Mathematics

# Evaluate the following integral:$\displaystyle \int { \cfrac { \sin { x } }{ \sqrt { 4\cos ^{ 2 }{ x } -1 } } } dx$

##### SOLUTION
$I = \displaystyle \int \dfrac{\sin x}{\sqrt{4 \cos^2 x - 1}} dx$

Let $2 \cos x = t$

$\dfrac{dt}{dx} = 2 (-\sin x)$

$\dfrac{-dt}{2} = \sin x dx$

$I = -\dfrac{1}{2} \displaystyle \int \dfrac{dt}{\sqrt{t^2 - 1^2}}$

$I = -\dfrac{1}{2} [\ln [t + \sqrt{t^2 - 1}] + c$

put the value of t, we get

$I = -\dfrac{1}{2} \ln [2 \cos x + \sqrt{(2 \cos x)^2 -1} ] + c$

$I = -\dfrac{1}{2} \ln [2 \cos x + \sqrt{4 \cos^2 x - 1} ] + c$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int\dfrac{(\cos x)^{n-1}}{(\sin x)^{n+1}}dx=$
• A. $\displaystyle \frac{-\cot^{n}x}{n+1}+c$
• B. $\displaystyle \frac{\cot^{n}x}{n}+c$
• C. $\displaystyle \frac{\cot^{n}x}{n+1}+c$
• D. $\displaystyle \frac{-\cot^{n}x}{n}+c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate:

$\int x.\sin{2x}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle\int _{ -1 }^{ 1 }{ x\left( 1-x \right) \left( 1+x \right) dx }$ is equal to
• A. $\dfrac { 1 }{ 3 }$
• B. $\dfrac { 2 }{ 3 }$
• C. $1$
• D. $-1$
• E. $0$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate the definite integral   $\displaystyle \int_1^2\frac {5x^2}{x^2+4x+3}$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$