Question

Let g(x) = $ \int_{{}}^{{}}{\dfrac{1+2\cos x}{{{\left( \cos x+2 \right)}^{2}}}dx} $ and g(0) = 0, then the value of $ 32g\left( \dfrac{\pi }{2} \right) $ is ?

Answer 1

Hint: Here, we will use the formula for integration by parts. We will first convert g(x) into a form in which integration by parts can be easily applied. We will then choose u and v according to the ILATE rule and proceed. Other formulas that will be used are $ \int_{{}}^{{}}{\cos xdx=\sin x+c} $ , $ \dfrac{d\left( \cos x \right)}{dx}=-\sin x $ and also $ \dfrac{d\left( \dfrac{1}{x} \right)}{dx}=-\dfrac{1}{{{x}^{2}}} $ .

Complete step-by-step answer:
Integration by parts formula is used for integrating the product of two functions. This method is to find the integrals by reducing them into standard forms. The formula for integrating by parts is given as:
 $ \int_{{}}^{{}}{u.vdx=u\int_{{}}^{{}}{vdx}-\int_{{}}^{{}}{\left( \dfrac{du}{dx}\int_{{}}^{{}}{vdx} \right)dx}}........\left( 1 \right) $
We take the functions as u and v using the ILATE rule, according to the ILATE rule preference order or taking u is as Inverse, Algebraic, Logarithm, Trigonometric and Exponent.
Here, the function given to us is:
 g(x) = $ \int_{{}}^{{}}{\dfrac{1+2\cos x}{{{\left( \cos x+2 \right)}^{2}}}dx} $
Since, we know the identity $ {{\cos }^{2}}x+{{\sin }^{2}}x=1 $ , we can replace 1 by $ {{\cos }^{2}}x+{{\sin }^{2}}x=1 $ .So, we get:
 g(x) = $ \int_{{}}^{{}}{\dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x+2\cos x}{{{\left( \cos x+2 \right)}^{2}}}dx} $
\[\begin{align}
  & \Rightarrow g\left( x \right)=\int_{{}}^{{}}{\dfrac{\cos x\left( \cos x+2 \right)+{{\sin }^{2}}x}{{{\left( \cos x+2 \right)}^{2}}}dx} \\
 & \Rightarrow g\left( x \right)=\int_{{}}^{{}}{\dfrac{\cos x\left( \cos x+2 \right)}{{{\left( \cos x+2 \right)}^{2}}}dx+\int_{{}}^{{}}{\dfrac{{{\sin }^{2}}x}{{{\left( \cos x+2 \right)}^{2}}}dx}} \\
\end{align}\]
 $ \Rightarrow g\left( x \right)=\int_{{}}^{{}}{\dfrac{\cos x}{\left( \cos x+2 \right)}dx+\int_{{}}^{{}}{\dfrac{{{\sin }^{2}}x}{{{\left( \cos x+2 \right)}^{2}}}dx}} $
Now, we will first find the integration of $ \int_{{}}^{{}}{\dfrac{\cos x}{\left( \cos x+2 \right)}dx} $ .
Now, we will use integration by parts. Let us consider the ILATE rule It gives the order in which the function is to be chosen as u or v for performing integration by parts. So, I stands for Inverse function, L for Logarithmic function, A for Algebraic function, T for trigonometric function and E for Exponential function. Here, $ \dfrac{1}{\cos x+2} $ and $ \cos x $ both are trigonometric functions. So, ILATE will not be applicable here. Therefore, we will choose u and v according to our convenience. So, let us take $ u=\dfrac{1}{\cos x+2} $ and $ v=\cos x $ .
So, using the formula given in equation (1) for integration by parts, we get:
 $ \int_{{}}^{{}}{\dfrac{\cos x}{\cos x+2}dx}=\dfrac{1}{\cos x+2}\int_{{}}^{{}}{\cos xdx}-\int_{{}}^{{}}{\dfrac{d\left( \dfrac{1}{\cos x+2} \right)}{dx}.\int_{{}}^{{}}{\cos xdx..............\left( 2 \right)}} $
Since, $ \int_{{}}^{{}}{\cos xdx}=\sin x+{{c}_{1}} $ .
And,
  $ \begin{align}
  & \dfrac{d\left( \dfrac{1}{\cos x+2} \right)}{dx}=\dfrac{-1}{{{\left( \cos x+2 \right)}^{2}}}.\dfrac{d}{dx}\left( \cos x+2 \right) \\
 & \Rightarrow \dfrac{d\left( \dfrac{1}{\cos x+2} \right)}{dx}=\dfrac{-1}{{{\left( \cos x+2 \right)}^{2}}}\left( -\sin x \right) \\
 & \Rightarrow \dfrac{d\left( \dfrac{1}{\cos x+2} \right)}{dx}=\dfrac{\sin x}{{{\left( \cos x+2 \right)}^{2}}}+{{c}_{2}} \\
\end{align} $
On putting these values in equation (2), we get:
\[\begin{align}
  & \int_{{}}^{{}}{\dfrac{\cos x}{\cos x+2}dx=\dfrac{1}{\left( \cos x+2 \right)}\left( \sin x \right)-\int_{{}}^{{}}{\dfrac{\sin x}{{{\left( \cos x+2 \right)}^{2}}}.\sin xdx+{{c}_{2}}}} \\
 & \Rightarrow \int_{{}}^{{}}{\dfrac{\cos x}{\cos x+2}dx=\dfrac{\sin x}{\left( \cos x+2 \right)}-\int_{{}}^{{}}{\dfrac{{{\sin }^{2}}x}{{{\left( \cos x+2 \right)}^{2}}}.dx+{{c}_{1}}+{{c}_{2}}}} \\
\end{align}\]
So, the value of g(x) will be:
 $ \begin{align}
  & g\left( x \right)=\dfrac{\sin x}{\left( \cos x+2 \right)}-\int_{{}}^{{}}{\dfrac{{{\sin }^{2}}x}{{{\left( \cos x+2 \right)}^{2}}}dx+}\int_{{}}^{{}}{\dfrac{{{\sin }^{2}}x}{{{\left( \cos x+2 \right)}^{2}}}dx+c} \\
 & \Rightarrow g\left( x \right)=\dfrac{\sin x}{\left( \cos x+2 \right)}+c \\
\end{align} $
Since, it is given that g(0) = 0. So, we have:
 $ \begin{align}
  & \dfrac{\sin 0}{\left( \cos 0+2 \right)}+c=0 \\
 & \Rightarrow \dfrac{0}{1+2}+c=0 \\
 & \Rightarrow c=0 \\
\end{align} $
So, we have:
 $ g\left( x \right)=\dfrac{\sin x}{\left( \cos x+2 \right)} $
Since, we have to find the value of $ 32g\left( \dfrac{\pi }{2} \right) $ , so on substituting x by $ \dfrac{\pi }{2} $ , we get:
 $ \Rightarrow g\left( \dfrac{\pi }{2} \right)=\dfrac{\sin \dfrac{\pi }{2}}{\left( \cos \dfrac{\pi }{2}+2 \right)}=\dfrac{1}{0+2}=\dfrac{1}{2} $
On multiplying both sides by 32, we get:
 $ \Rightarrow 32\times g\left( \dfrac{\pi }{2} \right)=32\times \dfrac{1}{2}=16 $
Hence, the value of $ 32g\left( \dfrac{\pi }{2} \right) $ is equal to 16.

Note: A common mistake that many students may do here is finding the value of $ \int_{{}}^{{}}{\dfrac{{{\sin }^{2}}x}{{{\left( \cos x+2 \right)}^{2}}}}dx $ . We have left it as it is while calculating $ \int_{{}}^{{}}{\dfrac{\cos x}{\cos x+2}dx} $ so that when we put the final value of $ \int_{{}}^{{}}{\dfrac{\cos x}{\cos x+2}} $ for finding g(x), then $ \int_{{}}^{{}}{\dfrac{{{\sin }^{2}}x}{{{\left( \cos x+2 \right)}^{2}}}}dx $ and $ -\int_{{}}^{{}}{\dfrac{{{\sin }^{2}}x}{{{\left( \cos x+2 \right)}^{2}}}}dx $ get cancelled. This will make the calculation easy.