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Question: If \(P=\int{{{\text{e}}^{ax}}\text{cosbx }dx} \) and \( Q=\int{{{\text{e}}^{ax}}\text{sinbx }dx} \) ...

If P=∫eaxcosbx dxP=\int{{{\text{e}}^{ax}}\text{cosbx }dx} and Q=∫eaxsinbx dxQ=\int{{{\text{e}}^{ax}}\text{sinbx }dx} , then tan⁡−1(QP)+tan⁡−1(ba){{\tan }^{-1}}\left( \dfrac{Q}{P} \right)+{{\tan }^{-1}}\left( \dfrac{b}{a} \right) is equal to:
(a) ax
(b) bx
(c) xa\dfrac{x}{a}
(d) xb\dfrac{x}{b}

Explanation

Solution

Hint: Start by applying by-parts as P=∫eaxcosbx dx=cos⁡bx∫eaxdx−∫d(cos⁡bx)dx(∫eaxdx)dxP=\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=∫eaxsinbx dxQ=\int{{{\text{e}}^{ax}}\text{sinbx }dx} , we get P=cos⁡bx×eaxa−ba×QP=\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⁡−1A+tan⁡−1B=tan⁡−1A+B1−AB{{\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=∫eaxcosbx dxP=\int{{{\text{e}}^{ax}}\text{cosbx }dx}
Now according to the rule of the integration by parts:
∫uvdx=u∫vdx−∫(dudx∫vdx)\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 eaxcosbx{{\text{e}}^{ax}}\text{cosbx} , v is eax{{e}^{ax}} and u is cos⁡bx\cos bx .
P=∫eaxcosbx dx=cos⁡bx∫eaxdx−∫d(cos⁡bx)dx(∫eaxdx)dxP=\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 d(cos⁡bx)dx=−sinb⁡xb and ∫eaxdx=eaxa\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×eaxa+∫basin⁡bx×eaxdxP=\cos bx\times \dfrac{{{e}^{ax}}}{a}+\int{\dfrac{b}{a}\sin bx\times {{e}^{ax}}}dx
⇒P=cos⁡bx×eaxa+ba×∫sin⁡bx×eaxdx\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=∫eaxsinbx dxQ=\int{{{\text{e}}^{ax}}\text{sinbx }dx} .
P=cos⁡bx×eaxa+ba×QP=\cos bx\times \dfrac{{{e}^{ax}}}{a}+\dfrac{b}{a}\times Q
⇒P−ba×Q=cos⁡bx×eaxa..........(i)\Rightarrow P-\dfrac{b}{a}\times Q=\dfrac{\cos bx\times {{e}^{ax}}}{a}..........(i)
Similarly, we will solve Q.
Q=∫eaxsinbx dx=sin⁡bx∫eaxdx−∫d(sin⁡bx)dx(∫eaxdx)dxQ=\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 d(sin⁡bx)dx=cosbxb and ∫eaxdx=eaxa\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×eaxa−∫bacos⁡bx×eaxdxQ=\sin bx\times \dfrac{{{e}^{ax}}}{a}-\int{\dfrac{b}{a}\cos bx\times {{e}^{ax}}}dx
⇒Q=sin⁡bx×eaxa−ba×∫cos⁡bx×eaxdx\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=∫eaxcosbx dxP=\int{{{\text{e}}^{ax}}\text{cosbx }dx} .
Q=sin⁡bx×eaxa−ba×PQ=\sin bx\times \dfrac{{{e}^{ax}}}{a}-\dfrac{b}{a}\times P
⇒Q+ba×P=sin⁡bx×eaxa..........(ii)\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

Q+ba×PP−ba×Q=sin⁡bx×eaxacos⁡bx×eaxa\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
⇒(QP+ba)P(1−ba×QP)P=sin⁡bxcos⁡bx\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 sin⁡xcos⁡x=tan⁡x\dfrac{\sin x}{\cos x}=\tan x .
QP+ba1−ba×QP=tan⁡bx\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⇒b=tan⁡−1tan⁡a\tan b=\tan a\Rightarrow b={{\tan }^{-1}}\tan a .
tan⁡−1(QP+ba1−ba×QP)=bx{{\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⁡−1A+tan⁡−1B=tan⁡−1A+B1−AB{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\dfrac{A+B}{1-AB} , we find that A=QPA=\dfrac{Q}{P} and B=baB=\dfrac{b}{a} .
tan⁡−1(QP)+tan⁡−1(ba)=bx{{\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⁡−1A+tan⁡−1B{{\tan }^{-1}}A+{{\tan }^{-1}}B is different for different cases and can be represented as:
{{\tan }^{-1}}A+{{\tan }^{-1}}B=\left\\{ \begin{aligned} & {{\tan }^{-1}}\dfrac{A+B}{1-AB}\text{ AB < 1} \\\ & \pi +{{\tan }^{-1}}\dfrac{A+B}{1-AB}\text{ AB > 1} \\\ \end{aligned} \right.