# Evaluate $\int\frac{\cos^2 x}{1-\sin x }dx$

$\int\frac{\cos^2 x}{1-\sin x} dx$ 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?

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

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: $$\require{cancel} \frac{\cos^2 x}{1 - \sin x}\cdot \frac{1 + \sin x}{1 + \sin x} = \frac{\cancel{\cos ^2 x}\cdot (1 + \sin x)}{\cancel{\cos^2 x}} = 1 + \sin x$$

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

• Shouldn't you at least explain?
– user197789
Jan 6, 2015 at 11:25
• @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. Jan 6, 2015 at 11:26
• got it with an operator thanks! Jan 6, 2015 at 11:29
• @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. Jan 6, 2015 at 15:27
• @Timbuc And you certainly have a point there.
– user197789
Jan 6, 2015 at 16:13

$$\int\frac{\cos^2x}{1-\sin x}dx=\int\frac{1-\sin^2x}{1-\sin x}dx=\int\frac{(1-\sin x)(1+\sin x)}{1-\sin x}dx=$$ $$=\int(1+\sin x)dx=\int1dx+\int\sin xdx=x-\cos x +C$$