Evaluate the given indefinite integral: $ I=\int{\dfrac{1+\cos x}{\sin x\cos x}dx} $ ?

Hint: We start solving the problem by dividing each term of the numerator with the denominator. We then make use of the results $ 2\sin x\cos x=\sin 2x $ and $ \dfrac{1}{\sin x}=\operatorname{cosec}x $ to proceed through the problem. We then make use of the result $ \int{\left( a+b \right)dx}=\int{adx}+\int{bdx} $ to proceed through the problem. We then make use of result \[\int{\operatorname{cosec}axdx}=\dfrac{\log \left| \operatorname{cosec}ax-\cot ax \right|}{a}+C\] and then make the necessary calculations to get the value of required answer.

Complete step by step answer:
According to the problem, we need to find the given indefinite integral: $ I=\int{\dfrac{1+\cos x}{\sin x\cos x}dx} $ .
So, we have $ I=\int{\dfrac{1+\cos x}{\sin x\cos x}dx} $ .
 $ \Rightarrow I=\int{\left( \dfrac{1}{\sin x\cos x}+\dfrac{\cos x}{\sin x\cos x} \right)dx} $ .
 $ \Rightarrow I=\int{\left( \dfrac{2}{2\sin x\cos x}+\dfrac{1}{\sin x} \right)dx} $ ---(1).
We know that $ 2\sin x\cos x=\sin 2x $ and $ \dfrac{1}{\sin x}=\operatorname{cosec}x $ , let us use these results in equation (1).
\[\Rightarrow I=\int{\left( \dfrac{2}{\sin 2x}+\operatorname{cosec}x \right)dx}\].
\[\Rightarrow I=\int{\left( 2\operatorname{cosec}2x+\operatorname{cosec}x \right)dx}\] ---(2).
We know that $ \int{\left( a+b \right)dx}=\int{adx}+\int{bdx} $ . Let us use this result in equation (2).
\[\Rightarrow I=\int{2\operatorname{cosec}2xdx}+\int{\operatorname{cosec}xdx}\] ---(3).
We know that \[\int{\operatorname{cosec}axdx}=\dfrac{\log \left| \operatorname{cosec}ax-\cot ax \right|}{a}+C\]. Let us use this result in equation (3).
\[\Rightarrow I=\dfrac{2\log \left| \operatorname{cosec}2x-\cot 2x \right|}{2}+\log \left| \operatorname{cosec}x-\cot x \right|+C\].
\[\Rightarrow I=\log \left| \operatorname{cosec}2x-\cot 2x \right|+\log \left| \operatorname{cosec}x-\cot x \right|+C\].
So, we have found the solution for the given indefinite integral as \[\log \left| \operatorname{cosec}2x-\cot 2x \right|+\log \left| \operatorname{cosec}x-\cot x \right|+C\].
 $ \therefore $ The solution for the given indefinite integral $ I=\int{\dfrac{1+\cos x}{\sin x\cos x}dx} $ is \[\log \left| \operatorname{cosec}2x-\cot 2x \right|+\log \left| \operatorname{cosec}x-\cot x \right|+C\].

Note:
We can also make use of the result $ \log a+\log b=\log ab $ to further simplify the obtained result. We should not forget to include the modulus symbol inside the logarithmic function we got in the answer. We should not forget to add the constant of integration while solving problems related to indefinite integrals. We can also solve this problem by making use of the results $ 1+\cos x=2{{\cos }^{2}}\dfrac{x}{2} $ , $ \sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2} $ and then converting the obtained result into the partial fractions to get the given answer. Similarly, we can expect problems to find the value of the indefinite integral $ I=\int{\dfrac{1+\sin x}{\sin x\cos x}dx} $ .