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Please help me find the following indefinite integral:

$$\int\dfrac{dx}{\sin x+\cos x-1}$$

I have tried many different substitutions to no avail. Any help is appreciated.

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    $\begingroup$ You could always try $t=\tan(x/2)$. $\endgroup$ – Angina Seng Jun 28 '17 at 5:40
  • $\begingroup$ u=tan(x/2) is the magic $\endgroup$ – Saketh Malyala Jun 28 '17 at 5:55
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Following Lord Shark the Unknown's hint, let $t=\tan(x/2)$ then $$dx=\frac{2dt}{1+t^2},\quad\cos(x)=\frac{1-t^2}{1+t^2},\quad\sin(x)=\frac{2t}{1+t^2}.$$ Hence $$\int\dfrac{1}{\sin\left(x\right)+\cos\left(x\right)-1}\,dx= \int\dfrac{1}{\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}-1}\cdot\frac{2dt}{1+t^2}=\int\dfrac{dt}{t(1-t)}\\=\int\left(\dfrac{1}{t}-\dfrac{1}{t-1}\right) dt. $$ Can you take it from here?

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HINT: multiplying numerator and denominator of your Integrand by $$\sin(x)+\cos(x)+1$$ we get $$\frac{\sin(x)+\cos(x)+1}{\sin(x)^2+\cos(x)^2+2\sin(x)\cos(x)-1}=\frac{\sin(x)+\cos(x)+1}{2\sin(x)\cos(x)}$$

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    $\begingroup$ shifty Sonnhardo, (+1) $\endgroup$ – tired Jun 28 '17 at 6:47
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Another way, using trigonometric identities

$$\sin{2x} = 2 \cdot \sin{x} \cdot \cos{x}$$ $$1 - \cos{x} = 2 \cdot \sin^2\frac{x}{2}$$

is as follows:

$$\int\dfrac{dx}{\sin x+\cos x-1}$$ $$= \int\dfrac{dx}{\sin x-(1 - \cos x)}$$ $$= \int\dfrac{dx}{2 \cdot \sin\frac{x}{2} \cdot \cos\frac{x}{2}-(2\cdot sin^2\frac{x}{2})}$$ $$= \int\dfrac{dx}{2 \cdot \sin\frac{x}{2}\cdot \big(\cos\frac{x}{2}-sin\frac{x}{2}\big)}$$ $$= \frac{1}{2}\int\dfrac{cosec\frac{x}{2} \cdot dx}{\big(\cos\frac{x}{2}-sin\frac{x}{2}\big)}$$ $$= \frac{1}{2}\int\dfrac{cosec^2\frac{x}{2} \cdot dx}{\big(\cot\frac{x}{2}-1\big)}$$ $$= \int\dfrac{cosec^2\frac{x}{2} \cdot \frac{dx}{2}}{\big(\cot\frac{x}{2}-1\big)}$$ $$ = -ln\bigg(cot\bigg(\frac{x}{2}\bigg) - 1\bigg) + C$$

Hence $$\int\dfrac{dx}{\sin x+\cos x-1}= \boxed{- ln\bigg(cot\bigg(\frac{x}{2}\bigg) - 1\bigg) + C}$$

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$$I = \int \frac{1}{\sin x + \cos x - 1}\, dx$$ Apply integral Subtitution $$u = \tan\left(\frac{x}{2}\right)$$ $$\sin(x) = \frac{2u}{1+u^2},\quad \cos(x) = \frac{1 - u^2}{1 + u^2}\, \quad dx = \frac{2}{1 + u^2}$$ It follows $$\int\frac{1}{u(-u + 1)}\, du = -\int\frac{1}{u(u - 1)}\, du $$ by partial fraction $$-\int\left(\frac{1}{(u - 1)}- \frac{1}{u}\right)\, du $$ $$-\left[\int\frac{1}{(u - 1)}\, du -\int \frac{1}{u}\, du\right] $$ $\int \frac{1}{u}\, du = \ln(u)+C $

$\int\frac{1}{(u - 1)}\, du = \ln(u - 1)+C$ $$-\left[\ln(u - 1) - \ln(u)\right]+C $$ $u = \tan\left(\frac{x}{2}\right)$ $$-\ln\big(\tan\left(\frac{x}{2}\right) - 1\big) + \ln\big(\tan\left(\frac{x}{2}\right)\big)+C $$

$$I= \ln\big(\tan\left(\frac{x}{2}\right)\big)-\ln\big(\tan\left(\frac{x}{2}\right) - 1\big) +C $$

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$$\int\frac{1}{\sin{x}+\cos{x}-1}dx=\int\frac{1}{2\sin\frac{x}{2}\cos\frac{x}{2}-2\sin^2\frac{x}{2}}dx=$$ $$=\int\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}dx=\int\frac{\sin^2\frac{x}{2}+\sin\frac{x}{2}\cos\frac{x}{2}-\sin\frac{x}{2}\cos\frac{x}{2}+\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}dx=$$ $$=\int\frac{\sin\frac{x}{2}\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)+\cos\frac{x}{2}\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}{2\sin\frac{x}{2}\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}dx=$$ $$=\int\left(\frac{\sin\frac{x}{2}+\cos\frac{x}{2}}{2\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}+\frac{\cos\frac{x}{2}}{2\sin\frac{x}{2}}\right)dx=$$ $$=\int\left(-\frac{d\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}{\cos\frac{x}{2}-\sin\frac{x}{2}}+\frac{d\left(\sin\frac{x}{2}\right)}{\sin\frac{x}{2}}\right)=\ln\left|\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}\right|+C$$

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\begin{align} I&=\int \frac{dx}{\sin x+\cos x-1}=\int \frac{\frac{dx}{\cos x-1}}{\frac{\sin x}{\cos x-1}+1}=\int \frac{-d\left(\frac{\sin x}{\cos x-1}+1\right)}{\frac{\sin x}{\cos x-1}+1}\\ &=-\ln \left|\frac{\sin x}{\cos x-1}+1\right|+C=\ln \left|\frac{\cos x-1}{\sin x+\cos x-1}\right|+C \end{align}

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