Question

Solve $\int_{0}^{\pi}x.sin^{3}x.\text{d}x$ [CEE 1993]
(A) $\dfrac{4\pi}{3}$
(B) $\dfrac{2\pi}{3}$
(C) 0
(D) None of these

Answer 1

Hint: Integration is the process of connecting the parts to create the whole. The region bounded by the graph of functions is defined and calculated using integration. Finding the area of the region under the curve is the process of integration. To do this, cover the area with as many little rectangles as possible, and then add up their areas. Finding an anti derivative of a function is the process of integration.



Formula Used:The integration is solved using the fundamental property $\int_{0}^{a}f(x)=\int_{0}^{a}f(a-x)$. It is a question of definite integration, so the upper and lower limits are substituted after solving the integration.



Complete step by step solution:The given integral is I = $\int_{0}^{\pi}x.sin^{3}x.\text{d}x$ --------- (1)
Here the given function is f(x) = $x.sin^{3}x$ and the total range over which the integral is to be done is 0 to $\\pi$, with a lower limit of 0 and an upper limit $\\pi$.
We know that the fundamental property of integrals is $\int_{0}^{a}f(x)=\int_{0}^{a}f(a-x)$.
Using this property, we get
Also, I = $\int_{0}^{\pi}(\pi-x).sin^{3}x.\text{d}x$ --------- (2)
Adding equations (1) and (2), we get
 2I = $\int_{0}^{\pi}\pi.sin^{3}x.\text{d}x$
We know that $\\sin^{3}x=(3sinx-sin3x)/4$, so we get
   2I = $\dfrac{\pi}{4}\int_{0}^{\pi}3\sin~x-\sin~3x.\text{d}x$
    = $\dfrac{\pi}{4}\left[-3cos~x+\dfrac{cos~3x}{3}\right]_{0}^{\pi}$
    = $\dfrac{\pi}{4}(-3(-1-1)-\dfrac{(-1-1)}{3})$
    = $\dfrac{\pi}{4}(6-2/3)$
    = $\dfrac{4\pi}{3}$
That is, 2I = $\dfrac{4\pi}{3}$
So, I= $\dfrac{2\pi}{3}$
Hence, the integration of $x.sin^{3}x$ over the integral 0 to $\\pi$ is $\dfrac{2\pi}{3}$.



Option ‘B’ is correct


Note: It should be kept in mind that changing the function also changes the upper and lower limit values after they are specified. Integration must be done with care. Trigonometric values should be carefully substituted.