MHB How Do You Evaluate and Differentiate Complex Trigonometric Functions?

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The discussion focuses on evaluating the integral of sin(2x)/(1 + cos^2(x)) and differentiating the function f(x) = sin^2(e^(sin^2(x))). For the integral, a substitution method is suggested, letting u = 1 + cos^2(x) to simplify the expression. The differentiation involves applying the chain rule, resulting in a complex expression involving sin and cos functions. Participants clarify the original questions to ensure accurate assistance. The thread provides valuable insights into tackling complex trigonometric functions through integration and differentiation techniques.
Ivan1
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Evaluate ∫[sin2x/(1+(cos)^2 x) dx]Differentiate f(x) = (sin)^2 (e^((sin^2) x))

Hello, I'm just really stumped with these review questions and i have a test coming up. For the first, I'm not too sure what to do since there is a sin2x in general and for the second i don't know how to deal the the exponential. Some help would be really appreciated!
 
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Hello Ivan, welcome to MHB! (Wave)

When you are integrating, it is always helpful to first see if you can find a function and its derivative in the integrand. In this case, rewriting the integrand produces exactly what we want:
$$\int \frac{\sin\left({2x}\right)}{1+\cos^2\left({x}\right)} \,dx=\int \frac{2\sin\left({x}\right)\cos\left({x}\right)}{1+1-\sin^2\left({x}\right)} \,dx$$

For your derivative, do you mean this?
$$\sin^2\left({e^{\sin^2x}x}\right)$$

Recall that given a function $e^{f(x)}$, its derivative is $e^{f(x)}\cdot f'(x)$
 
Ivan said:
Evaluate ∫[sin2x/(1+(cos)^2 x) dx]Differentiate f(x) = (sin)^2 (e^((sin^2) x))
Just to be clear, are your questions
[math]\int \frac{\sin(2x)}{1 + \cos^2(x)}~dx[/math]

and
[math]f(x) = \sin^2 \left ( e^{\sin^2(x)} \right )[/math]

-Dan
 
topsquark said:
Just to be clear, are your questions
[math]\int \frac{\sin(2x)}{1 + \cos^2(x)}~dx[/math]

and
[math]f(x) = \sin^2 \left ( e^{\sin^2(x)} \right )[/math]

-Dan

Yes
 
Let's look at the first problem, and use a method similar to that suggested by Rido12:

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

Now, if we let:

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

then we find:

$$du=2\cos(x)(-\sin(x))\,dx=-\sin(2x)\,dx$$

Now, if we write the integral as:

$$-\int\frac{-\sin(2x)\,dx}{1+\cos^2(x)}$$

What does it become when we use our $u$-substitution?
 
topsquark said:
[math]f(x) = \sin^2 \left ( e^{\sin^2(x)} \right )[/math]
Do this with the chain rule. I'll separate the different derivatives with [ ] symbols.
[math]f'(x) = \left [ 2~sin \left ( e^{\sin^2(x)} \right )~cos \left ( e^{\sin^2(x)} \right ) \right ] \cdot \left [ e^{\sin^2(x)} \right ] \cdot \left [ 2~sin(x)~cos(x) \right ] [/math]

-Dan
 
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