup$ For $(i)$, since $0\le x\le\dfrac\pi2$, I think we can use the facts that $\cos x\le1$ and $\sin x +\cos x\ge1$. Divide them by $\cos x$ and combine. $\endgroup$
    – Tunk-Fey
    Jun 14, 2014 at 22:19
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Using separation of variables and a little more rearranging (you almost had it), we get the following:

$$\int \cos^2(x)dx = \int \frac{-1}{y^2}dy$$

I simply took what you already had and divided both sides by $-y^2$ and $\sec^2(x)$. Now it's just a matter of good old integration. :)

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