
What is the integral of $\int{\left[ x.\cos \left( {{x}^{2}} \right) \right]\left( dx \right)?}$
Answer
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Hint: We solve this question by using the substitution method of solving integrals. We substitute the value of ${{x}^{2}}$ as a variable t. We then find the value of $dx$ and substitute for it in the integral. Then we simplify using the basic integration formula to solve the given integral. After simplifying and obtaining the answer, we substitute back for t as ${{x}^{2}}.$
Complete step by step solution:
In order to solve this question, let us consider the given integral in the question.
$= \int{\left[ x.\cos \left( {{x}^{2}} \right) \right]\left( dx \right)}$
Let us consider the value ${{x}^{2}}$ as a variable t. Let us assume,
$= t={{x}^{2}}\ldots \left( 1 \right)$
We differentiate this above equation and we know the differentiation of ${{x}^{2}}$ with respect to x is 2x.
$= dt=2xdx$
We divide both sides of the equation by 2.
$= \dfrac{1}{2}dt=xdx$
We substitute the value of ${{x}^{2}}$ as t and the value of $xdx$ by $\dfrac{1}{2}dt$ in the integral. Before that, we rearrange the terms,
$= \int{\left[ \cos \left( {{x}^{2}} \right) \right]x.dx}$
Now, we make the above said substitutions,
$= \int{\cos \left( t \right).\dfrac{1}{2}dt}$
Taking the constant $\dfrac{1}{2}$ outside the integral,
$= \dfrac{1}{2}\int{\cos \left( t \right)dt}$
We know the integral of a $\cos t$ is given by $\sin t$ and we add a constant of integration after it.
$= \dfrac{1}{2}\int{\cos \left( t \right)dt}=\dfrac{1}{2}\sin t+C$
Now, we substitute back for the variable t in terms of x given by the equation 1.
$= \dfrac{1}{2}\sin \left( {{x}^{2}} \right)+C$
Hence, the integral of the above function is found to be $\dfrac{1}{2}\sin \left( {{x}^{2}} \right)+C,$ where C is the constant of integration.
Note:
We need to know the basic formulae for the integration of basic functions. We can also solve this question by taking integration by parts for the two terms $x$ and $\cos \left( {{x}^{2}} \right)$ and simplifying it using the basic integration formulae. It is important to know the integration formula for trigonometric functions to solve this question.
Complete step by step solution:
In order to solve this question, let us consider the given integral in the question.
$= \int{\left[ x.\cos \left( {{x}^{2}} \right) \right]\left( dx \right)}$
Let us consider the value ${{x}^{2}}$ as a variable t. Let us assume,
$= t={{x}^{2}}\ldots \left( 1 \right)$
We differentiate this above equation and we know the differentiation of ${{x}^{2}}$ with respect to x is 2x.
$= dt=2xdx$
We divide both sides of the equation by 2.
$= \dfrac{1}{2}dt=xdx$
We substitute the value of ${{x}^{2}}$ as t and the value of $xdx$ by $\dfrac{1}{2}dt$ in the integral. Before that, we rearrange the terms,
$= \int{\left[ \cos \left( {{x}^{2}} \right) \right]x.dx}$
Now, we make the above said substitutions,
$= \int{\cos \left( t \right).\dfrac{1}{2}dt}$
Taking the constant $\dfrac{1}{2}$ outside the integral,
$= \dfrac{1}{2}\int{\cos \left( t \right)dt}$
We know the integral of a $\cos t$ is given by $\sin t$ and we add a constant of integration after it.
$= \dfrac{1}{2}\int{\cos \left( t \right)dt}=\dfrac{1}{2}\sin t+C$
Now, we substitute back for the variable t in terms of x given by the equation 1.
$= \dfrac{1}{2}\sin \left( {{x}^{2}} \right)+C$
Hence, the integral of the above function is found to be $\dfrac{1}{2}\sin \left( {{x}^{2}} \right)+C,$ where C is the constant of integration.
Note:
We need to know the basic formulae for the integration of basic functions. We can also solve this question by taking integration by parts for the two terms $x$ and $\cos \left( {{x}^{2}} \right)$ and simplifying it using the basic integration formulae. It is important to know the integration formula for trigonometric functions to solve this question.
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