Intertwined Integrals

Calculus Level 1

If an antiderivative of $f\left( x \right)$ is ${ e }^{ x }$ and that of $g\left( x \right)$ is $\cos { x }$ , then $\displaystyle \int { f\left( x \right) \cos { x } \ dx } +\int { g\left( x \right) { e }^{ x } \ dx }$ is equal to

{ e }^{ x }\cos { x } +C

f\left( x \right) -g\left( x \right) +C

f\left( x \right) +g\left( x \right) +C

f\left( x \right) .g\left( x \right) +C

3 solutions

Caleb Townsend
Apr 6, 2015

$f(x) = \frac{d}{dx} e^x = e^x \\ g(x) = \frac{d}{dx} \cos(x) = -\sin(x) \\ \int f(x) \cos(x) dx = \int e^x\cos(x)dx \\ = e^x\cos(x) - \int -e^x\sin(x) dx$ Now substituting this into the problem, $\int f(x)\cos(x) dx + \int g(x)e^x dx = \\ \int e^x \cos(x) dx + \int -\sin(x) e^x dx \\ = e^x\cos(x) - \int -e^x\sin(x) dx + \int -e^x\sin(x) dx \\ = \boxed{e^x\cos(x) + C}$

Stewart Gordon
May 14, 2015

The integral is just $\int e^x (\cos x - \sin x) dx$ Since $e^x$ is its own derivative, and $-\sin x$ is the derivative of $\cos x$ , the integrand is a result of applying the product rule.

Hjalmar Orellana Soto
Aug 11, 2015

$\displaystyle \int f(x) cosx dx + \displaystyle \int g(x) e^{x} dx$ $=e^{x}cosx - \displaystyle \int e^{x}g(x) dx + \displaystyle \int g(x) e^{x} dx$ $=e^{x}cosx +C$

Intertwined Integrals

3 solutions

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