Question

If $P=\int{{{\text{e}}^{ax}}\text{cosbx }dx} $ and $ Q=\int{{{\text{e}}^{ax}}\text{sinbx }dx} $ , then $ {{\tan }^{-1}}\left( \dfrac{Q}{P} \right)+{{\tan }^{-1}}\left( \dfrac{b}{a} \right) $ is equal to:
(a) ax
(b) bx
(c) $ \dfrac{x}{a} $
(d) $ \dfrac{x}{b} $

Answer 1

Hint: Start by applying by-parts as \[P=\int{{{\text{e}}^{ax}}\text{cosbx }dx}=\cos bx\int{{{e}^{ax}}}dx-\int{\dfrac{d\left( \cos bx \right)}{dx}\left( \int{{{e}^{ax}}}dx \right)}dx\] . When you will simplify and substitute $ Q=\int{{{\text{e}}^{ax}}\text{sinbx }dx} $ , we get \[P=\cos bx\times \dfrac{{{e}^{ax}}}{a}-\dfrac{b}{a}\times Q\] . Rearrange it and let it be equation (i). Repeat the same with Q and let it be equation (ii). Now divide both the equations and use the formula $ {{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A+B}{1-AB} $ to get the answer.

Complete step-by-step answer:
Let us start with the integral given in the above question.
 $ P=\int{{{\text{e}}^{ax}}\text{cosbx }dx} $
Now according to the rule of the integration by parts:
 $ \int{uvdx=u\int{v}}dx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)} $ . From the two functions, u and v are decided using the ILATE preference rule. According to the ILATE rule, the preference order for u in decreasing order is Inverse, Logarithmic, Algebraic, Trigonometric and Exponential. For $ {{\text{e}}^{ax}}\text{cosbx} $ , v is $ {{e}^{ax}} $ and u is $ \cos bx $ .
\[P=\int{{{\text{e}}^{ax}}\text{cosbx }dx}=\cos bx\int{{{e}^{ax}}}dx-\int{\dfrac{d\left( \cos bx \right)}{dx}\left( \int{{{e}^{ax}}}dx \right)}dx\]
Now, we know that $ \dfrac{d\left( \cos bx \right)}{dx}=-\dfrac{\operatorname{sinb}x}{b}\text{ and }\int{{{e}^{ax}}dx}=\dfrac{{{e}^{ax}}}{a} $ . So, if we put this in our integral and simplify, we get
\[P=\cos bx\times \dfrac{{{e}^{ax}}}{a}+\int{\dfrac{b}{a}\sin bx\times {{e}^{ax}}}dx\]
\[\Rightarrow P=\cos bx\times \dfrac{{{e}^{ax}}}{a}+\dfrac{b}{a}\times \int{\sin bx\times {{e}^{ax}}}dx\]
Now, we know that $ Q=\int{{{\text{e}}^{ax}}\text{sinbx }dx} $ .
\[P=\cos bx\times \dfrac{{{e}^{ax}}}{a}+\dfrac{b}{a}\times Q\]
\[\Rightarrow P-\dfrac{b}{a}\times Q=\dfrac{\cos bx\times {{e}^{ax}}}{a}..........(i)\]
Similarly, we will solve Q.
\[Q=\int{{{\text{e}}^{ax}}\text{sinbx }dx}=\sin bx\int{{{e}^{ax}}}dx-\int{\dfrac{d\left( \sin bx \right)}{dx}\left( \int{{{e}^{ax}}}dx \right)}dx\]
Now, we know that $ \dfrac{d\left( \sin bx \right)}{dx}=\dfrac{cosbx}{b}\text{ and }\int{{{e}^{ax}}dx}=\dfrac{{{e}^{ax}}}{a} $ . So, if we put this in our integral and simplify, we get
\[Q=\sin bx\times \dfrac{{{e}^{ax}}}{a}-\int{\dfrac{b}{a}\cos bx\times {{e}^{ax}}}dx\]
\[\Rightarrow Q=\sin bx\times \dfrac{{{e}^{ax}}}{a}-\dfrac{b}{a}\times \int{\cos bx\times {{e}^{ax}}}dx\]
Now, we know that $ P=\int{{{\text{e}}^{ax}}\text{cosbx }dx} $ .
\[Q=\sin bx\times \dfrac{{{e}^{ax}}}{a}-\dfrac{b}{a}\times P\]
\[\Rightarrow Q+\dfrac{b}{a}\times P=\dfrac{\sin bx\times {{e}^{ax}}}{a}..........(ii)\]
Now we will divide equation (ii), by equation (i). On doing so, we get

\[\dfrac{Q+\dfrac{b}{a}\times P}{P-\dfrac{b}{a}\times Q}=\dfrac{\dfrac{\sin bx\times {{e}^{ax}}}{a}}{\dfrac{\cos bx\times {{e}^{ax}}}{a}}\]
Now we will take P common from numerator and denominator. On doing so, we get
\[\Rightarrow \dfrac{\left( \dfrac{Q}{P}+\dfrac{b}{a} \right)P}{\left( 1-\dfrac{b}{a}\times \dfrac{Q}{P} \right)P}=\dfrac{\sin bx}{\cos bx}\]
We know that $ \dfrac{\sin x}{\cos x}=\tan x $ .
\[\dfrac{\dfrac{Q}{P}+\dfrac{b}{a}}{1-\dfrac{b}{a}\times \dfrac{Q}{P}}=\tan bx\]
Now, we will use the fact that \[\tan b=\tan a\Rightarrow b={{\tan }^{-1}}\tan a\] .
\[{{\tan }^{-1}}\left( \dfrac{\dfrac{Q}{P}+\dfrac{b}{a}}{1-\dfrac{b}{a}\times \dfrac{Q}{P}} \right)=bx\]
Now, if we compare this with the formula $ {{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A+B}{1-AB} $ , we find that $ A=\dfrac{Q}{P} $ and $ B=\dfrac{b}{a} $ .
\[{{\tan }^{-1}}\left( \dfrac{Q}{P} \right)+{{\tan }^{-1}}\left( \dfrac{b}{a} \right)=bx\]
Hence, the answer to the above question is option (b).

Note: See while you use integral by-parts, you don’t have to put the constant of integral I each and every step, but you just have to put a constant term at the end of the integrated answer to compensate for all the constant terms that would have appeared. Also, you need to remember all the basic formulas that we use for indefinite integrals as they are used in definite integrations as well. Also, the formula $ {{\tan }^{-1}}A+{{\tan }^{-1}}B $ is different for different cases and can be represented as:
   $ {{\tan }^{-1}}A+{{\tan }^{-1}}B=\left\{ \begin{align}
  & {{\tan }^{-1}}\dfrac{A+B}{1-AB}\text{ AB < 1} \\
 & \pi +{{\tan }^{-1}}\dfrac{A+B}{1-AB}\text{ AB > 1} \\
\end{align} \right. $