Mathematics

# $\displaystyle \int_{0}^{\pi/2} \sqrt{1 - sin \,2 \,x}dx$ is equal to

$2 (\sqrt{2} - 1)$

##### SOLUTION
Let $I = \displaystyle \int_{0}^{\pi/2} \sqrt{1 - sin \,2x \,dx}$
$= \displaystyle \int_{0}^{\pi/4} \sqrt{(cos \,x - sin \,x)^2} dx + \displaystyle \int_{\pi/4}^{\pi/2} \sqrt{(sin x - cos x)^2} dx$
$= [sin x + cos x ]_{0}^{\pi/4} + [-cos x - sin x]_{\pi/4}^{\pi/2}$
$= \dfrac{1}{\sqrt 2} + \dfrac{1}{\sqrt 2} - 0 - 1 + \left ( -0 - 1 + \dfrac{1}{\sqrt 2} + \dfrac{1}{\sqrt 2} \right )$
$= 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

Q1 Multiple Correct Hard
Let $f(x) = 7\tan^8x+7\tan^6x-3\tan^4x-3\tan^2x$ for all $x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, then the correct expression(s) is (are)
• A. $\displaystyle \int_0^{\dfrac{\pi}{4}}xf(x)dx = \dfrac{1}{6}$
• B. $\displaystyle \int_0^{\dfrac{\pi}{4}}xf(x)dx =1$
• C. $\displaystyle \int_0^{\dfrac{\pi}{4}}xf(x)dx = \dfrac{1}{12}$
• D. $\displaystyle \int_0^{\dfrac{\pi}{4}}f(x)dx = 0$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate $\int {\dfrac {\sin x}{\sin (x+a)}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate the following integral:
$\displaystyle\int^a_0\dfrac{x}{\sqrt{a^2+x^2}}dx$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the following integrals
$\displaystyle\int {\dfrac{{{x^2}}}{{{{\left( {{a^2} - {x^2}} \right)}^{3/2}}}}dx}$

Evaluate the integral $\displaystyle\int_{-3}^{3}|x+1|\ dx$.