Mathematics

# Integrate $\displaystyle \int _{ 0 }^{ 2\pi }{ \sqrt {1+\sin x }}dx$

$4$

##### SOLUTION
$\int_0^{2\pi } {\sqrt {1 + \sin x} } dx$
$= \int_0^{2\pi } {\sqrt {{{\sin }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2} + 2\sin \frac{x}{2}\cos } \frac{x}{2}} dx$    (${\therefore {{\sin }^2}x + {{\cos }^2}x = 1}$ and $\sin 2x = 2\sin x\cos x$)
$= \int_0^{2\pi } {\sqrt {{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}} dx}$
$= \int_0^{2\pi } {\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)} dx$
$= \left[ { - 2\cos \frac{x}{2} + 2\sin \frac{x}{2}} \right]_0^{2\pi }$
$= 2\left[ {\sin \frac{x}{2} - \cos \frac{x}{2}} \right]_0^{2\pi }$
$= 2\left[ {1 + 1} \right]$
$= 4$

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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
Solve:$\int \dfrac{x^2}{\sqrt{1-x}}.dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Evalaute the integral
$\displaystyle \int_{0}^{\pi}xf(\sin x)dx$
• A. $2\pi$
• B. $\displaystyle \pi\int_{0}^{\pi}f(\cos x)dx$
• C. $\displaystyle \pi\int_{0}^{\pi}f(\sin x) dx$
• D. $\displaystyle \pi\int_{0}^{\pi/2}f(\cos x)dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate: $\displaystyle\int \dfrac {dx}{x(1 + x^{2})}$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $\displaystyle \int_{-1}^{-4} f(x) dx = 4$ and $\displaystyle \int_2^{-4} (3 - f(x)) dx = 7$, then the value of $\displaystyle \int_{-2}^1 f(-x) dx$, is
• A. 2
• B. -3
• C. 5
• D. none of these

$\displaystyle\int\dfrac{x}{\sqrt{3x^2+4}}dx$.