Question

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

Answer 1

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.