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 :
$\smallint {\text{ }}\left[ {uv} \right] {\text{ }}dx{\text{ }} = {\text{ }}u{\text{ }} \times {\text{ }}\smallint {\text{ }}\left[ v \right] {\text{ }}dx{\text{ }} - {\text{ }}\smallint \left[ {{\text{ }}\left( {{\text{ }}\dfrac{d}{{dx}}{\text{ }}u} \right){\text{ }} \times {\text{ }}\smallint {\text{ }}} \right[v\left] {\text{ }} \right] {\text{ }}dx$