Solveeit Logo
Login

Question

Question: How do you find the antiderivative of \(\int{{{x}^{2}}\cos xdx}\)....

How do you find the antiderivative of ∫x2cos⁡xdx\int{{{x}^{2}}\cos xdx}.

Explanation

Solution

In the problem we have two functions which are in multiplication one is x2{{x}^{2}} and the second one is cos⁡x\cos x. In integration we have the uvuv formula as ∫uv=u∫v−∫(u′)∫v\int{uv}=u\int{v}-\int{\left( {{u}^{'}} \right)\int{v}}. So, we will use the ILATE rule and determine the values of uu and vv. After getting these values we will use the uvuv formula and do simplification. Now we will get another equation which is in the same form. So, we will again use the uvuv formula and simplify the obtained equation. Now we will get the required result.

Complete step by step answer:
Given that, ∫x2cos⁡xdx\int{{{x}^{2}}\cos xdx}.
In the above equation we have two functions, one is x2{{x}^{2}} which is an algebraic function and the second one is cos⁡x\cos x which is a trigonometric function. By using the ILATE formula
u=x2u={{x}^{2}}, v=cos⁡xv=\cos x.
Applying uvuv formula in the given equation, then we will get
∫x2cos⁡xdx=x2∫cos⁡xdx−∫((x2)′∫cos⁡xdx)dx\int{{{x}^{2}}\cos xdx}={{x}^{2}}\int{\cos xdx}-\int{\left( {{\left( {{x}^{2}} \right)}^{'}}\int{\cos xdx} \right)dx}
We know that (x2)′=2x{{\left( {{x}^{2}} \right)}^{'}}=2x, ∫cos⁡xdx=sin⁡x+C\int{\cos xdx}=\sin x+C, then we will get
⇒∫x2cos⁡xdx=x2(sin⁡x)−∫2xsin⁡xdx ⇒∫x2cos⁡xdx=x2sin⁡x−2∫xsin⁡xdx....(i) \begin{aligned} & \Rightarrow \int{{{x}^{2}}\cos xdx}={{x}^{2}}\left( \sin x \right)-\int{2x\sin xdx} \\\ & \Rightarrow \int{{{x}^{2}}\cos xdx}={{x}^{2}}\sin x-2\int{x\sin xdx}....\left( \text{i} \right) \\\ \end{aligned}
In the above equation, we have ∫xsin⁡xdx\int{x\sin xdx} which is similar to the uvuv rule. So again, using the ILATE formula then we will get
u=xu=x, v=sin⁡xv=\sin x.
Now the value of ∫xsin⁡xdx\int{x\sin xdx} is given by
∫xsin⁡xdx=x∫sin⁡x−∫((x)′∫sin⁡xdx)dx\int{x\sin xdx}=x\int{\sin x}-\int{\left( {{\left( x \right)}^{'}}\int{\sin xdx} \right)dx}
We have x′=1{{x}^{'}}=1, ∫sin⁡xdx=−cos⁡x+C\int{\sin xdx}=-\cos x+C, then we will get
∫xsin⁡xdx=−xcos⁡x−∫(−cos⁡x)dx ⇒∫xsin⁡xdx=−xcos⁡x+∫cos⁡xdx \begin{aligned} & \int{x\sin xdx}=-x\cos x-\int{\left( -\cos x \right)dx} \\\ & \Rightarrow \int{x\sin xdx}=-x\cos x+\int{\cos xdx} \\\ \end{aligned}
We have ∫cos⁡xdx=sin⁡x+C\int{\cos xdx}=\sin x+C, then we will get
⇒∫xsin⁡xdx=−xcos⁡x+sin⁡x+C\Rightarrow \int{x\sin xdx}=-x\cos x+\sin x+C
Substituting the above value in the equation (i)\left( \text{i} \right), then we will get
⇒∫x2cos⁡xdx=x2sin⁡x−2(−xcos⁡x+sin⁡x)+C\Rightarrow \int{{{x}^{2}}\cos xdx}={{x}^{2}}\sin x-2\left( -x\cos x+\sin x \right)+C
Simplifying the above equation by applying multiplication distribution law, then we will get
⇒∫x2cos⁡xdx=x2sin⁡x+2xcos⁡x−2sin⁡x+C\Rightarrow \int{{{x}^{2}}\cos xdx}={{x}^{2}}\sin x+2x\cos x-2\sin x+C
Taking sin⁡x\sin x common from the terms x2sin⁡x−2sin⁡x{{x}^{2}}\sin x-2\sin x, then we will get
⇒∫x2cos⁡xdx=(x2−2)sin⁡x+2xcos⁡x+C\Rightarrow \int{{{x}^{2}}\cos xdx}=\left( {{x}^{2}}-2 \right)\sin x+2x\cos x+C

Hence the value of ∫x2cos⁡xdx\int{{{x}^{2}}\cos xdx} is (x2−2)sin⁡x+2xcos⁡x+C\left( {{x}^{2}}-2 \right)\sin x+2x\cos x+C.

Note: In the above problem we have used the uvuv, ILATE formulas several times. We need to follow some rules while using the uvuv rule which is denoted by ILATE which indicates the order of giving priority for a function. It states that the order of the priority of functions as Inverse, Logarithmic, Algebraic, Trigonometric, Exponential. From the above priority table, we will choose the functions uu and vv in uvuv rule.