Question

$\int\frac{x^2 + cos^2x}{(1+x^2)sin^2x}dx$ is equals to

• A. $tan^{-1}x + cotx + c$
• B. $tan^{-1}x$
• C. $cot^{-1}x - tanx + c$
• D. $-tan^{-1}$

Answer 1

Answer: (A)

$\tan^{-1}x + \cot x + c$

Answer 2

The given integral is: $I = \int\frac{x^2 + \cos^2x}{(1+x^2)\sin^2x}dx$

Step 1: Rewrite the numerator. We know the identity $\cos^2x = 1 - \sin^2x$. Substitute this into the numerator: $x^2 + \cos^2x = x^2 + (1 - \sin^2x) = (x^2 + 1) - \sin^2x$

Step 2: Substitute the rewritten numerator into the integral. $I = \int\frac{(x^2 + 1) - \sin^2x}{(1+x^2)\sin^2x}dx$

Step 3: Split the integrand into two separate fractions. $I = \int\left(\frac{x^2 + 1}{(1+x^2)\sin^2x} - \frac{\sin^2x}{(1+x^2)\sin^2x}\right)dx$

Step 4: Simplify each term. The first term simplifies by canceling $(x^2+1)$: $\frac{x^2 + 1}{(1+x^2)\sin^2x} = \frac{1}{\sin^2x}$ The second term simplifies by canceling $\sin^2x$: $\frac{\sin^2x}{(1+x^2)\sin^2x} = \frac{1}{1+x^2}$ So the integral becomes: $I = \int\left(\frac{1}{\sin^2x} - \frac{1}{1+x^2}\right)dx$

Step 5: Use trigonometric identity for the first term. Recall that $\frac{1}{\sin^2x} = \csc^2x$. $I = \int\left(\csc^2x - \frac{1}{1+x^2}\right)dx$

Step 6: Integrate term by term. The standard integral of $\csc^2x$ is $-\cot x$. The standard integral of $\frac{1}{1+x^2}$ is $\tan^{-1}x$. Therefore, $I = -\cot x - \tan^{-1}x + C$

Step 7: Compare the result with the given options. Our calculated result is $-\cot x - \tan^{-1}x + C$. Let's check the options: (a) $\tan^{-1}x + \cot x + c$ (b) $\tan^{-1}x$ (c) $\cot^{-1}x - \tan x + c$ (d) $-tan^{-1}x$

Our result is $-( \tan^{-1}x + \cot x) + C$. Option (a) is $(\tan^{-1}x + \cot x) + C$. There is a sign difference between our derived answer and option (a). However, option (a) contains the correct functional forms ($\tan^{-1}x$ and $\cot x$). Given that this is a multiple-choice question and option (a) is the only one with these correct functions, it is highly probable that there is a sign error in the question or the option itself. If we were to differentiate option (a), we would get the negative of the original integrand. $\frac{d}{dx}(\tan^{-1}x + \cot x + C) = \frac{1}{1+x^2} - \csc^2x = \frac{1}{1+x^2} - \frac{1}{\sin^2x} = \frac{\sin^2x - (1+x^2)}{(1+x^2)\sin^2x} = \frac{\sin^2x - 1 - x^2}{(1+x^2)\sin^2x} = \frac{-\cos^2x - x^2}{(1+x^2)\sin^2x} = -\frac{x^2+\cos^2x}{(1+x^2)\sin^2x}$ Since the derivative of option (a) is the negative of the integrand, the integral of the given expression would be the negative of option (a). However, among the given choices, option (a) is the most appropriate if we assume a sign error in the question or options.