Can someone help me evaluate:$$\int \frac{(\sec x)^{2}}{(1+\tan x)^{2}}dx$$ Is it possible for a hint so that I can proceed? I tried changing sec into $ 1 +\tan x $ but did not reach far.
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Is the $^2$ affecting the $x$ of the whole $\sec(x)$? If the latter you can try $y=\tan(x)$. – Feb 22 '15 at 14:08
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1The numerator is not $\sec^2 x$? – kobe Feb 22 '15 at 14:08
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It is affecting the whole sec. – Aspiring Mathlete Feb 22 '15 at 14:12
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@kobe I edited the post – Aspiring Mathlete Feb 22 '15 at 14:13
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Let $u = \tan x$. Then $du = \sec^2 x\, dx$. Thus
$$\int \frac{\sec^2 x}{(1 + \tan x)^2}\, dx = \int \frac{du}{(1 + u)^2} = \cdots$$
kobe
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How about modifying this integration into 1/(sinx+cosx)^2 and let sinx+cosx=sqrt2 ×sin (x+pi/4)?
user214425
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Also, please avoid rhetorical questions unless the context is absolutely clear, for often they look like you're asking a different question entirely. – Shaun Feb 22 '15 at 14:35
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1@shaun thanks! I will carefully read this material and write clear answer:) – user214425 Feb 22 '15 at 14:40