
Which of the following is the integration of \[\int{\dfrac{1-\sin 2x}{x+{{\cos }^{2}}x}dx}\]?
A. \[-\log \left| x+{{\cos }^{2}}x \right|+C\]
B. \[\log \left| x+{{\cos }^{2}}x \right|+C\]
C. \[-\log \left| x+{{\sin }^{2}}x \right|+C\]
D. \[\log \left| x+{{\sin }^{2}}x \right|+C\]
Answer
543.3k+ views
Hint: In the given question, we have been asked to integrate the given expression. In order to solve the question, we integrate the given function by using the basic concept of integration. First we need to substitute \[t=x+{{\cos }^{2}}x\] then differentiate it with respect to ‘x’ and put these values in the given integral. Later we will need to integrate the simplified expression using a suitable integration formula and we will get our required answer.
Complete step by step solution:
We have given that,
\[\int{\dfrac{1-\sin 2x}{x+{{\cos }^{2}}x}dx}\]
Let I be the integration of the given equation.
Therefore, we can write the integration as,
\[\Rightarrow I=\int{\dfrac{1-\sin 2x}{x+{{\cos }^{2}}x}dx}\]
Substituting \[t=x+{{\cos }^{2}}x\]
We have,
\[t=x+{{\cos }^{2}}x\]
Differentiate with respect to ‘x’, we get
Using the differential formula of \[{{\cos }^{2}}x=-\sin 2x\]
\[\dfrac{dt}{dx}=1-\sin 2x\]
Simplifying the above,
\[\left( 1-\sin 2x \right)dx=dt\]
Substituting these values in the given integral, we obtained
\[\int{\dfrac{dt}{t}}\]
This is the standard integral, we get
\[\int{\dfrac{dt}{t}=\log t+C}\]
Undo the substitution i.e. \[t=x+{{\cos }^{2}}x\] in the above expression, we get
\[\int{\dfrac{dt}{t}=\log \left| x+{{\cos }^{2}}x \right|+C}\]
Therefore,
\[\int{\dfrac{1-\sin 2x}{x+{{\cos }^{2}}x}dx}=\log \left| x+{{\cos }^{2}}x \right|+C\]
hence the option (B) is correct.
Note: Here, we need to remember that we have to put the constant term C after the integration equation and the value of the given constant i.e. C, it can be any value 0 equal to zero also. While we substitute any value to solve the integration, do not forget to undo the substitution or replace the values of substitution. In order to solve the question that is given above, students need to know the basic formula of integration and they should very well keep all the standard integral into their mind because sometimes the given integration is the standard integral and we do not need to solve the question further and directly write the resultant integral.
Complete step by step solution:
We have given that,
\[\int{\dfrac{1-\sin 2x}{x+{{\cos }^{2}}x}dx}\]
Let I be the integration of the given equation.
Therefore, we can write the integration as,
\[\Rightarrow I=\int{\dfrac{1-\sin 2x}{x+{{\cos }^{2}}x}dx}\]
Substituting \[t=x+{{\cos }^{2}}x\]
We have,
\[t=x+{{\cos }^{2}}x\]
Differentiate with respect to ‘x’, we get
Using the differential formula of \[{{\cos }^{2}}x=-\sin 2x\]
\[\dfrac{dt}{dx}=1-\sin 2x\]
Simplifying the above,
\[\left( 1-\sin 2x \right)dx=dt\]
Substituting these values in the given integral, we obtained
\[\int{\dfrac{dt}{t}}\]
This is the standard integral, we get
\[\int{\dfrac{dt}{t}=\log t+C}\]
Undo the substitution i.e. \[t=x+{{\cos }^{2}}x\] in the above expression, we get
\[\int{\dfrac{dt}{t}=\log \left| x+{{\cos }^{2}}x \right|+C}\]
Therefore,
\[\int{\dfrac{1-\sin 2x}{x+{{\cos }^{2}}x}dx}=\log \left| x+{{\cos }^{2}}x \right|+C\]
hence the option (B) is correct.
Note: Here, we need to remember that we have to put the constant term C after the integration equation and the value of the given constant i.e. C, it can be any value 0 equal to zero also. While we substitute any value to solve the integration, do not forget to undo the substitution or replace the values of substitution. In order to solve the question that is given above, students need to know the basic formula of integration and they should very well keep all the standard integral into their mind because sometimes the given integration is the standard integral and we do not need to solve the question further and directly write the resultant integral.
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