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Solve \[\int{\dfrac{\cos 2x}{{{\left( \cos x+\sin x \right)}^{2}}}dx}\]

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Last updated date: 13th Jun 2024
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Answer
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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.