Mathematics

# Integrate the function    $\displaystyle \frac {1}{1-\tan x}$

##### SOLUTION
Let $\displaystyle I=\int \frac {1}{1-\tan x}dx$
$\displaystyle=\int \frac {\cos x}{\cos x- \sin x}dx$
$\displaystyle=\frac {1}{2}\in \frac {2 \cos x}{\cos x-\sin x}dx$
$\displaystyle=\frac {1}{2}\int \frac {(\cos x-\sin x)+(\cos x+ \sin x)}{(\cos x-\sin x)}dx$
$\displaystyle=\frac {1}{2}\int 1 dx+\frac {1}{2}\int \frac {\cos x+\sin x}{\cos x-\sin x}dx$
$\displaystyle=\frac {x}{2}+\frac {1}{2}\int \frac {\cos x+\sin x}{\cos x-\sin x}dx$
Put $\cos x-\sin x=t\Rightarrow (-\sin x-\cos x)dx=dt$
$\therefore \displaystyle I=\frac {x}{2}+\frac {1}{2}\int \frac {-(dt)}{t}$
$\displaystyle=\frac {x}{2}-\frac {1}{2}\log |t|+C$
$\displaystyle =\frac {x}{2}-\frac {1}{2}\log |\cos x- \sin x|+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Subjective Medium
Evaluate:
$\displaystyle\int x^2\tan(x^3)\ dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
$\displaystyle\int_{0}^{1}\dfrac{1-x^{2}}{(1+x^{2})^{2}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Evaluate using limit of sum:
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• A. $26$
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1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard

Let $f(x)$ be a function satisfying $f'(x)=f(x)=e^x$ with $f(\mathrm{0})=1$ and $g(x)$ be a function that satisfies $f(x)+g(x)=x^{2}$. Then, the value of the integral $\displaystyle \int_{0}^{1}f(x)g(x)dx$ is
• A. $e+\displaystyle \frac{e^{2}}{2}-\frac{3}{2}$
• B. $e+\displaystyle \frac{e^{2}}{2}+\frac{5}{2}$
• C. $e-\displaystyle \frac{e^{2}}{2}-\frac{5}{2}$
• D. $e-\displaystyle \frac{e^{2}}{2}-\frac{3}{2}$

If $\displaystyle \int \dfrac{e^x - 1}{e^x + 1}dx =f(x) + c$ then $f(x) =$