Question

Evaluate $\int{\left( \dfrac{2+\sin x}{1+\cos x} \right)}{{e}^{\dfrac{x}{2}}}dx$ \[\]

Answer 1

Hint: Use trigonometric identities to simplify the fraction inside the bracket involving sine and cosine. You will get an integrand involving secant and tangent functions which you can integrate by parts.

Complete step-by-step answer:
We begin by denoting the given fractional expression inside the bracket of integrand as $A=\dfrac{2+\sin x}{1+\cos x}$.

Let us first simplify the numerator and denominator of using twice angle trigonometric formulas. We know from sine and cosine twice angles formula that that $\sin 2\theta =2\sin \theta \cos \theta $ and $\cos 2\theta =2{{\cos }^{2}}\theta -1\Rightarrow \dfrac{1}{2}\left( 1+\cos 2\theta \right)=2{{\cos }^{2}}\theta $ . We also know the trigonometric relation between sine and cosine that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$. Let us take $\theta =\dfrac{x}{2}$ and use $A$. So we have
\[ \begin{align}
  & A=\dfrac{2+\sin x}{1+\cos x} \\
 & =\dfrac{1+1+\sin x}{1+\cos x} \\
 & =\dfrac{1}{1+\cos x}+\dfrac{1+\sin x}{1+\cos x} \\
 & =\dfrac{1}{2{{\cos }^{2}}\dfrac{x}{2}}+\dfrac{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}} \\
\end{align} \]

Let us use the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and we have
\[ \begin{align}
  & A=\dfrac{1}{2{{\cos }^{2}}\dfrac{x}{2}}+\dfrac{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}{2{{\cos }^{2}}\dfrac{x}{2}} \\
 & =\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\dfrac{1}{2}{{\left( \dfrac{\sin \dfrac{x}{2}+\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}^{2}} \\
 & =\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\dfrac{1}{2}{{\left( \tan \dfrac{x}{2}+1 \right)}^{2}} \\
 & =\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\dfrac{1}{2}\left( {{\tan }^{2}}\dfrac{x}{2}+1+2\tan \dfrac{x}{2} \right) \\
\end{align}\]
 Now let us use the trigonometric relation between tangent and secant that is ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ and we get

\[ \begin{align}
  & A=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\dfrac{1}{2}\left( {{\sec }^{2}}\dfrac{x}{2}+2\tan \dfrac{x}{2} \right) \\
 & ={{\sec }^{2}}\dfrac{x}{2}+\tan \dfrac{x}{2} \\
\end{align}\]
Now we shall evaluate the integral denoting as $I$ and by putting the values of $A$,
\[\begin{align}
  & I=\int{\left( \dfrac{2+\sin x}{1+\cos x} \right)}{{e}^{\dfrac{x}{2}}}dx \\
 & =\int{\dfrac{{{e}^{\dfrac{x}{2}}}}{2}{{\sec }^{2}}\dfrac{x}{2}}dx+\int{{{e}^{\dfrac{x}{2}}}\tan \dfrac{x}{2}}dx=\int{{{e}^{\dfrac{x}{2}}}{{\sec }^{2}}\dfrac{x}{2}}dx+{{I}_{1}}....(1) \\
\end{align}\]
Here we have denoted ${{I}_{1}}=\int{\left( \tan \dfrac{x}{2} \right)}{{e}^{\dfrac{x}{2}}}dx$ and evaluate it using integration by parts \[\]
Evaluation of ${{I}_{1}}$ :
We know that if there two single variable real valued integrable functions say $u$ and $v$then we integrate them by parts taking $u$ as first function and $v$as second function using the formula
\[\int{\left( uv \right)}dx=u\int{vdx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}}dx\]
The choice of first function depends upon how many times differentiating the function will make zero. So the rule that is used when we are integrating by parts is called ILATE, an acronym for inverse, algebraic, logarithm, trigonometric and finally exponent. It means we have to choose the first function in the order of ILATE. \[\]
We see that in the integration of $I=\int{\left( \tan \dfrac{x}{2} \right)}{{e}^{\dfrac{x}{2}}}dx$two functions are involved trigonometric and exponential . We use the ILATE rule and take tangent function as first function and exponential as second function. Then we integrate by parts,
\[ \begin{align}
  & {{I}_{1}}=\int{\left( \tan \dfrac{x}{2} \right)}{{e}^{\dfrac{x}{2}}}dx \\
 & =\tan \dfrac{x}{2}\int{{{e}^{\dfrac{x}{2}}}dx}-\int{\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}\int{{{e}^{\dfrac{x}{2}}}dx}} \\
 & =2\tan \dfrac{x}{2}{{e}^{\dfrac{x}{2}}}-2\times \dfrac{1}{2}\int{{{e}^{\dfrac{x}{2}}}{{\sec }^{2}}\dfrac{x}{2}dx}+c \\
 & =2\tan \dfrac{x}{2}{{e}^{\dfrac{x}{2}}}-\int{{{e}^{\dfrac{x}{2}}}{{\sec }^{2}}\dfrac{x}{2}dx}+c \\
\end{align}\]
Where $c$ is a real number and constant of integration. We put back the value of ${{I}_{1}}$ in equation and get
\[\begin{align}
  & I=\int{{{e}^{\dfrac{x}{2}}}{{\sec }^{2}}\dfrac{x}{2}}dx+{{I}_{1}} \\
 & =\int{{{e}^{\dfrac{x}{2}}}{{\sec }^{2}}\dfrac{x}{2}}dx+2\tan \dfrac{x}{2}{{e}^{\dfrac{x}{2}}}-\int{{{e}^{\dfrac{x}{2}}}{{\sec }^{2}}\dfrac{x}{2}dx} \\
 & =2\tan \dfrac{x}{2}{{e}^{\dfrac{x}{2}}} \\
\end{align}\]
So the value of the integral is $2\tan \dfrac{x}{2}{{e}^{\dfrac{x}{2}}}$.

Note: While solving integrating by parts for products of two different functions we need to keep in mind the ILATE rule otherwise we won’t be able to integrate. You can also use the formula $\int{{{e}^{x}}\left[ f\left( x \right)+{{f}^{'}}\left( x \right) \right]}dx={{e}^{x}}f\left( x \right)+c$ in this problem.