Mathematics

If an antiderivative of $$\displaystyle f ( x )$$ is $$e ^ { x }$$ and that of $$\displaystyle g ( x )$$ is $$\cos x ,$$ then $$\int f ( x ) \cos x d x + \int g ( x ) e ^ { x } d x =$$


ANSWER

$$e ^ { x } \cos x + c$$


SOLUTION
$$\displaystyle\int \underset{I}{f(x)}\underset{II}{\cos x}dx+\displaystyle\int \underset{I}{g(x)}\underset{II}{e^x}d(x)$$
Using by parts
$$f(x)\displaystyle\int \cos xdx-\displaystyle\int f'(x)\cdot \sin xdx+g(x)\displaystyle\int e^x-\displaystyle\int g'(x)e^x$$
$$\displaystyle\int \underset{II}{f(x)}\underset{I}{\cos x}dx+\displaystyle\underset{II}{g(x)}\underset{I}{e^x}dx$$
Using by parts
$$\cos x\displaystyle\int f(x)+\displaystyle\int \sin x(\displaystyle\int f(x)dx)dx+e^x\displaystyle\int g(x)-\displaystyle\int e^x(\displaystyle\int g(x)dx)$$
$$=\cos xe^x+\displaystyle\int \sin x e^xdx+e^x\cos x-\displaystyle\int e^x\cos x dx$$
$$=2\cos x e^x+\displaystyle\int \sin x e^x dx+\displaystyle\int e^x\cos x-\displaystyle\int e^x\cos x dx$$
$$=2\cos x e^x+\displaystyle\int \sin x e^xdx+\displaystyle\int \underset{II}{e^x}\underset{I}{(-\cos x)}dx$$
$$=2\cos x e^x+\displaystyle\int \sin x e^xdx+(-\cos x)e^x-\displaystyle\int \sin x e^x dx$$
$$=\cos x e^x+c$$.
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Single Correct Medium Published on 17th 09, 2020
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