- #1
juantheron
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$\displaystyle \int_{0}^{\frac{\pi}{2}}\sqrt{\sin x}dx.\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin x}}dx =$
A definite integral is a mathematical concept that represents the area under a curve on a given interval. It is denoted by ∫f(x)dx where f(x) is the function and dx represents the infinitesimal change in x.
To calculate a definite integral, you first need to find the indefinite integral of the function. Then, you plug in the upper and lower limits of the interval into the indefinite integral and subtract the result. This gives you the value of the definite integral.
The function $\sqrt{\sin x}$ is a trigonometric function that represents the square root of the sine of x. It is a periodic function with a period of 2π and has a range of [0,1].
The definite integral of $\sqrt{\sin x}$ is important because it allows us to calculate the area under the curve of the function on a given interval. This is useful in many real-life applications, such as calculating the work done by a varying force or finding the displacement of a moving object.
The value of the definite integral of $\sqrt{\sin x}$ cannot be expressed in terms of elementary functions. It can only be approximated using numerical methods or expressed in terms of special functions.