How can i find the integration of this example $\int \frac{\sin x}{\sin x - \cos x } dx$ I tried first add cos and then substracting cos but then what about $\int \frac{\cos x}{\sin x - \cos x } dx\ ?$
Integration Example
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0the ans. in my book will not match when i take t=tan(x/2) or convert it into double angle Is there any easy way to solve this? – 2012-06-25
3 Answers
Multiply and divide Numerator by 2 to get $2\sin(x)$ and write it as $(\sin(x)+\cos(x)) + (\sin(x)-\cos(x))$ then divide each term by denominator.Second term would be 1(you can integrate it easily),for first term, put $(\sin(x)-\cos(x))$ as $z$ then, $(\sin(x)+\cos(x))dx$ would become $dz$ and your first integral will look like $\int {\frac{dz}{z}}$, which is $ln(z)+c$.
Let $f(a) = \int \frac{\sin(ax)}{\sin(x) - \cos(x)}dx$. Differentiating throughout by a, we get $f'(a) = a\int \frac{\cos(ax)}{\sin(x) - \cos(x)}dx$ Therefore, $af(a) - f'(a) = a\int \frac{\sin(ax) - \cos(ax)}{\sin(x) - \cos(x)}dx$ Substituting $a = 1$, we get $f(1) - f'(1) = x + C_1$ Also, $af(a) + f'(a) = a\int \frac{\sin(ax) + \cos(ax)}{\sin(x) - \cos(x)}dx$ Substituting $a = 1$ gives, $f(1) + f'(1) = \int \frac{\sin(x) + \cos(x)}{\sin(x) - \cos(x)}dx = \ln(\sin(x) - \cos(x)) + C_2$ The integral was simply solved using the substitution $t = \sin(x) - \cos(x)$. Solving the above two equations for $f(1)$ gives, $2f(1) = x + \ln(\sin(x) - \cos(x)) + C_1 + C_2$ Therefore, $f(1) = \frac{1}{2}(x + \ln(\sin(x) - \cos(x)) + C_1 + C_2)$ which is what we sought.
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11+ for the big hammer :D – 2012-06-25
Now let $ I = \int \frac{\cos x}{\cos x + \sin x} \, \mathrm{d}x \quad \text{and} \quad J = \int \frac{\sin x}{\cos x + \sin x} \, \mathrm{d}x$ Notice that \begin{align*} I + J & = \int \frac{\cos x + \sin x}{\cos x + \sin x} \, \mathrm{d}x = x + \mathcal{C} \\ I - J & = \int \frac{(\sin x + \cos x)'}{\cos x + \sin x} \, \mathrm{d}x = \ln\bigl( \cos x + \sin x \bigr) + \mathcal{C} \end{align*} Solving the system yields $I = \frac{1}{2}x + \frac{1}{2}\ln\bigl( \cos x + \sin x \bigr) + \mathcal{C}$ and $J = \frac{1}{2}x - \frac{1}{2}\ln\bigl( \cos x + \sin x \bigr) + \mathcal{C}$ as wanted.