Tan^2x wrt cos^2x

E power sin^2x+sin^3x+sin^4x+............+infinite

To tackle the expression you've provided, we need to break it down into manageable parts. The expression consists of two main components: the derivative of \( \tan^2 x \) with respect to \( \cos^2 x \), and the infinite series \( e^{\sin^2 x} + \sin^3 x + \sin^4 x + \ldots \). Let's analyze each part step by step.

Finding the Derivative of \( \tan^2 x \) with Respect to \( \cos^2 x \)

First, we need to differentiate \( \tan^2 x \) with respect to \( \cos^2 x \). To do this, we can use the chain rule. Recall that:

• \( \tan x = \frac{\sin x}{\cos x} \)
• Therefore, \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \)

Now, let’s differentiate \( \tan^2 x \) with respect to \( x \):

Using the chain rule:

Let \( u = \tan^2 x \) and \( v = \cos^2 x \). We need \( \frac{du}{dv} \), which can be found using the derivatives:

1. Differentiate \( u \) with respect to \( x \):

Using the derivative of \( \tan^2 x \):

\( \frac{du}{dx} = 2\tan x \sec^2 x \)

2. Differentiate \( v \) with respect to \( x \):

\( \frac{dv}{dx} = -2\cos x \sin x = -\sin(2x) \)

Now, applying the chain rule:

\( \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2\tan x \sec^2 x}{-\sin(2x)} \)

Substituting \( \tan x = \frac{\sin x}{\cos x} \) and \( \sec^2 x = 1 + \tan^2 x \), we can simplify further if needed.

Evaluating the Infinite Series

Next, let’s consider the infinite series \( e^{\sin^2 x} + \sin^3 x + \sin^4 x + \ldots \). The first term, \( e^{\sin^2 x} \), is straightforward, but the series \( \sin^3 x + \sin^4 x + \ldots \) can be expressed as a geometric series.

Notice that:

• The series \( \sin^3 x + \sin^4 x + \ldots \) can be factored as \( \sin^3 x (1 + \sin x + \sin^2 x + \ldots) \)
• The inner series \( 1 + \sin x + \sin^2 x + \ldots \) is a geometric series with the first term \( 1 \) and common ratio \( \sin x \), valid for \( |\sin x| < 1 \).

The sum of an infinite geometric series is given by:

\( S = \frac{a}{1 - r} \) where \( a \) is the first term and \( r \) is the common ratio. Thus, we have:

\( S = \frac{1}{1 - \sin x} \)

Putting it all together, the infinite series becomes:

\( e^{\sin^2 x} + \sin^3 x \cdot \frac{1}{1 - \sin x} \)

Final Expression

Combining both parts, we arrive at the final expression:

\( \frac{2\tan x \sec^2 x}{-\sin(2x)} + e^{\sin^2 x} + \frac{\sin^3 x}{1 - \sin x} \)

This expression encapsulates the derivative of \( \tan^2 x \) with respect to \( \cos^2 x \) and the evaluation of the infinite series. Each component plays a crucial role in understanding the overall behavior of the expression as \( x \) varies.