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Question: Evaluate the given integral: \(\int{\text{xsi}{{\text{n}}^{2}}\text{x }dx}\) ....

Evaluate the given integral: ∫xsin2x dx\int{\text{xsi}{{\text{n}}^{2}}\text{x }dx} .

Explanation

Solution

Hint: Start by applying by-parts as ∫xsin2x dx=x∫sin⁡2xdx−∫dxdx(∫sin⁡2xdx)dx\int{\text{xsi}{{\text{n}}^{2}}\text{x }dx}=x\int{{{\sin }^{2}}x}dx-\int{\dfrac{dx}{dx}\left( \int{{{\sin }^{2}}x}dx \right)}dx . For finding the integral of sin⁡2x{{\sin }^{2}}x with respect to x, use the formula sin⁡2x=1−cos⁡2x2{{\sin }^{2}}x=\dfrac{1-\cos 2x}{2} followed by the use of the formulas ∫sin⁡xdx=−cos⁡x+c and ∫cos⁡xdx=sin⁡x\int{\sin xdx=-\cos x+c\text{ and }\int{\cos xdx}=\sin x} to reach the answer.

Complete step-by-step solution -
Let us start with the integral given in the above question.
∫xsin2x dx\int{\text{xsi}{{\text{n}}^{2}}\text{x }dx}
Now according to the rule of the integration by parts:
∫uvdx=u∫vdx−∫(dudx∫vdx)\int{uvdx=u\int{v}}dx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)} . From the two functions, u and v are decided using the ILATE preference rule. According to the ILATE rule, the preference order for u in decreasing order is Inverse, Logarithmic, Algebraic, Trigonometric and Exponential. For xsin2x\text{xsi}{{\text{n}}^{2}}\text{x} , u is x and v is sin⁡2x{{\sin }^{2}}x .
∫xsin2x dx=x∫sin⁡2xdx−∫dxdx(∫sin⁡2xdx)dx\int{\text{xsi}{{\text{n}}^{2}}\text{x }dx}=x\int{{{\sin }^{2}}x}dx-\int{\dfrac{dx}{dx}\left( \int{{{\sin }^{2}}x}dx \right)}dx
Now, we know that dxdx=1 and sin⁡2x=1−cos⁡2x2\dfrac{dx}{dx}=1\text{ and }{{\sin }^{2}}x=\dfrac{1-\cos 2x}{2} . So, if we put this in our integral and simplify, we get
∫xsin2x dx=x∫(1−cos⁡2x2)dx−∫(∫(1−cos⁡2x2)dx)dx\int{\text{xsi}{{\text{n}}^{2}}\text{x }dx}=x\int{\left( \dfrac{1-\cos 2x}{2} \right)}dx-\int{\left( \int{\left( \dfrac{1-\cos 2x}{2} \right)}dx \right)}dx
⇒∫xsin2x dx=x(∫12dx−∫cos⁡2x2dx)−∫(∫12dx−∫cos⁡2x2dx)dx\Rightarrow \int{\text{xsi}{{\text{n}}^{2}}\text{x }dx}=x\left( \int{\dfrac{1}{2}dx}-\int{\dfrac{\cos 2x}{2}dx} \right)-\int{\left( \int{\dfrac{1}{2}dx}-\int{\dfrac{\cos 2x}{2}dx} \right)}dx
Now, we know that ∫cos⁡2xdx=sin⁡2x2+c\int{\cos 2x}dx=\dfrac{\sin 2x}{2}+c and the integral of a constant term is equal to x+c.
∫xsin2x dx=x(12x−sin⁡2x4)−∫(12x−sin⁡2x4)dx\int{\text{xsi}{{\text{n}}^{2}}\text{x }dx}=x\left( \dfrac{1}{2}x-\dfrac{\sin 2x}{4} \right)-\int{\left( \dfrac{1}{2}x-\dfrac{\sin 2x}{4} \right)}dx
We also know that ∫sin2xdx=−cos⁡2x2+c\int{sin2x}dx=-\dfrac{\cos 2x}{2}+c and ∫xdx=x22+c\int{x}dx=\dfrac{{{x}^{2}}}{2}+c . So, our integral comes out to be:
∫xsin2x dx=x(12x−sin⁡2x4)−(12×x22−(−cos⁡2x2)4)+c\int{\text{xsi}{{\text{n}}^{2}}\text{x }dx}=x\left( \dfrac{1}{2}x-\dfrac{\sin 2x}{4} \right)-\left( \dfrac{1}{2}\times \dfrac{{{x}^{2}}}{2}-\dfrac{\left( -\dfrac{\cos 2x}{2} \right)}{4} \right)+c
⇒∫xsin2x dx=12x2−xsin⁡2x4−x24−cos⁡2x8+c\Rightarrow \int{\text{xsi}{{\text{n}}^{2}}\text{x }dx}=\dfrac{1}{2}{{x}^{2}}-\dfrac{x\sin 2x}{4}-\dfrac{{{x}^{2}}}{4}-\dfrac{\cos 2x}{8}+c
The c has been added to compensate for all the constant terms that would have appeared when we put the formulas of different integrals.
So, we can say that the value of ∫xsin2x dx=12x2−xsin⁡2x4−x24−cos⁡2x8+c\int{\text{xsi}{{\text{n}}^{2}}\text{x }dx}=\dfrac{1}{2}{{x}^{2}}-\dfrac{x\sin 2x}{4}-\dfrac{{{x}^{2}}}{4}-\dfrac{\cos 2x}{8}+c .

Note: See while you use integral by-parts, you don’t have to put the constant of integral I each and every step, but you just have to put a constant term at the end of the integrated answer to compensate for all the constant terms that would have appeared. Also, you need to remember all the basic formulas that we use for indefinite integrals as they are used in definite integrations as well. Also, be careful about the formula of integral of sin2x, as there is a negative sign in the result which is generally missed by the students.