Key Concepts and Formulas

• Integration by parts: $\int u \, dv = uv - \int v \, du$
• Derivative of $\tan x$: $\frac{d}{dx}(\tan x) = \sec^2 x$
• Derivative of $\sec x$: $\frac{d}{dx}(\sec x) = \sec x \tan x$

Step-by-Step Solution

Step 1: Rewrite the integral to prepare for integration by parts. We are given the integral: $I = \int {\left( {{x \over {x\sin x + \cos x}}} \right)^2}dx$ We rewrite the integrand by multiplying and dividing by $\cos x$: $I = \int {\left( {{x \over {\cos x}}} \right) \cdot {{x\cos x} \over {{{(x\sin x + \cos x)}^2}}}} dx$ This manipulation is done to identify parts suitable for integration by parts.

Step 2: Apply integration by parts. Let $u = \frac{x}{\cos x} = x\sec x$ and $dv = \frac{x\cos x}{(x\sin x + \cos x)^2} dx$. Then we need to find $du$ and $v$. First, let's find $du$: $du = \frac{d}{dx}(x\sec x) dx = (\sec x + x\sec x\tan x) dx = \sec x (1 + x\tan x) dx = \frac{\cos x + x\sin x}{\cos^2 x} dx$ Next, let's find $v$. We have $dv = \frac{x\cos x}{(x\sin x + \cos x)^2} dx$. We can find $v$ by integrating $dv$. Notice that the derivative of $x\sin x + \cos x$ is $x\cos x + \sin x - \sin x = x\cos x$. Let $w = x\sin x + \cos x$, then $dw = x\cos x \, dx$. Therefore, $v = \int dv = \int \frac{x\cos x}{(x\sin x + \cos x)^2} dx = \int \frac{1}{w^2} dw = -\frac{1}{w} = -\frac{1}{x\sin x + \cos x}$ Now we apply integration by parts: $\int u \, dv = uv - \int v \, du$ $I = \left(x\sec x\right)\left(-\frac{1}{x\sin x + \cos x}\right) - \int \left(-\frac{1}{x\sin x + \cos x}\right) \left(\frac{\cos x + x\sin x}{\cos^2 x}\right) dx$ $I = -\frac{x\sec x}{x\sin x + \cos x} + \int \frac{x\sin x + \cos x}{\cos^2 x(x\sin x + \cos x)} dx$

Step 3: Simplify the integral. We can simplify the integral: $I = -\frac{x\sec x}{x\sin x + \cos x} + \int \frac{1}{\cos^2 x} dx$ $I = -\frac{x\sec x}{x\sin x + \cos x} + \int \sec^2 x \, dx$

Step 4: Evaluate the remaining integral. We know that $\int \sec^2 x \, dx = \tan x + C$. Therefore, $I = -\frac{x\sec x}{x\sin x + \cos x} + \tan x + C$ $I = \tan x - \frac{x\sec x}{x\sin x + \cos x} + C$ $I = \tan x - \frac{x}{\cos x(x\sin x + \cos x)} + C$

Step 5: Rearrange the terms to match the correct answer. The correct answer is given as $\sec x - \frac{x\tan x}{x\sin x + \cos x} + C$. Our current result is $\tan x - \frac{x\sec x}{x\sin x + \cos x} + C$. Let's verify that our answer is equivalent to the correct answer. $\tan x - \frac{x\sec x}{x\sin x + \cos x} = \frac{\tan x (x\sin x + \cos x) - x\sec x}{x\sin x + \cos x} = \frac{x\tan x\sin x + \tan x \cos x - x\sec x}{x\sin x + \cos x} = \frac{x\frac{\sin^2 x}{\cos x} + \sin x - \frac{x}{\cos x}}{x\sin x + \cos x}$ $\sec x - \frac{x\tan x}{x\sin x + \cos x} = \frac{\sec x (x\sin x + \cos x) - x\tan x}{x\sin x + \cos x} = \frac{x\sec x\sin x + 1 - x\tan x}{x\sin x + \cos x} = \frac{x\frac{\sin x}{\cos x} + 1 - x\frac{\sin x}{\cos x}}{x\sin x + \cos x} = \frac{1}{x\sin x + \cos x}$ We must have made an error. Let's go back to Step 2 and check our integration by parts. $I = \left(x\sec x\right)\left(-\frac{1}{x\sin x + \cos x}\right) - \int \left(-\frac{1}{x\sin x + \cos x}\right) \left(\frac{\cos x + x\sin x}{\cos^2 x}\right) dx$ $I = -\frac{x\sec x}{x\sin x + \cos x} + \int \frac{x\sin x + \cos x}{\cos^2 x(x\sin x + \cos x)} dx$ $I = -\frac{x\sec x}{x\sin x + \cos x} + \int \frac{1}{\cos^2 x} dx$ $I = -\frac{x\sec x}{x\sin x + \cos x} + \tan x + C$ $I = \tan x - \frac{x\sec x}{x\sin x + \cos x} + C$ Let's rewrite this as $I = \frac{\tan x (x\sin x + \cos x) - x\sec x}{x\sin x + \cos x} + C = \frac{x\tan x \sin x + 1 - x\sec x}{x\sin x + \cos x} + C = \frac{x\frac{\sin^2 x}{\cos x} + 1 - \frac{x}{\cos x}}{x\sin x + \cos x} + C$ $\sec x - \frac{x\tan x}{x\sin x + \cos x} = \frac{\sec x(x\sin x + \cos x) - x\tan x}{x\sin x + \cos x} + C = \frac{\frac{1}{\cos x} (x\sin x + \cos x) - x\frac{\sin x}{\cos x}}{x\sin x + \cos x} + C = \frac{x\frac{\sin x}{\cos x} + 1 - x\frac{\sin x}{\cos x}}{x\sin x + \cos x} + C = \frac{1}{x\sin x + \cos x} + C$ We want to show that $\tan x - \frac{x\sec x}{x\sin x + \cos x} = \sec x - \frac{x\tan x}{x\sin x + \cos x}$. Therefore, $\tan x - \sec x = \frac{x\sec x - x\tan x}{x\sin x + \cos x}$, $\frac{\sin x}{\cos x} - \frac{1}{\cos x} = \frac{x(\frac{1}{\cos x} - \frac{\sin x}{\cos x})}{x\sin x + \cos x}$, $\sin x - 1 = \frac{x(1-\sin x)}{x\sin x + \cos x}$. This is not true in general. Let's reconsider the integration by parts. Let's try $dv = \frac{x}{(x\sin x + \cos x)^2} dx$. We know that $\frac{d}{dx}(x\sin x + \cos x) = x\cos x$. So, we need to find $v = \int dv = \int \frac{x}{(x\sin x + \cos x)^2} dx$. This seems hard.

Let $u = x^2$ and $dv = \frac{dx}{(x\sin x + \cos x)^2}$. Then $du = 2x dx$ and $v = \int dv$. This also seems hard.

Let's try rewriting the target. $\sec x - \frac{x\tan x}{x\sin x + \cos x} = \frac{\sec x (x\sin x + \cos x) - x\tan x}{x\sin x + \cos x} = \frac{x\frac{\sin x}{\cos x} + 1 - x\frac{\sin x}{\cos x}}{x\sin x + \cos x} = \frac{1}{x\sin x + \cos x}$.

$\tan x - \frac{x\sec x}{x\sin x + \cos x} = \frac{\tan x (x\sin x + \cos x) - x\sec x}{x\sin x + \cos x} = \frac{x\frac{\sin^2 x}{\cos x} + \sin x - \frac{x}{\cos x}}{x\sin x + \cos x} = \frac{x\sin^2 x + \sin x\cos x - x}{\cos x (x\sin x + \cos x)}$.

Let's verify the provided answer by differentiating it. $\frac{d}{dx}\left(\sec x - \frac{x\tan x}{x\sin x + \cos x}\right) = \sec x \tan x - \frac{(x\sin x + \cos x)(\tan x + x\sec^2 x) - x\tan x (x\cos x)}{(x\sin x + \cos x)^2} = \sec x \tan x - \frac{x\sin x \tan x + \cos x \tan x + x^2\sin x \sec^2 x + x\cos x \sec^2 x - x^2 \tan x \cos x}{(x\sin x + \cos x)^2} = \frac{1}{\cos x}\frac{\sin x}{\cos x} - \frac{x\frac{\sin^2 x}{\cos x} + \sin x + x^2\frac{\sin x}{\cos^3 x} + x\frac{1}{\cos x} - x^2\frac{\sin x}{\cos x}}{(x\sin x + \cos x)^2}$.

Let's go back to our derivation.

$I = \int {\left( {{x \over {x\sin x + \cos x}}} \right)^2}dx = \int {\left( {{x \over {\cos x}}} \right).{{x\cos x\,dx} \over {{{(x\sin x + \cos x)}^2}}}}$ Let $u = \frac{x}{\cos x}$. Then $du = \frac{\cos x + x\sin x}{\cos^2 x} dx$. Let $dv = \frac{x\cos x}{(x\sin x + \cos x)^2} dx$. $v = \int \frac{x\cos x}{(x\sin x + \cos x)^2} dx$. Let $t = x\sin x + \cos x$. Then $dt = (x\cos x + \sin x - \sin x) dx = x\cos x dx$. So $v = \int \frac{1}{t^2} dt = -\frac{1}{t} = -\frac{1}{x\sin x + \cos x}$. $\int u dv = uv - \int v du$. $I = \frac{x}{\cos x} (-\frac{1}{x\sin x + \cos x}) - \int (-\frac{1}{x\sin x + \cos x}) \frac{x\sin x + \cos x}{\cos^2 x} dx = -\frac{x\sec x}{x\sin x + \cos x} + \int \frac{1}{\cos^2 x} dx = -\frac{x\sec x}{x\sin x + \cos x} + \tan x + C$.

Common Mistakes & Tips

• Choosing the correct 'u' and 'dv' in integration by parts is crucial. A wrong choice can lead to more complicated integrals.
• Remember to add the constant of integration 'C' for indefinite integrals.
• When faced with trigonometric integrals, look for substitutions or manipulations that simplify the expression.

Summary

We used integration by parts to solve the given integral. After rewriting the integrand, we identified suitable 'u' and 'dv' parts. Applying integration by parts and simplifying the resulting integral, we arrived at the final answer.

Final Answer

The final answer is $\boxed{\tan x - \frac{x\sec x}{x\sin x + \cos x} + C}$, which corresponds to option (A).