Question

$\int e^{x}(2e^{x}+\sin x+\cos x+2)dx$ is equal to

• A. $e^{x}(e^{x}+\cos x+2)+C$
• B. $e^{x}(e^{x}+\sin x+2x)+C$
• C. $e^{x}(e^{x}+\sin x+2)+C$
• D. $e^{x}(e^{x}+\sin x+1)+C$
• E. $e^{x}(e^{x}-\sin x+2)+C$

Hint:

Math Tip: Whenever you see an $e^x$ multiplying a mix of trigonometric functions, immediately look for the $f(x) + f'(x)$ pattern. It saves you from having to perform Integration by Parts twice!

Answer 1

The Correct Option is C

Concept:
Calculus - Integration using Standard Forms.
The integral $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.
Step 1: Distribute the exponential term.
Multiply $e^x$ into the bracket: $$ \int (2e^{2x} + e^x\sin x + e^x\cos x + 2e^x) dx $$
Step 2: Group the terms for integration.
Split the integral into two distinct parts: the purely exponential part and the trigonometric part. $$ \int (2e^{2x} + 2e^x) dx + \int e^x(\sin x + \cos x) dx $$
Step 3: Integrate the exponential group.
Integrate the terms $2e^{2x}$ and $2e^x$ directly: $$ \int 2e^{2x} dx + \int 2e^x dx = \frac{2e^{2x}}{2} + 2e^x = e^{2x} + 2e^x $$
Step 4: Integrate the trigonometric group.
Identify the function and its derivative in the expression $e^x(\sin x + \cos x)$:

• Let $f(x) = \sin x$
• Then $f'(x) = \cos x$

Apply the standard formula $\int e^x [f(x) + f'(x)] dx = e^x f(x)$: $$ \int e^x(\sin x + \cos x) dx = e^x \sin x $$
Step 5: Combine the results and factor.
Add the integrated parts together and add the constant of integration $C$: $$ e^{2x} + 2e^x + e^x \sin x + C $$ Factor out the common term $e^x$ to match the given options: $$ e^x (e^x + 2 + \sin x) + C $$ $$ e^x (e^x + \sin x + 2) + C $$