Mathematics

# The value of the integral $\displaystyle\int^1_0\dfrac{dx}{x^2+2x\cos \alpha +1}$, where $0 < \alpha < \dfrac{\pi}{2}$, is equal to

$\dfrac{\alpha}{2\sin \alpha}$

##### SOLUTION
$\displaystyle\int^1_0\dfrac{dx}{x^2+2x\cos \alpha +1}$

$\implies \displaystyle\int^1_0\dfrac{dx}{x^2+2x\cos \alpha +\cos^2 \alpha+\sin^2 \alpha}$

$\implies \displaystyle \int_0^1 \cfrac{dx}{(x+ \cos \alpha)^2 + \sin^2 \alpha}$

$\implies \displaystyle \cfrac{1}{\sin \alpha} \Bigg[\tan^{-1} \cfrac{x+\cos x}{\sin x}\Bigg]_0^1$   $\left[\because \int \cfrac{1}{x^2+a^2}=\dfrac{1}{a}tan^{-1}(\dfrac{x}{a}) \right]$

$\implies \displaystyle \cfrac{1}{\sin \alpha} \Bigg[\tan^{-1} \cfrac{1+\cos \alpha}{\sin \alpha} - \tan^{-1} \cfrac{\cos \alpha}{\sin \alpha}\Bigg]$

$\implies \cfrac{1}{\sin \alpha} \Bigg[ \tan^{-1} \cfrac{\cfrac{1+\cos \alpha}{\sin \alpha} - \cfrac{\cos \alpha}{\sin \alpha}}{1+\cfrac{\cos \alpha + \cos^2 \alpha}{\sin^2 \alpha}}\Bigg]$

$\implies \cfrac{1}{\sin \alpha} \tan^{-1} \cfrac{\sin \alpha}{1+\cos \alpha}$

$\implies \cfrac{1}{\sin \alpha} \tan^{-1} \tan \cfrac{\alpha}{2}$

$\implies \cfrac{\alpha}{2 \sin \alpha}$

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

#### Realted Questions

Q1 Subjective Medium
Evaluate
$\displaystyle \int^{2}_{0}\dfrac {1}{\sqrt {3+2x-x^{2}}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Evaluate $\int \dfrac{1}{x^2+2x+2} dx$
• A. $\tan^{-1}(x+2)+C$
• B. $\tan^{-1}(x+3)+C$
• C. None of these
• D. $\tan^{-1}(x+1)+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The anti derivative of $\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)$ equals
• A. $\dfrac{1}{3}x^{\dfrac{1}{3}}+2x^{\dfrac{1}{2}}+C$
• B. $\dfrac{2}{3}x^{\dfrac{2}{3}}+\dfrac{1}{2}x^{2}+C$
• C. $\dfrac{2}{3}x^{\dfrac{3}{2}}+\dfrac{1}{2}x^{\dfrac{1}{2}}+C$
• D. $\dfrac{2}{3}x^{\dfrac{3}{2}}+2x^{\dfrac{1}{2}}+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int\{\frac{(\log \mathrm{x}-1)}{(1+(\log \mathrm{x})^{2}}\}^{2}$ dx is equal to

• A. $\dfrac{logx}{(logx)^{2}+1}+c$
• B. $\dfrac{x}{x^{2}+1}+c$
• C. $\dfrac{xe^{x}}{1+x^{2}}+c$
• D. $\dfrac{x}{(logx)^{2}+1}+c$

Let g(x) =$\displaystyle \int_{0}^{x}f\left ( t \right )dt,$ where f is a function