Question

$\int e^x \cos x \, dx =$

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

Hint:

For integrals of the form $\int e^{ax} \cos(bx) \, dx$, you can use the direct formula: $$\frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) + c$$ In this case, $a=1$ and $b=1$, which gives $\frac{e^x}{2}(\cos x + \sin x)$ immediately.

Answer 1

The Correct Option is A

Step 1: First Integration by Parts: Let $I = \int e^x \cos x \, dx$. Using the ILATE rule, let $u = \cos x$ and $dv = e^x \, dx$. Then $du = -\sin x \, dx$ and $v = e^x$. Using $\int u \, dv = uv - \int v \, du$: $$I = e^x \cos x - \int e^x (-\sin x) \, dx$$ $$I = e^x \cos x + \int e^x \sin x \, dx$$

Step 2: Second Integration by Parts: Now, integrate $\int e^x \sin x \, dx$. Let $u = \sin x$ and $dv = e^x \, dx$. Then $du = \cos x \, dx$ and $v = e^x$. $$\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx$$ $$\int e^x \sin x \, dx = e^x \sin x - I$$

Step 3: Solve for $I$: Substitute this back into the equation for $I$: $$I = e^x \cos x + (e^x \sin x - I)$$ $$2I = e^x (\cos x + \sin x)$$ $$I = \frac{1}{2} e^x (\cos x + \sin x) + c$$