Question

If $\int \frac{sin^4 x cos^2 x + cos^4 x sin^2 x}{(sin^5 x + cos^3 x sin^2 x + sin^3 x cos^2 x + cos^5 x)^2} dx = \frac{-1}{\alpha (1+tan^{\beta} x)} + C$, where $\alpha, \beta \in R$ then $\alpha + \beta$ is equal to _______.

Answer 1

6

Answer 2

The problem asks us to evaluate a definite integral and then determine the values of $\alpha$ and $\beta$ by comparing our result with a given form. Finally, we need to find the sum $\alpha + \beta$.

Let the given integral be $I$. $I = \int \frac{\sin^4 x \cos^2 x + \cos^4 x \sin^2 x}{(\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x)^2} dx$

Step 1: Simplify the numerator. The numerator (N) is $\sin^4 x \cos^2 x + \cos^4 x \sin^2 x$. We can factor out $\sin^2 x \cos^2 x$: $N = \sin^2 x \cos^2 x (\sin^2 x + \cos^2 x)$ Since $\sin^2 x + \cos^2 x = 1$, the numerator simplifies to: $N = \sin^2 x \cos^2 x$

Step 2: Simplify the term inside the parenthesis in the denominator. Let $T = \sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x$. Group terms: $T = (\sin^5 x + \sin^3 x \cos^2 x) + (\cos^3 x \sin^2 x + \cos^5 x)$ Factor out common terms from each group: $T = \sin^3 x (\sin^2 x + \cos^2 x) + \cos^3 x (\sin^2 x + \cos^2 x)$ Since $\sin^2 x + \cos^2 x = 1$: $T = \sin^3 x (1) + \cos^3 x (1)$ $T = \sin^3 x + \cos^3 x$ So, the denominator (D) is $(\sin^3 x + \cos^3 x)^2$.

Step 3: Rewrite the integral with the simplified numerator and denominator. The integral becomes: $I = \int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$

Step 4: Transform the integrand for substitution. To make the integral solvable, we can divide both the numerator and the denominator by a suitable power of $\cos x$. Dividing by $\cos^6 x$ (which is $(\cos^3 x)^2$) will convert the expression into terms of $\tan x$ and $\sec x$.

Numerator transformation: $\frac{\sin^2 x \cos^2 x}{\cos^6 x} = \frac{\sin^2 x}{\cos^4 x} = \frac{\sin^2 x}{\cos^2 x} \cdot \frac{1}{\cos^2 x} = \tan^2 x \sec^2 x$

Denominator transformation: $\frac{(\sin^3 x + \cos^3 x)^2}{\cos^6 x} = \left(\frac{\sin^3 x + \cos^3 x}{\cos^3 x}\right)^2 = \left(\frac{\sin^3 x}{\cos^3 x} + \frac{\cos^3 x}{\cos^3 x}\right)^2 = (\tan^3 x + 1)^2$

Now, the integral is: $I = \int \frac{\tan^2 x \sec^2 x}{(\tan^3 x + 1)^2} dx$

Step 5: Use substitution to evaluate the integral. Let $u = \tan^3 x + 1$. Differentiate $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(\tan^3 x + 1) = 3 \tan^2 x \cdot \frac{d}{dx}(\tan x) = 3 \tan^2 x \sec^2 x$ From this, we get $du = 3 \tan^2 x \sec^2 x dx$, or $\tan^2 x \sec^2 x dx = \frac{1}{3} du$.

Substitute $u$ and $du$ into the integral: $I = \int \frac{1}{u^2} \cdot \frac{1}{3} du$ $I = \frac{1}{3} \int u^{-2} du$

Step 6: Integrate $u^{-2}$. $I = \frac{1}{3} \left( \frac{u^{-2+1}}{-2+1} \right) + C$ $I = \frac{1}{3} \left( \frac{u^{-1}}{-1} \right) + C$ $I = -\frac{1}{3u} + C$

Step 7: Substitute back $u = \tan^3 x + 1$. $I = -\frac{1}{3(\tan^3 x + 1)} + C$ This can be written as: $I = -\frac{1}{3(1 + \tan^3 x)} + C$

Step 8: Compare the result with the given form. The given form is $\frac{-1}{\alpha (1+\tan^{\beta} x)} + C$. By comparing our result $I = -\frac{1}{3(1 + \tan^3 x)} + C$ with the given form, we can identify the values of $\alpha$ and $\beta$: $\alpha = 3$ $\beta = 3$

Step 9: Calculate $\alpha + \beta$. $\alpha + \beta = 3 + 3 = 6$

The final answer is $\boxed{6}$.

The final answer is $\boxed{6}$.

Subject, Chapter, and Topic:

Subject: Mathematics Chapter: Integrals Topic: Integration by Substitution

Difficulty Level:

Medium

Question Type:

Integer