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Question: If anti-derivative of f(x) is \[{e^x}\] and that of g(x) is \[{\rm{cosx}}\], then \[\int {f(x)} \cos...

If anti-derivative of f(x) is ex{e^x} and that of g(x) is cosx{\rm{cosx}}, then ∫f(x)cos⁡xdx+∫g(x)exdx\int {f(x)} \cos xdx + \int {g(x){e^x}dx} is equal to
A.f(x).g(x)+c\B.f(x)+g(x)+cC.excos⁡x+c\D.f(x)−g(x)+c\begin{array}{l}A.f(x).g(x) + c\\\B.f(x) + g(x) + c\\\C.{e^x}\cos x + c\\\D.f(x) - g(x) + c\end{array}

Explanation

Solution

Hint: We know that anti-derivative of any function means the integration of the function. So, by using differentiation we will find the function and to integration we will use the property of integration as below,
∫ex(f(x)+f′(x)dx)=exf(x)+c\int {{e^x}\left( {f(x) + f'(x)dx} \right)} = {e^x}f(x) + c
Here, f′(x)f'(x) is derivative of f(x).

Complete step-by-step answer:
We have been given that, anti-derivative of f(x) is ex{e^x} and that of g(x) is cosx{\rm{cosx}}, then we have to find ∫f(x)cos⁡xdx+∫g(x)exdx\int {f(x)} \cos xdx + \int {g(x){e^x}dx} .
We know that anti-derivative of any function means the integration of that function.
⇒∫f(x)dx=ex\Rightarrow \int {f(x)dx = {e^x}}
We have to remember that the derivative of ex{e^x} is ex{e^x}. Now, on differentiating both sides with respect to ‘x’, we get,
On differentiating both sides with respect to ‘x’, we get,
⇒f(x)=ex\Rightarrow f(x) = {e^x}
Again, we have ∫g(x)dx=cos⁡x\int {g(x)dx = \cos x}
We have to remember that the derivative of cosx = -sinx. Now, on differentiating both sides with respect to ‘x’, we get,
⇒g(x)=−sin⁡x\Rightarrow g(x) = - \sin x
Now, let us consider ∫f(x)cos⁡xdx+∫g(x)exdx\int {f(x)} \cos xdx + \int {g(x){e^x}dx}
On substituting value of f(x) and g(x), we get,
⇒∫excosxdx+∫(−sin⁡x)exdx ⇒∫excosxdx+∫exsin⁡xdx ⇒∫(excosx−exsin⁡x)dx\begin{array}{l} \Rightarrow \int {{e^x}} cosxdx + \int {\left( { - \sin x} \right)} {e^x}dx\\\ \Rightarrow \int {{e^x}} cosxdx + \int {{e^x}} \sin xdx\\\ \Rightarrow \int {\left( {{e^x}cosx - {e^x}\sin x} \right)dx} \end{array}
On taking ex{e^x} as common, we get,
⇒∫ex(cosx−sin⁡x)dx\Rightarrow \int {{e^x}\left( {cosx - \sin x} \right)dx}
We can use the property of integration i.e.
∫ex(f(x)+f′(x))=exf(x)+c\int {{e^x}\left( {f(x) + f'(x)} \right)} = {e^x}f(x) + c
Where, f′(x)f'(x) is derivative of f(x).
⇒∫ex(cosx−sin⁡x)dx=excos⁡x+c\Rightarrow \int {{e^x}\left( {cosx - \sin x} \right)dx} = {e^x}\cos x + c
Therefore, the correct option is C.

Note: If we do not remember the property of integration which we use in the above question to find the integral then, we have to use the formula of integration which is given below,
∫u.vdx=u∫vdx−∫dudx∫vdxdx\int {u.vdx = u\int {vdx - \int {\dfrac{{du}}{{dx}}} } } \int {vdxdx}
Here, u and v are two functions of ‘x’. This will definitely take more time and decrease our efficiency in competitive examinations. So, try to remember the properties related with integration.