Proving The Definite Integral Of Trigonometric Functions

Hey guys! Let's dive into a fascinating integral problem today. We're going to explore how to prove a definite integral that looks a bit intimidating at first glance. Don't worry, we'll break it down step-by-step and make it super clear. Our mission is to prove the following:

$$\int_0^{\frac{\pi}{2}} \frac{\sin(2x)\cos(x)-a\sin(2x)\sin(x)\tan(x)}{\sqrt{1+a^2\tan^2(x)}} dx=-\frac{2}{3}\cdot \frac{a-1}{a+1}$$

This integral involves trigonometric functions and a parameter a, so it's a great exercise in applying various integration techniques and trigonometric identities. We will explore the detailed, step-by-step solution to tackle this integral, ensuring clarity and understanding at every stage.

Breaking Down the Integral

Initial Simplification

To start, let's simplify the integrand. The first step in tackling this intricate integral is to simplify the integrand using trigonometric identities. Remember, guys, the key to solving complex integrals often lies in making them more manageable through simplification. We'll leverage our knowledge of trigonometric relationships to break down the expression inside the integral. Our integrand is:

$$\frac{\sin(2x)\cos(x)-a\sin(2x)\sin(x)\tan(x)}{\sqrt{1+a^2\tan^2(x)}}$$

We know that $\sin(2x) = 2\sin(x)\cos(x)$ and $\tan(x) = \frac{\sin(x)}{\cos(x)}$. Let's substitute these into the integrand:

\frac{2\sin(x)\cos(x)\cos(x) - a(2\sin(x)\cos(x))\sin(x)\frac{\sin(x)}{\cos(x)}}{\sqrt{1+a^2\frac{\sin^2(x)}{\cos^2(x)}}}}

Simplify the numerator:

\frac{2\sin(x)\cos^2(x) - 2a\sin^3(x)}{\sqrt{1+a^2\frac{\sin^2(x)}{\cos^2(x)}}}}

Now, let's work on the denominator. Multiply the numerator and denominator by $\cos(x)$ to get rid of the fraction inside the square root:

\frac{2\sin(x)\cos^2(x) - 2a\sin^3(x)}{\sqrt{\frac{\cos^2(x) + a^2\sin^2(x)}{\cos^2(x)}}}}

This simplifies to:

\frac{2\sin(x)\cos^2(x) - 2a\sin^3(x)}{\frac{\sqrt{\cos^2(x) + a^2\sin^2(x)}}{\cos(x)}}}

Which further simplifies to:

$$\frac{(2\sin(x)\cos^2(x) - 2a\sin^3(x))\cos(x)}{\sqrt{\cos^2(x) + a^2\sin^2(x)}}$$

Factor out $2\sin(x)$ from the numerator:

$$\frac{2\sin(x)(cos^2(x) - a\sin^2(x))\cos(x)}{\sqrt{\cos^2(x) + a^2\sin^2(x)}}$$

Substitution Strategy

Now, we need to figure out the best substitution strategy. The goal here is to simplify the integral into a form that we can recognize and solve. Looking at the simplified integrand:

$$\frac{2\sin(x)(\cos^2(x) - a\sin^2(x))\cos(x)}{\sqrt{\cos^2(x) + a^2\sin^2(x)}}$$

The square root in the denominator looks like a good candidate for substitution. Let's try the following substitution:

$$u = \cos^2(x) + a^2\sin^2(x)$$

This substitution seems promising because its derivative will involve $\sin(x)\cos(x)$, which we have in the numerator. Now, let's find the derivative of u with respect to x:

$$\frac{du}{dx} = -2\cos(x)\sin(x) + 2a^2\sin(x)\cos(x)$$

Factor out $2\sin(x)\cos(x)$:

$$\frac{du}{dx} = 2\sin(x)\cos(x)(a^2 - 1)$$

Now, we can express dx in terms of du:

$$dx = \frac{du}{2\sin(x)\cos(x)(a^2 - 1)}$$

Transforming the Integrand

Next, we need to rewrite the numerator in terms of u. Notice that $\cos^2(x) - a\sin^2(x)$ appears in the numerator. We can rewrite the substitution equation as:

$$u = \cos^2(x) + a^2\sin^2(x) = \cos^2(x) + a^2(1 - \cos^2(x)) = \cos^2(x) + a^2 - a^2\cos^2(x)$$

We want to express $\cos^2(x) - a\sin^2(x)$ in terms of u. Let's rewrite $\sin^2(x)$ as $1 - \cos^2(x)$:

$$\cos^2(x) - a\sin^2(x) = \cos^2(x) - a(1 - \cos^2(x)) = \cos^2(x) - a + a\cos^2(x) = (1+a)\cos^2(x) - a$$

From the substitution equation, we have:

$$u = \cos^2(x) + a^2(1 - \cos^2(x)) = \cos^2(x) + a^2 - a^2\cos^2(x)$$

Rearrange to solve for $\cos^2(x)$:

$$\cos^2(x)(1 - a^2) = u - a^2$$

$$\cos^2(x) = \frac{u - a^2}{1 - a^2}$$

Substitute this back into the expression for $\cos^2(x) - a\sin^2(x)$:

$$(1+a)\cos^2(x) - a = (1+a)\frac{u - a^2}{1 - a^2} - a = \frac{(1+a)(u - a^2)}{1 - a^2} - a$$

Simplify the fraction:

$$\frac{(1+a)(u - a^2)}{(1 - a)(1 + a)} - a = \frac{u - a^2}{1 - a} - a = \frac{u - a^2 - a(1 - a)}{1 - a} = \frac{u - a^2 - a + a^2}{1 - a} = \frac{u - a}{1 - a}$$

Changing the Limits of Integration

Don't forget to change the limits of integration! When we make a substitution in a definite integral, we need to adjust the limits of integration to match the new variable. Our original limits were $x = 0$ and $x = \frac{\pi}{2}$.

For $x = 0$:

$$u = \cos^2(0) + a^2\sin^2(0) = 1^2 + a^2(0) = 1$$

For $x = \frac{\pi}{2}$:

$$u = \cos^2(\frac{\pi}{2}) + a^2\sin^2(\frac{\pi}{2}) = 0 + a^2(1) = a^2$$

So our new limits of integration are from $u = 1$ to $u = a^2$.

Putting It All Together

Now, let's substitute everything back into the integral. We have:

$$\int_0^{\frac{\pi}{2}} \frac{2\sin(x)(\cos^2(x) - a\sin^2(x))\cos(x)}{\sqrt{\cos^2(x) + a^2\sin^2(x)}} dx$$

Substitute $u = \cos^2(x) + a^2\sin^2(x)$, $dx = \frac{du}{2\sin(x)\cos(x)(a^2 - 1)}$, and $\cos^2(x) - a\sin^2(x) = \frac{u - a}{1 - a}$:

$$\int_1^{a^2} \frac{2\sin(x)(\frac{u - a}{1 - a})\cos(x)}{\sqrt{u}} \cdot \frac{du}{2\sin(x)\cos(x)(a^2 - 1)}$$

Cancel out the $2\sin(x)\cos(x)$ terms:

$$\int_1^{a^2} \frac{\frac{u - a}{1 - a}}{\sqrt{u}} \cdot \frac{du}{a^2 - 1}$$

Simplify the constants:

$$\frac{1}{(1 - a)(a^2 - 1)} \int_1^{a^2} \frac{u - a}{\sqrt{u}} du$$

Notice that $a^2 - 1 = (a - 1)(a + 1)$, so we can rewrite the constant term:

$$\frac{1}{(1 - a)(a - 1)(a + 1)} \int_1^{a^2} \frac{u - a}{\sqrt{u}} du = \frac{-1}{(a - 1)^2(a + 1)} \int_1^{a^2} \frac{u - a}{\sqrt{u}} du$$

Evaluating the Simplified Integral

Now, let's evaluate the simplified integral:

$$\int_1^{a^2} \frac{u - a}{\sqrt{u}} du = \int_1^{a^2} (u^{1/2} - a u^{-1/2}) du$$

Integrate term by term:

$$\left[ \frac{2}{3}u^{3/2} - 2au^{1/2} \right]_1^{a^2}$$

Evaluate at the limits of integration:

$$\left( \frac{2}{3}(a^2)^{3/2} - 2a(a^2)^{1/2} \right) - \left( \frac{2}{3}(1)^{3/2} - 2a(1)^{1/2} \right)$$

Simplify:

$$\frac{2}{3}a^3 - 2a^2 - \frac{2}{3} + 2a$$

Final Simplification and the Grand Finale!

Now, let's put everything back together. We had:

$$\frac{-1}{(a - 1)^2(a + 1)} \int_1^{a^2} \frac{u - a}{\sqrt{u}} du$$

Substitute the result of the integral:

$$\frac{-1}{(a - 1)^2(a + 1)} \left( \frac{2}{3}a^3 - 2a^2 - \frac{2}{3} + 2a \right)$$

Multiply through by the constant:

$$\frac{-\frac{2}{3}a^3 + 2a^2 + \frac{2}{3} - 2a}{(a - 1)^2(a + 1)}$$

Factor out $\frac{2}{3}$ from the numerator:

$$\frac{\frac{2}{3}(-a^3 + 3a^2 - 3a + 1)}{(a - 1)^2(a + 1)}$$

Notice that $-a^3 + 3a^2 - 3a + 1 = -(a - 1)^3$, so we can simplify further:

$$\frac{\frac{2}{3}(-(a - 1)^3)}{(a - 1)^2(a + 1)} = \frac{-\frac{2}{3}(a - 1)^3}{(a - 1)^2(a + 1)}$$

Cancel out $(a - 1)^2$:

$$\frac{-\frac{2}{3}(a - 1)}{a + 1} = -\frac{2}{3} \cdot \frac{a - 1}{a + 1}$$

And there we have it! We've successfully proven that:

$$\int_0^{\frac{\pi}{2}} \frac{\sin(2x)\cos(x)-a\sin(2x)\sin(x)\tan(x)}{\sqrt{1+a^2\tan^2(x)}} dx=-\frac{2}{3}\cdot \frac{a-1}{a+1}$$

Conclusion

Wow, guys, what a journey! We started with a seemingly complex integral and, through careful simplification, strategic substitution, and a bit of algebraic manipulation, we arrived at the solution. Remember, the key to mastering integrals is practice and patience. Keep exploring, keep simplifying, and you'll conquer any integral that comes your way!

This problem showcases the power of trigonometric identities and strategic substitutions in integral calculus. By breaking down the problem into smaller, manageable steps, we were able to navigate through the complexities and arrive at the elegant final result. Remember, every challenging integral is an opportunity to sharpen your skills and deepen your understanding. Happy integrating!