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Question

Question: Evaluate the given Integral \[\int {x{{\csc }^2}xdx} \]....

Evaluate the given Integral xcsc2xdx\int {x{{\csc }^2}xdx} .

Explanation

Solution

Here, we will first use the Integration by Parts formula and rewrite the given integrand. We will then use the suitable integration formula to find the integral of the given function. Integration is defined as the summation of all the discrete data.

Formula used:
We will use the following formulas:
1. Integration by Parts: uvdx=uvvdu\int {uvdx = uv - \int {vdu} }
2. Derivative Formula: ddx(C)=1\dfrac{d}{{dx}}\left( C \right) = 1
3. Integral Formula: csc2xdx=cotx\int {{{\csc }^2}xdx = - \cot x} , andcotxdx=cosxsinxdx=ln(sinx)+C\int {\cot xdx = \int {\dfrac{{\cos x}}{{\sin x}}dx} = \ln \left( {\sin x} \right) + C}

Complete Step by Step Solution:
We are given an Integral function xcsc2xdx\int {x{{\csc }^2}xdx} .
Now, we will find the integral function by using Integration by Parts uvdx=uvvdu\int {uvdx = uv - \int {vdu} } for the given Integral function, we get u=xu = x according to ILATE rule and v=csc2xv = {\csc ^2}x.
Now, we will differentiate the variableuu, so we get
du=dxdu = dx…………………………………………………….(1)\left( 1 \right)
We will now integrate the variable vv using the formula csc2xdx=cotx\int {{{\csc }^2}xdx = - \cot x} , so we get
csc2xdx=cotx\int {{{\csc }^2}xdx = - \cot x} ……………………………………………………………………(2)\left( 2 \right)
By substituting equation (1)\left( 1 \right) and (2)\left( 2 \right) in the integration by parts formula uvdx=uvvdu\int {uvdx = uv - \int {vdu} } , we get
xcsc2xdx=xcotxcotxdx\int {x{{\csc }^2}xdx = - x\cot x - \int { - \cot xdx} } …………………………………………..(3)\left( 3 \right)
Now, by rewriting the equation (3)\left( 3 \right) in the integral function, we get
xcsc2xdx=xcotx+cotxdx\Rightarrow \int {x{{\csc }^2}xdx} = - x\cot x + \int {\cot xdx}
Substituting cotx=cosxsinx\cot x = \dfrac{{\cos x}}{{\sin x}} in the above equation, we get
xcsc2xdx=xcotx+cosxsinxdx\Rightarrow \int {x{{\csc }^2}xdx} = - x\cot x + \int {\dfrac{{\cos x}}{{\sin x}}dx}
Now, we will integrate the function by using the integral formula cosxsinxdx=ln(sinx)+C\int {\dfrac{{\cos x}}{{\sin x}}dx = \ln \left( {\sin x} \right) + C} .
xcsc2xdx=xcotx+[ln(sinx)]+C\Rightarrow \int {x{{\csc }^2}xdx} = - x\cot x + \left[ {\ln \left( {\sin x} \right)} \right] + C

Therefore, the value of xcsc2xdx\int {x{{\csc }^2}xdx} is xcotx+ln(sinx)+C - x\cot x + \ln \left( {\sin x} \right) + C.

Note:
We know that Integration is the process of adding small parts to find the whole parts. While performing the Integration by Parts, the first function is selected according to ILATE rule where Inverse Trigonometric function, followed by Logarithmic function, Arithmetic Function, Trigonometric Function and at last Exponential Function. Integration by Parts is applicable only when the integrand is a product of two Functions. Whenever the integration is done with no limits, then an Arbitrary constant should be added at the last step of the Integration.