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$\int\frac{\cos^2 x}{1-\sin x} dx $ can someone explain me how to solve this one and please show your complete solution? So am I supposed to make the numerator $1+sinx$? but I think that doesn't help. Should I do long division?

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    $\begingroup$ "I think that doesn't help": did you even try ? $\endgroup$
    – user65203
    Jan 6, 2015 at 11:24

3 Answers 3

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We know that $\displaystyle \cos^2 x = 1 - \sin^2 x = (1-\sin x)(1+ \sin x)$

Hence, $$ \require{cancel} \begin{align} \int\frac{\cos^2x}{1-\sin x}\mathrm dx &= \int \frac{\cancel{(1-\sin x)}(1+\sin x)}{\cancel{1- \sin x}}\mathrm dx\\ &= \int \mathrm dx + \int \sin x\\ &= x - \cos x + \color{gray}{\mathcal C} \end{align}$$


Aliter:: The same method in another approach: $$\require{cancel} \frac{\cos^2 x}{1 - \sin x}\cdot \frac{1 + \sin x}{1 + \sin x} = \frac{\cancel{\cos ^2 x}\cdot (1 + \sin x)}{\cancel{\cos^2 x}} = 1 + \sin x$$

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$\int \frac{1-\sin ^2 x}{1-\sin x}dx=\int (1+\sin x) dx=x-\cos x+C_1$.

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  • $\begingroup$ Shouldn't you at least explain? $\endgroup$
    – user197789
    Jan 6, 2015 at 11:25
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    $\begingroup$ @SixthOfFour What exactly is there to explain?? And even if there is something, this is so basic that it can safely be left to the OP to think about. $\endgroup$
    – Timbuc
    Jan 6, 2015 at 11:26
  • $\begingroup$ got it with an operator thanks! $\endgroup$
    – Mickey
    Jan 6, 2015 at 11:29
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    $\begingroup$ @SixthOfFour I think you may have a point there, yet I think anyone dealing with integrals (or antiderivatives) has already done a good deal of basic algebra and trigonometry, so hints should be enough. Anymore, as in this case,is suspicious of being "do my homework for me", imo. $\endgroup$
    – Timbuc
    Jan 6, 2015 at 15:27
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    $\begingroup$ @Timbuc And you certainly have a point there. $\endgroup$
    – user197789
    Jan 6, 2015 at 16:13
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$$\int\frac{\cos^2x}{1-\sin x}dx=\int\frac{1-\sin^2x}{1-\sin x}dx=\int\frac{(1-\sin x)(1+\sin x)}{1-\sin x}dx=$$ $$=\int(1+\sin x)dx=\int1dx+\int\sin xdx=x-\cos x +C$$

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