Evaluvate the following intregals: 1 1- dx - Vidyadip

Question

Evaluvate the following intregals:
$\int\frac{1}{1-\tan\text{x}}\text{ dx}$

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Answer

Let $\text{I}=\int\frac{1}{1-\tan\text{x}}\text{ dx}$
$=\int\frac{1}{1-\frac{\sin\text{x}}{\cos\text{x}}}\ \text{dx}$
$=\int\frac{\cos\text{x}}{\cos\text{x}-\sin\text{x}}\ \text{dx}$
Let $\cos\text{x}=\lambda\frac{\text{d}}{\text{dx}}(\cos\text{x}-\sin\text{x})+\mu(\cos\text{x}-\sin\text{x})+\text{v}$
$=\lambda\frac{\text{d}}{\text{dx}}(-\sin\text{x}-\cos\text{x})+\mu(\cos\text{x}-\sin\text{x})+\text{v}$
$\cos\text{x}=\sin(-\lambda-\mu)+\cos\text{x}(-\lambda+\mu)+\text{v}$
Compairing the cooefficients of $\cos\text{x}\ \&\sin\text{x}$ on the both the sides,
$-\lambda-\mu=0\ ...(1)$
$-\lambda+\mu=1\ ...(2)$
$\text{v}=0\ ...(3)$
Equation (1), (2), (3) gives
$\lambda=-\frac{1}{2},\mu=\frac{1}{2},\text{v}=0$
$\text{I}=\int\frac{-\frac{1}{2}(-\sin\text{x}-\cos\text{x})+\frac{1}{2}(\cos\text{x}-\sin\text{x}_)}{(\cos\text{x}-\sin\text{x})}\ \text{dx}$
$=\frac{1}{2}\int\frac{(\cos\text{x}+\sin\text{x})}{(\sin\text{x}-\cos\text{x})}\ \text{dx}+\frac{1}{2}\int\ \text{dx}$
$\text{I}=-\frac{1}{2}\log|\cos\text{x}-\sin\text{x}|+\frac{1}{2}\text{x}+\text{C}$
$\text{I}=\frac{1}{2}\text{x}-\frac{1}{2}\log|\cos\text{x}-\sin\text{x}|+\text{C}$

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