# 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 ? – Yves Daoust Jan 6 '15 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 '15 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. – Timbuc Jan 6 '15 at 11:26
• got it with an operator thanks! – Mickey Jan 6 '15 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. – Timbuc Jan 6 '15 at 15:27
• @Timbuc And you certainly have a point there. – user197789 Jan 6 '15 at 16:13

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