Mathematics

# Solve - $\int {\dfrac{{\cos x}}{{1 - \cos x}}} dx$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int\frac{dx}{2-3\cos 2x}=$
• A. $\displaystyle \frac{1}{\sqrt{5}}\log|\frac{\sqrt{5}tanx-1}{\sqrt{5}tanx+1}|+c$
• B. $\displaystyle \frac{-1}{2\sqrt{5}}\log|\frac{\sqrt{5}tanx-1}{\sqrt{5}tanx+1}|+c$
• C. $-\dfrac{1}{\sqrt{5}}\log|\displaystyle \frac{\sqrt{5}tanx-1}{\sqrt{5}tanx+1}|+c$
• D. $\displaystyle \frac{1}{2\sqrt{5}}\log|\frac{\sqrt{5}tanx-1}{\sqrt{5}tanx+1}|+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int\dfrac{dx}{\sqrt{1-x^2}-1}=$
• A. $\dfrac{1+\sqrt{1-x^2}}{x}-2tan^{-1}\sqrt{\dfrac{1+x}{1-x}}+C$
• B. $\dfrac{1+\sqrt{1-x^2}}{x}+2tan^{-1}\sqrt{\dfrac{1+x}{1-x}}+C$
• C. $\dfrac{1+\sqrt{1-x^2}}{x}+2tan^{-1}\sqrt{\dfrac{1-x}{1+x}}+C$
• D. $\dfrac{1+\sqrt{1-x^2}}{x}-2tan^{-1}\sqrt{\dfrac{1-x}{1+x}}+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of $\int \sqrt { \dfrac { e^ x\quad -\quad 1 }{ e^ x\quad +\quad 1 } } dx$ is equal to
• A. $\ell n\left( e^ x+\sqrt { e^ 2x\quad -1 } \right) - sec^ -1(e^ x)+c$
• B. $\ell n\left( e^ x+\sqrt { e^ 2x\quad -1 } \right) +sec^ -1(e^ x)+c$
• C. $\ell n\left( e^ x-\sqrt { e^ 2x\quad -1 } \right) -sec^ -1(e^ x)+c$
• D. $\ell n\left( e^ x+\sqrt { e^ 2x\quad -1 } \right) -sin^ -1(e^ {-x})+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Find solution in terms of indefinite integration, using substitution
$\int_0^1 {x\cos \left( {{{\tan }^{ - 1}}x} \right)} \,dx$

$\int { \cfrac { f'(x) }{ f(x) } dx } =\log { [f(x)] } +c$