The value of $ \int{\frac{2+\sin x}{1+\cos x}}\,{{e}^{x/2}}dx $ is
- $ 2.{{e}^{x/2}}\,\tan \frac{x}{2}+C $
- $ {{e}^{x/2}}\,\tan \,x+C $
- $ \frac{1}{2}{{e}^{x/2}}\,\sin x+C $
- $ \frac{1}{2}\,\,{{e}^{x/2}}\,\sin \frac{x}{2}+C $
The Correct Option is A
Solution and Explanation
$ \Rightarrow $ $ l=\int{\frac{2+\frac{2\,\tan \,x/2}{1+{{\tan }^{2}}x/2}}{1+\frac{1-{{\tan }^{2}}x/2}{1+{{\tan }^{2}}x/2}}}\,.\,\,{{e}^{-x/2}}dx $
$ \Rightarrow $ $ l=\frac{2{{\tan }^{2}}\frac{x}{2}+2+2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}-{{\tan }^{2}}\frac{x}{2}+1}\,\,.\,{{e}^{x/2}}\,dx $
$ \Rightarrow $ $ l=2\int{\frac{{{\tan }^{2}}\frac{x}{2}+\tan \frac{x}{2}+1}{2}}.{{e}^{x/2}}\,\,dx $
$ \Rightarrow $ $ l=\int{{{\tan }^{2}}\frac{x}{2}.{{e}^{x/2}}dx+\int{\tan \frac{x}{2}.{{e}^{x/2}}\,dx}} $
$ +\int{{{e}^{x/2}}\,\,dx} $
$ \Rightarrow $ $ l=\int{\underset{II}{\mathop{{{\sec }^{2}}}}\,}\,x/2.{{e}^{x/2}}\,dx $
$ -\int{{{e}^{x/2}}\,dx+\int{\tan \frac{x}{2}.{{e}^{x/2}}\,dx+\int{{{e}^{x/2}}\,dx}}} $
$ \Rightarrow $ $ l=2{{e}^{x/2}}.\tan \frac{x}{2}-\int{\frac{1}{2}{{e}^{x/2}}.\tan \frac{x}{2}.2dx} $
$ +\int{\tan \,\frac{x}{2}.\,{{e}^{x/2}}\,dx+C} $
$ \Rightarrow $ $ l=2{{e}^{x/2}}.\tan \frac{x}{2}-\int{{{e}^{x/2}}.\tan \frac{x}{2}\,dx} $
$ +\int{{{e}^{x/2}}.\tan \frac{x}{2}dx+C} $
$ \Rightarrow $ $ l=2{{e}^{x/2}}.\tan \frac{x}{2}+C $
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Concepts Used:
Methods of Integration
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
Integration by Parts:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
Method of Integration Using Partial Fractions:
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Integration by Substitution Method
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
Integration by Decomposition
Reverse Chain Rule
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C