Question

How do you find the integral of $\int{48\left( \dfrac{{{\left( \sin \left( \sqrt{x} \right) \right)}^{3}}}{\sqrt{x}} \right)}$ ?

Answer 1

In the given question, we have been asked to integrate the given constant. In order to solve the question, we integrate the numerical by using the basic concept of integration. First we need to take out the constant part out of the integration. Later we will need to integrate the variable part using a suitable integration formula i.e. here we will need to use a substitution method to solve the integration and we will get our required answer.

Complete step by step answer:
We have given,
$\Rightarrow \int{48\left( \dfrac{{{\sin }^{3}}\left( \sqrt{x} \right)}{\sqrt{x}} \right)}$
Let I be the integration of the given equation.
Therefore, we can write the integration as,
$\Rightarrow I=\int{48\left( \dfrac{{{\sin }^{3}}\left( \sqrt{x} \right)}{\sqrt{x}} \right)}$
As we know that,
Integration of any constant ‘k’,
$\Rightarrow \int{kf(t)dt}=k\int{f(t)dt}$
Therefore,
Taking the constant part out of the integration, we get
$\Rightarrow I=48\int{\left( \dfrac{{{\sin }^{3}}\left( \sqrt{x} \right)}{\sqrt{x}} \right)}$
Now,let

\Rightarrow \dfrac{1}{2\sqrt{x}}dx=dt\\\ \Rightarrow dx=2tdt$$ Substituting the above values, we get $$\Rightarrow I=48\int{\left( \dfrac{{{\sin }^{3}}\left( t \right)}{t}2tdt \right)}$$ Cancelling out the common terms, we obtained $$I=48\int{\left( {{\sin }^{3}}\left( t \right)tdt \right)}$$ As we know that, Using the formula of integration of trigonometric, $$\sin 3x=3\sin x-4{{\left( \sin x \right)}^{3}}\\\ \Rightarrow 4{{\left( \sin x \right)}^{3}}=3\sin x-\sin 3x\\\ \Rightarrow {{\left( \sin x \right)}^{3}}=\dfrac{3}{4}\sin x-\dfrac{1}{4}\sin 3x$$ Therefore, $$\Rightarrow I=48\int{\dfrac{3\sin t-\sin 3t}{4}dt}$$ Simplifying the above expression, we get $$\Rightarrow I=12\int{3\sin t-\sin 3tdt}$$ $$\Rightarrow I=12\left( 3\int{\sin tdt}-\int{\sin 3tdt} \right)$$ Using the integration formula, we obtained $$\Rightarrow I=12\left( 3\left( -\cos t \right)-\dfrac{\left( -\cos 3t \right)}{3} \right)+C$$ $$\Rightarrow I=4\cos \left( 3t \right)-36\cos t+C$$ Undo the substitution in the above expression, we get $$\Rightarrow I=4\cos \left( 3\sqrt{x} \right)-36\cos \sqrt{x}+C$$ $$\therefore \int{48\left( \dfrac{{{\left( \sin \left( \sqrt{x} \right) \right)}^{3}}}{\sqrt{x}} \right)}=4\cos \left( 3\sqrt{x} \right)-36\cos \sqrt{x}+C$$ **Hence,the integral of $$\int{48\left( \dfrac{{{\left( \sin \left( \sqrt{x} \right) \right)}^{3}}}{\sqrt{x}} \right)}$$ is $4\cos \left( 3\sqrt{x} \right)-36\cos \sqrt{x}+C$.** **Note:** While solving the questions of integration, students always need to remember that they must not forget to put the constant term C after the integration of an equation. The value of this constant can be anything as it can be any whole number or integer or it can also be zero. The question given above its just a simple integration of a constant for that we should always remember to take out the constant term out of the integration and integrate the remaining integral by using the basic concepts or methods of integration. We should remember the property or the formulas of integration, this would make it easier to solve the question.