Question

The integral $\int \frac{sin^2xcos^2x}{(sin^3x + cos^3x)sin^2x + sin^3xcos^2x + cos^3x)} dx$ is

• A. $\frac{1}{3(1 + tan^3x)} + c$, where c is a constant of integration.
• B. $\frac{-1}{3(1 + tan^3x)} + c$, where c is a constant of integration.
• C. $\frac{1}{1 + cot^3x} + c$, where c is a constant of integration.
• D. $\frac{-1}{1 + cos^3x} + c$, where c is a constant of integration.

Answer 1

Answer: (B)

$\frac{-1}{3(1 + tan^3x)} + c$, where c is a constant of integration.

Answer 2

Let the given integral be $I$.

$I = \int \frac{\sin^2x\cos^2x}{(\sin^3x + \cos^3x)\sin^2x + \sin^3x\cos^2x + \cos^3x)} dx$.

It appears that there is a typo in the denominator, and it was intended to be $(\sin^3x + \cos^3x)^2$. Assuming this corrected denominator, the solution proceeds as above.

Let the integral be $I = \int \frac{\sin^2x\cos^2x}{(\sin^3x + \cos^3x)^2} dx$.

Divide the numerator and the denominator by $\cos^6x$:

$I = \int \frac{\frac{\sin^2x\cos^2x}{\cos^6x}}{\frac{(\sin^3x + \cos^3x)^2}{\cos^6x}} dx = \int \frac{\frac{\sin^2x}{\cos^4x}}{(\frac{\sin^3x}{\cos^3x} + \frac{\cos^3x}{\cos^3x})^2} dx$

$I = \int \frac{\tan^2x \sec^2x}{(\tan^3x + 1)^2} dx$.

Let $u = 1 + \tan^3x$.

Differentiating $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(1 + \tan^3x) = 0 + 3\tan^2x \frac{d}{dx}(\tan x) = 3\tan^2x \sec^2x$.

So, $du = 3\tan^2x \sec^2x dx$.

The integral becomes:

$I = \int \frac{1}{(\tan^3x + 1)^2} (\tan^2x \sec^2x dx)$

$I = \int \frac{1}{u^2} \frac{du}{3} = \frac{1}{3} \int u^{-2} du$.

Integrating $u^{-2}$ with respect to $u$:

$\int u^{-2} du = \frac{u^{-2+1}}{-2+1} + c = \frac{u^{-1}}{-1} + c = -\frac{1}{u} + c$.

Substitute back $u = 1 + \tan^3x$:

$I = \frac{1}{3} \left(-\frac{1}{1 + \tan^3x}\right) + c = -\frac{1}{3(1 + \tan^3x)} + c$.