Answer
Verified
409.2k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry 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
Write a letter to the principal requesting him to grant class 10 english CBSE