If an antiderivative of f(x) is ex and that of g(x) is cos x, then ∫ f(x) cos x d x+∫ g(x) ex d x is equal to where C is constant of integration.

Question

If an antiderivative of f(x) is ex and that of g(x) is cos x, then ∫ f(x) cos x d x+∫ g(x) ex d x is equal to where C is constant of integration. (A) f(x) g(x)+C (B) f(x)+g(x)+C (C)  e x cos x + C (D) f(x)-g(x)+C

Tardigrade Admin · Accepted Answer

∵ ∫ f(x) d x=ex ⇒ f(x)=ex ∫ g(x) d x= cos x ⇒ g(x)=- sin x text Given expression =∫ ex cos x d x+∫- sin x ex d x =∫ ex( cos x+(- sin x)) d x=ex cos x+C

Q. If an antiderivative of $f (x)$ is $e^{x}$ and that of $g (x)$ is $cos x$ , then $\int f (x) cos x d x + \int g (x) e^{x} d x$ is equal to
where $C$ is constant of integration.

$f (x) g (x) + C$

$f (x) + g (x) + C$

$e^{x} cos x + C$

$f (x) - g (x) + C$

$∵ \int f (x) d x = e^{x} \Rightarrow f (x) = e^{x}$
$\int g (x) d x = cos x \Rightarrow g (x) = - sin x$
$Given expression = \int e^{x} cos x d x + \int - sin x e^{x} d x$
$= \int e^{x} (cos x + (- sin x)) d x = e^{x} cos x + C$

Q. If an antiderivative of f(x) is ex and that of g(x) is cosx, then ∫f(x)cosxdx+∫g(x)exdx is equal to where C is constant of integration.

f(x)g(x)+C

f(x)+g(x)+C

excosx+C

f(x)−g(x)+C