Answer
Verified
415.5k+ views
Hint: This question is from the topic of integration. In this question, we will first solve the term \[\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}\] in simple form. After that, we will do the integration. We are going to use a substitution method for the integration. We will put the value of \[1+\sin 2x\] as t. After that, we will write the integration in terms of t. After solving the further solution, we will get the value of integration. After that, we will replace t by \[1+\sin 2x\]. After that, we will get our exact answer.
Complete step by step solution:
Let us solve this question.
In this question, we have asked to solve \[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}\].
As we know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\], so we can write the term \[\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}\] as
\[\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\dfrac{\cos 2x}{{{\left( \cos x \right)}^{2}}+{{\left( \sin x \right)}^{2}}+2\cos x\sin x}\]
As we know that \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\], so we can write the above equation as
\[\Rightarrow \dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\dfrac{\cos 2x}{1+2\cos x\sin x}\]
As we know that \[\sin 2x=2\cos x\sin x\], so we can write the above equation as
\[\Rightarrow \dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\dfrac{\cos 2x}{1+\sin 2x}\]
Now, we can write the integration as
\[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\int{\dfrac{\cos 2x}{1+\sin 2x}dx}\]
Now, using the substitution method we will solve this integration.
Let \[t=1+\sin 2x\]
Then, we will differentiate the term t with respect to x.
\[\dfrac{dt}{dx}=\dfrac{d}{dx}\left( 1+\sin 2x \right)=\dfrac{d}{dx}(1)+\dfrac{d}{dx}(\sin 2x)=0+\left( \cos 2x \right)\dfrac{d}{dx}\left( 2x \right)\]
Using chain rule in the above equation, we get
\[\dfrac{dt}{dx}=\left( \cos 2x \right)\left( 2 \right)=2\cos 2x\]
The above equation can also be written as
\[\Rightarrow \dfrac{1}{2}dt=\cos 2xdx\]
So, in the integration, we can write
\[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\int{\dfrac{\cos 2x}{1+\sin 2x}dx}=\int{\dfrac{1}{t}\times \dfrac{1}{2}dt}=\dfrac{1}{2}\int{\dfrac{1}{t}dt}\]
As we know that integration of \[\dfrac{1}{t}\] with respect of t is\[\ln t+C\], where C is any constant, so we can write
\[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\ln t+C\]
Now, putting the value of t in the above equation, we get
\[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\ln \left( 1+\sin 2x \right)+C\]
In the above, we have found that \[1+\sin 2x\] can also be written as \[{{\left( \cos x+\sin x \right)}^{2}}\], so we can write
\[\Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\ln {{\left( \cos x+\sin x \right)}^{2}}+C\]
As we know that \[{{\ln }_{b}}{{a}^{n}}=n{{\ln }_{b}}a\], so we can write
\[\Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=2\times \dfrac{1}{2}\ln \left| \cos x+\sin x \right|+C=\ln \left| \cos x+\sin x \right|+C\]
Hence, we have solved the integration.
So, we can write
\[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\ln \left| \cos x+\sin x \right|+C\]
Note:
We should have a better knowledge in the topic of integration to solve this type of question easily. We should remember the following formulas to solve this type of question:
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
\[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]
\[\sin 2x=2\cos x\sin x\]
\[\dfrac{d}{dx}\sin x=\cos x\]
\[\dfrac{d}{dx}\dfrac{1}{x}=\ln x\]
\[{{\ln }_{b}}{{a}^{n}}=n{{\ln }_{b}}a\]
We have used chain rule here, so remember that. The chain rule helps us to differentiate the composite functions like\[f\left( g\left( x \right) \right)\]. So, \[\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)\times g'\left( x \right)\]. Here, f and g are two different functions, and \[f'\] and\[g'\] are differentiation of f and g respectively.
Complete step by step solution:
Let us solve this question.
In this question, we have asked to solve \[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}\].
As we know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\], so we can write the term \[\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}\] as
\[\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\dfrac{\cos 2x}{{{\left( \cos x \right)}^{2}}+{{\left( \sin x \right)}^{2}}+2\cos x\sin x}\]
As we know that \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\], so we can write the above equation as
\[\Rightarrow \dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\dfrac{\cos 2x}{1+2\cos x\sin x}\]
As we know that \[\sin 2x=2\cos x\sin x\], so we can write the above equation as
\[\Rightarrow \dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}=\dfrac{\cos 2x}{1+\sin 2x}\]
Now, we can write the integration as
\[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\int{\dfrac{\cos 2x}{1+\sin 2x}dx}\]
Now, using the substitution method we will solve this integration.
Let \[t=1+\sin 2x\]
Then, we will differentiate the term t with respect to x.
\[\dfrac{dt}{dx}=\dfrac{d}{dx}\left( 1+\sin 2x \right)=\dfrac{d}{dx}(1)+\dfrac{d}{dx}(\sin 2x)=0+\left( \cos 2x \right)\dfrac{d}{dx}\left( 2x \right)\]
Using chain rule in the above equation, we get
\[\dfrac{dt}{dx}=\left( \cos 2x \right)\left( 2 \right)=2\cos 2x\]
The above equation can also be written as
\[\Rightarrow \dfrac{1}{2}dt=\cos 2xdx\]
So, in the integration, we can write
\[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\int{\dfrac{\cos 2x}{1+\sin 2x}dx}=\int{\dfrac{1}{t}\times \dfrac{1}{2}dt}=\dfrac{1}{2}\int{\dfrac{1}{t}dt}\]
As we know that integration of \[\dfrac{1}{t}\] with respect of t is\[\ln t+C\], where C is any constant, so we can write
\[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\ln t+C\]
Now, putting the value of t in the above equation, we get
\[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\ln \left( 1+\sin 2x \right)+C\]
In the above, we have found that \[1+\sin 2x\] can also be written as \[{{\left( \cos x+\sin x \right)}^{2}}\], so we can write
\[\Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\dfrac{1}{2}\ln {{\left( \cos x+\sin x \right)}^{2}}+C\]
As we know that \[{{\ln }_{b}}{{a}^{n}}=n{{\ln }_{b}}a\], so we can write
\[\Rightarrow \int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=2\times \dfrac{1}{2}\ln \left| \cos x+\sin x \right|+C=\ln \left| \cos x+\sin x \right|+C\]
Hence, we have solved the integration.
So, we can write
\[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}=\ln \left| \cos x+\sin x \right|+C\]
Note:
We should have a better knowledge in the topic of integration to solve this type of question easily. We should remember the following formulas to solve this type of question:
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
\[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]
\[\sin 2x=2\cos x\sin x\]
\[\dfrac{d}{dx}\sin x=\cos x\]
\[\dfrac{d}{dx}\dfrac{1}{x}=\ln x\]
\[{{\ln }_{b}}{{a}^{n}}=n{{\ln }_{b}}a\]
We have used chain rule here, so remember that. The chain rule helps us to differentiate the composite functions like\[f\left( g\left( x \right) \right)\]. So, \[\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)\times g'\left( x \right)\]. Here, f and g are two different functions, and \[f'\] and\[g'\] are differentiation of f and g respectively.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
What percentage of the solar systems mass is found class 8 physics CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Why is there a time difference of about 5 hours between class 10 social science CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE