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Question

Question: What is the integral of \[|\sin (x)|\] ?...

What is the integral of sin(x)|\sin (x)| ?

Explanation

Solution

In this question, we need to find the integral of sin(x)|sin(x)| . The modulus is nothing but it is an absolute value of any number or any function. Integration is nothing but its derivative is equal to its original function. Integration is also known as antiderivative. The inverse of differentiation is known as integral. The symbol `\int’ is the sign of the integration. The process of finding the integral of the given function is known as integration. The modulus of sin(x)\sin (x) is nothing but it’s absolute value can never be greater than 11. Since the given expression is in the modulus bracket , the expression can be written as ±\pm the given function thus we get two cases. Then we need to integrate both the cases separately.

Complete step-by-step solution:
Given, sin(x)|\sin (x)|
This function can be written as sin(x)\sin (x) if sinx  0\sin x\ \geq \ 0 and sin(x)- \sin (x) if sinx<0\sin x < 0 (since the function is in the modulus bracket).
Let us consider the given function as II.
 I=sin(x)\Rightarrow \ I = |\sin (x)|
Case 1 )
I=sin(x) if sinx  0I = \sin (x)\ if\ \sin x\ \geq \ 0
On integrating,
We get,
  I=sin(x)dx\Rightarrow \ \ I = \int\sin\left( x \right){dx}
We know that the integral of sin x=cosxsin\ x = - \cos x
 I=cos(x) +c\Rightarrow \ I = - \cos\left( x \right)\ + c if xx belongs to [2nπ,(2n+1)π]\lbrack 2n\pi,(2n + 1)\pi\rbrack and nZn \in Z
Where cc is the constant of integration.
Case 2 )
I=sinxI = - \sin x if sinx<0\sin x < 0
On integrating,
We get,
  I=sin(x)dx=sin(x)dx\Rightarrow \ \ I = \int - \sin\left( x \right){dx} = - \int\sin\left( x \right){dx}
We know that the integral of sin x=cosxsin\ x = - \cos x
 I=(cosx)+c\Rightarrow \ I = - ( - \cos x) + c
Where cc is the constant of integration.
 I=cosx+c\Rightarrow \ I = \cos x + c if xx belongs to [(2n1)π, 2nπ]\lbrack(2n – 1)\pi,\ 2n\pi\rbrack and nZn \in Z
Thus we get the integral of sin(x) \left| \sin\left( x \right) \right|\ is ±cosx +c \pm \cos x\ + c\
Final answer :
The integral of sin(x)\left| \sin\left( x \right) \right| is ±cosx +c \pm \cos x\ + c\

Note: The anti-derivative of the function is also known as the inverse of the derivative of the function . The concept used in this question is integration method, that is integration of the hyperbolic function . Since this is an indefinite integral we have to add an arbitrary constant ‘cc’. cc is called the constant of integration. The variable xx in dxdx is known as the variable of integration or integrator. The range of sin(x)\sin (x) is 1  sin x  1- 1\ \leq \ sin\ x\ \leq \ 1. Mathematically, integrals are also used to find many useful quantities such as areas, volumes, displacement, etc…