Question

Solve the following integration using various formulas and identities of integration $\int{\dfrac{\cos x}{1+\cos x}dx}$.

Answer 1

Hint: We will add and subtract 1 in numerator as $\int{\dfrac{\cos x+1-1}{1+\cos x}dx}$ and then separate it as $\int{\dfrac{1+\cos x}{1+\cos x}dx-\int{\dfrac{1}{1+\cos x}dx}}$ and then solve accordingly. We will also use few trigonometric formula such as $\cos 2\theta =2{{\cos }^{2}}\theta -1$ and $\cos 2\theta +1=2{{\cos }^{2}}\theta $.

Complete step-by-step answer:
We have given that to integrate $\int{\dfrac{\cos x}{1+\cos x}dx}$. We will add and subtract 1 in the numerator we get $\int{\dfrac{\cos x+1-1}{1+\cos x}dx}$. Now we split the integration into two simplified integration and solve them, independently, $\int{\dfrac{1+\cos x}{1+\cos x}dx-\int{\dfrac{1}{1+\cos x}dx}}$.
On further solving we get $\int{1dx-\int{\dfrac{1}{1+\cos x}dx}}$. Now we know that $\cos 2\theta =2{{\cos }^{2}}\theta -1$ and $\cos 2\theta +1=2{{\cos }^{2}}\theta $ on replacing $\theta $ with $\dfrac{x}{2}$, we get $\cos 2\dfrac{x}{2}=2{{\cos }^{2}}\dfrac{x}{2}-1$, further simplifying $\operatorname{cosx}+1=2{{\cos }^{2}}\dfrac{x}{2}$. So, on putting $1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$, we get $\int{\left( 1 \right)dx-\int{\dfrac{1}{2{{\cos }^{2}}\dfrac{x}{2}}dx}}$.
We know that $\cos \theta =\dfrac{1}{\sec x}$, thus ${{\cos }^{2}}\dfrac{x}{2}$ can be written as $\dfrac{1}{{{\sec }^{2}}\dfrac{x}{2}}$ , we get $\int{\left( 1 \right)dx-\dfrac{1}{2}\int{{{\sec }^{2}}\dfrac{x}{2}dx}}$.
We know that $\int{{{\sec }^{2}}\left( ax+b \right)dx=\dfrac{\tan \left( ax+b \right)}{a}}+c$, we get = $x-\dfrac{1}{2}\dfrac{\tan \left( \dfrac{x}{2} \right)}{\dfrac{1}{2}}+c$ simplifying further, we get our final answer as = $x-\tan \dfrac{x}{2}+c$.

Note: Usually students make mistakes in the last step in the integration of $\int{{{\sec }^{2}}\dfrac{x}{2}}$. Most of the student directly integrate $\int{{{\sec }^{2}}\dfrac{x}{2}}$ as $\tan \dfrac{x}{2}+c$, which is not correct. The correct integration of $\int{{{\sec }^{2}}\dfrac{x}{2}}$ is$\dfrac{\tan \dfrac{x}{2}}{\dfrac{1}{2}}+c$. Also, student may forget the sub trigonometric formulas like $\cos 2\theta =2{{\cos }^{2}}\theta -1$ thus, it is recommended to memorize all the formulas of trigonometry before solving such questions.