Question

$\int (\frac{sinx+sin2x+sin3x}{cosx+cos2x+cos3x})^2dx$

Answer 1

$\frac{1}{2}\tan(2x) - x + C$

Answer 2

The problem asks to evaluate the integral $\int (\frac{\sin x+\sin 2x+\sin 3x}{\cos x+\cos 2x+\cos 3x})^2dx$.

Step 1: Simplify the trigonometric expression inside the integral. The expression is $E = \frac{\sin x+\sin 2x+\sin 3x}{\cos x+\cos 2x+\cos 3x}$. We can rearrange the terms in the numerator and denominator and use the sum-to-product trigonometric identities: $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ $\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$

Numerator: $\sin x + \sin 2x + \sin 3x = (\sin x + \sin 3x) + \sin 2x$ $= 2 \sin\left(\frac{x+3x}{2}\right)\cos\left(\frac{x-3x}{2}\right) + \sin 2x$ $= 2 \sin(2x)\cos(-x) + \sin 2x$ $= 2 \sin(2x)\cos x + \sin 2x$ (since $\cos(-x) = \cos x$) $= \sin 2x (2\cos x + 1)$

Denominator: $\cos x + \cos 2x + \cos 3x = (\cos x + \cos 3x) + \cos 2x$ $= 2 \cos\left(\frac{x+3x}{2}\right)\cos\left(\frac{x-3x}{2}\right) + \cos 2x$ $= 2 \cos(2x)\cos(-x) + \cos 2x$ $= 2 \cos(2x)\cos x + \cos 2x$ $= \cos 2x (2\cos x + 1)$

Now, substitute these back into the expression $E$: $E = \frac{\sin 2x (2\cos x + 1)}{\cos 2x (2\cos x + 1)}$ Assuming $2\cos x + 1 \neq 0$, we can cancel the common term $(2\cos x + 1)$: $E = \frac{\sin 2x}{\cos 2x} = \tan 2x$

Step 2: Integrate the simplified expression. The integral becomes: $I = \int (\tan 2x)^2 dx = \int \tan^2(2x) dx$

We use the trigonometric identity $\tan^2 \theta = \sec^2 \theta - 1$: $I = \int (\sec^2(2x) - 1) dx$

Now, integrate term by term: $I = \int \sec^2(2x) dx - \int 1 dx$

For the first integral, $\int \sec^2(2x) dx$, we use a substitution. Let $u = 2x$, so $du = 2 dx$, which means $dx = \frac{1}{2} du$. $\int \sec^2(2x) dx = \int \sec^2(u) \frac{1}{2} du = \frac{1}{2} \int \sec^2(u) du$ We know that $\int \sec^2 u du = \tan u + C$. So, $\frac{1}{2} \tan u + C = \frac{1}{2} \tan(2x) + C_1$.

For the second integral, $\int 1 dx = x + C_2$.

Combining the results: $I = \frac{1}{2} \tan(2x) - x + C$ where $C = C_1 + C_2$ is the constant of integration.

The final answer is $\boxed{\frac{1}{2}\tan(2x) - x + C}$.

Explanation of the solution:

1. Simplify the integrand: Use sum-to-product formulas for trigonometric functions to simplify the numerator and denominator. $\sin x + \sin 3x = 2\sin(2x)\cos x$ $\cos x + \cos 3x = 2\cos(2x)\cos x$ This leads to $\frac{\sin 2x(2\cos x+1)}{\cos 2x(2\cos x+1)} = \tan 2x$.
2. Rewrite the integral: The integral becomes $\int \tan^2(2x) dx$.
3. Apply trigonometric identity: Use $\tan^2 \theta = \sec^2 \theta - 1$ to rewrite the integrand as $\sec^2(2x) - 1$.
4. Integrate: Integrate term by term. $\int \sec^2(2x) dx = \frac{1}{2}\tan(2x)$ (using substitution $u=2x$). $\int 1 dx = x$.
5. Combine results: The final integral is $\frac{1}{2}\tan(2x) - x + C$.

Answer:

The integral is $\frac{1}{2}\tan(2x) - x + C$.