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Question: The primitive of the function \(x\left| {\cos x} \right|\), when \(\dfrac{\pi }{2} < x < \pi \) give...

The primitive of the function x∣cos⁡x∣x\left| {\cos x} \right|, when π2<x<π\dfrac{\pi }{2} < x < \pi given by
A. Cos x + x Sinx + c
B. Cos x − x sinx + c
C. xsinx − cosx + c
D. None of these + c

Explanation

Solution

By checking out the question, we get to know that the primitive of the function is integral to the function. Which means we need to integrate the given function. Secondly, divide the function according to the behaviour at different intervals as the Modulus function works differently for different intervals before integrating. Hence, derive the function without modulus and then integrate it to obtain the primitive relation of the given function.

Complete step by step solution: Let’s begin with the given function x∣cos⁡x∣x\left| {\cos x} \right| in the given interval (π2,π)\left( {\dfrac{\pi }{2},\pi } \right).
As we know that, the modulus function results from the output into positive while breaking modulus we need to take care of Cos function, which will always result in positive.
As you know that Cos function is positive in (0,π2)\left( {0,\dfrac{\pi }{2}} \right) and negative in(π2,π)\left( {\dfrac{\pi }{2},\pi } \right) . We are going to open the function with a negative sign which will over all result the output in positive value
f(x)=x∣cosx∣f\left( x \right) = x\left| {cosx} \right|
=−xcos⁡x= - x\cos x in the interval π2\dfrac{\pi }{2} toπ\pi .
We need primitive of the function, which is integration with respect to function variable
∫f(x)    dx=∫−x  cos⁡x    dx\int {f\left( x \right)} \;\;dx = \int { - x\;\cos x\;\;dx}
Here we are going to treat x as one function and cosx as another one as we know that the integral of two functions let’s suppose U and V in the form of multiple.
i.e.∫u.v    dx\int {u.v\;\;dx} is equal to u∫v    dx−∫dudx∫vdx    dxu\int {v\;\;dx - \int {\dfrac{{du}}{{dx}}\int {vdx} \;\;dx} }
Similarly in this condition, we are going to integrate.
−∫xcos⁡x    dx=−[x∫cos⁡xdx−∫dxdx−∫cos⁡xdx    dx]- \int {x\cos x\;\;dx = - \left[ {x\int {\cos xdx - \int {\dfrac{{dx}}{{dx}} - \int {\cos xdx\;\;dx} } } } \right]}
=−[xsin⁡x−∫sin⁡x    dx]= - \left[ {x\sin x - \int {\sin x\;\;dx} } \right]
as we know that integral of function cos is positive sin function.
Further
=−xsin⁡x+∫sin⁡x    dx= - x\sin x + \int {\sin x\;\;dx}
=−xsin⁡x−cos⁡x+c= - x\sin x - \cos x + c
Here, we also know that the integral of sinx is negative cos x, and the constant c is applied as some variable might be present which gets zero in differentiation.

The correct option is B.

Note: Always try to divide functions carefully to integrate. If u have multiple functions to integrate, divide it into 2 and later apply the same to integrate and differentiate. Students may think why is it so important? Because sometimes if we choose wrongly then the integration sign will never be removed. exp(sinx) is an example of such a function.