Evaluate the following functions : 1 1- x d x - Vidyadip

Question

Evaluate the following functions : $\int \frac{1}{1-\tan x} \cdot d x$

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Answer

$
\begin{aligned}
& I=\int \frac{1}{1-\frac{\sin x}{\cos x}} \cdot d x \\
&=\int \frac{\cos x}{\cos x-\sin x} \\
&=\int \frac{\cos x}{\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos x-\frac{1}{\sqrt{2}} \cos x\right)} \cdot d x \\
&=\frac{1}{\sqrt{2}} \int \frac{\cos x}{\cos \frac{\pi}{4} \cos x-\sin \frac{\pi}{4} \sin x} \cdot d x \\
&=\frac{1}{\sqrt{2}} \int \frac{\cos x}{\cos \left(x+\frac{\pi}{4}\right)} \cdot d x \\
& \text { put } x+\frac{\pi}{4}=t \quad \therefore x=t-\frac{\pi}{4}
\end{aligned}
$
Differentiating both sides
$
1 \cdot d x=1 \cdot d t
$
$
\begin{aligned}
& =\frac{1}{\sqrt{2}} \int \frac{\cos \left(t-\frac{\pi}{4}\right)}{\cos t} \cdot d t \\
& =\frac{1}{\sqrt{2}} \int \frac{\cos t \cdot \cos \frac{\pi}{4}+\sin t \cdot \sin \frac{\pi}{4}}{\cos t} \cdot d t \\
& =\frac{1}{\sqrt{2}} \int\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \tan t\right] \cdot d t \\
& =\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}[t+\log (\sec t)]+c \\
& =\frac{1}{2}\left[x+\frac{\pi}{4}+\log \sec \left(x+\frac{\pi}{4}\right)\right]+c
\end{aligned}
$

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