Question

Find the value of the integral $ \dfrac{{(1 + 2\cos x)}}{{{{(2 + \cos x)}^2}}}dx $ from $ 0 $ to \[\dfrac{\pi }{2}\].

Answer 1

Hint: We have to integrate the given function $ \dfrac{{(1 + 2\cos x)}}{{{{(2 + \cos x)}^2}}}dx $ for the values of the integrals $ 0 $ to $ \dfrac{\pi }{2} $ with respect to $ x $ .We solve this question using the substitution method of integration . We simply put the value of \[\left( {{\text{ }}cos{\text{ }}x{\text{ }} + {\text{ }}2{\text{ }}} \right){\text{ }} = {\text{ }}\dfrac{1}{t}\] . Then differentiate the substituted function with respect to $ x $ . Put the value of limit in $ x $ $ $ and then compute the new limit . As we change the integral with respect to a variable then the limit of the integral also changes .

Complete step-by-step answer:
Given : \[\mathop \smallint \limits_0^{\dfrac{\pi }{2}} \dfrac{{(1 + 2cosx)}}{{{{(2 + cosx)}^2}}}dx\]
Let $ I = \mathop \smallint \limits_0^{\dfrac{\pi }{2}} \dfrac{{(1 + 2cosx)}}{{{{(2 + cosx)}^2}}}dx $
We have to integrate $ I $ with respect to
Put \[\left( {{\text{ }}cos{\text{ }}x{\text{ }} + {\text{ }}2{\text{ }}} \right){\text{ }} = {\text{ }}\dfrac{1}{t}\]
As $ si{n^2}x + co{s^2}x = 1 $
 \[\sin x = \sqrt {(1 - {{(\dfrac{1}{t} - 2)}^2})} \]
Differentiate $ t $ with respect to, we get
( derivative of \[cos{\text{ }}x{\text{ }} = {\text{ }} - {\text{ }}sin{\text{ }}x\] )
( derivative of \[constant\] \[ = {\text{ }}0\] )
( derivative $ {x^n} = n{x^{(n - 1)}} $ )
 $ - \sin xdx = \dfrac{{ - 1}}{{{t^2}}}dt $
Putting values of limit in $ x $ we get the new limits
When \[x = {\text{ }}\dfrac{\pi }{2}\] then \[t{\text{ }} = {\text{ }}\dfrac{1}{2}\]
When \[x{\text{ }} = {\text{ }}0\] then \[t{\text{ }} = {\text{ }}\dfrac{1}{3}\]
Now , the integral becomes
 \[I = \mathop \smallint \limits_{\dfrac{1}{3}}^{\dfrac{1}{2}} [\dfrac{{(1 + 2 \times (\dfrac{1}{t} - 2))}}{{\dfrac{1}{{{t^2}}} \times {t^2} \times \sqrt {(1 - {{(\dfrac{1}{t} - 2)}^2})} }}] dt\]
On further simplifying , we get
 $ I = \mathop \smallint \limits_{\dfrac{1}{3}}^{\dfrac{1}{2}} [\dfrac{{(2 - 3t)}}{{\sqrt {( - 1 - 3{t^2} + 4t)} }}] dt $
Put $ ( - 1 - 3{t^2} + 4t) = z $
Differentiate $ z $ with respect to , we get
( derivative $ {x^n} = n{x^{(n - 1)}} $ )
Putting values of limit in $ x $ we get the new limits
When \[\;t{\text{ }} = {\text{ }}\dfrac{1}{2}\] then \[z{\text{ }} = {\text{ }}\dfrac{1}{4}\]
When \[\;t{\text{ }} = {\text{ }}\dfrac{1}{3}\] then \[z{\text{ }} = {\text{ }}0\]
Now , the integral becomes
 $ I = \mathop \smallint \limits_0^{\dfrac{1}{4}} (\dfrac{1}{2})[\dfrac{1}{{\sqrt z }}] dz $
Integrating the integral we get
 $ I = \mathop [\limits_0^{\dfrac{1}{4}} \sqrt z ] $
Putting the values of limit in the integral , we get
 \[I{\text{ }} = {\text{ }}\dfrac{1}{2}\]
Thus , integral $ \dfrac{{(1 + 2\cos x)}}{{{{(2 + \cos x)}^2}}}dx $ from $ 0 $ to $ \dfrac{\pi }{2} $ \[ = {\text{ }}\dfrac{1}{2}\]
So, the correct answer is “Option B”.

Note: As the question was of definite integral that’s why we have not added integration constant . If the question would be of indefinite integral then we would add an integral constant to the final answer . We use the formula of By-Parts to integrate two functions of a single variable x by taking one functions as u and second function as v and then applying the formula :
 \[\smallint {\text{ }}\left[ {uv} \right] {\text{ }}dx{\text{ }} = {\text{ }}u{\text{ }} \times {\text{ }}\smallint {\text{ }}\left[ v \right] {\text{ }}dx{\text{ }} - {\text{ }}\smallint \left[ {{\text{ }}\left( {{\text{ }}\dfrac{d}{{dx}}{\text{ }}u} \right){\text{ }} \times {\text{ }}\smallint {\text{ }}} \right[v\left] {\text{ }} \right] {\text{ }}dx\]