im still stuck. i need help.
evaluate the integral
∫sinx /(1+x^2) dx from -1 to 1
The integrand is an odd function of x, therefore the integral will be zero.
As an exercise try to prove that if
f(-x) = -f(x) that the integral from
-r to r of f(x) dx is zero, e.g. by splitting it from -r to zero and from zero to r and then by substituting
x =-t in the first integral, that inttegral then becomes minus the second integral, so they cancel.
To evaluate the integral ∫(sinx / (1+x^2)) dx from -1 to 1, you can use the method of integration called "integration by parts". Here is how you can approach it:
First, let's define u and dv for integration by parts. In this case, u = sinx and dv = dx / (1 + x^2).
To find du, differentiate u with respect to x, which gives du = cosx dx.
To find v, integrate dv with respect to x, which gives v = arctan(x).
Now, we can apply the formula for integration by parts:
∫(u dv) = uv - ∫(v du)
Using the formula, we can rewrite the original integral as:
∫(sinx / (1+x^2)) dx = sinx * arctan(x) - ∫(arctan(x) cosx) dx
Next, we need to evaluate the integral ∫(arctan(x) cosx) dx. This integral requires a different integration technique called integration by substitution.
- Start by letting u = arctan(x). Then, differentiate u with respect to x to get du/dx = 1 / (1 + x^2).
- Rearrange this equation to get dx = (1 + x^2) du.
- Substitute these values into the integral, giving:
∫(arctan(x) cosx) dx = ∫(u * cosx * (1 + x^2) du)
Now, we have a new integral to evaluate:
∫(u * cosx * (1 + x^2) du
Expand the integrand:
= ∫(u * (cosx + x^2 * cosx)) du
Now, we can integrate term by term:
= ∫(u * cosx) du + ∫(u * x^2 * cosx) du
∫(u * cosx) du can be evaluated using integration by parts. Let's define u and dv for this integral.
Let v = sinx and du = du (the differential of u).
Then, dv = cosx dx and the integral becomes:
∫(u * cosx) du = u * sinx - ∫sinx du = u * sinx - ∫sinx du = u * sinx + ∫sinx du
We can rewrite the term as:
∫sinx du = -cosx + C (where C is the constant of integration)
Substituting back into the original integral, we have:
∫(u * cosx) du = u * sinx - cosx + C
Now, let's evaluate the second integral:
∫(u * x^2 * cosx) du
We can use integration by parts again. Let's define u and dv for this integral.
Let u = x^2 and dv = u * cosx du.
Then, du = 2x dx and v can be found by integrating dv:
v = ∫(u * cosx) du = x^2 * sinx - 2 ∫(x * sinx) dx
Using the integration by parts formula for the second integral:
∫(x * sinx) dx = -x * cosx + ∫cosx dx = -x * cosx + sinx + C
Putting it all together, we have:
∫(u * x^2 * cosx) du = x^2 * sinx - 2 * (-x * cosx + sinx + C)
= x^2 * sinx + 2x * cosx - 2sinx - 2C
Now, we can substitute back the values of u and v, and integrate the remaining term:
∫(sinx / (1+x^2)) dx = sinx * arctan(x) - (u * sinx + ∫sinx du) - x^2 * sinx - 2x * cosx + 2sinx + 2C
At this point, you can simplify the expression and then substitute the limits of integration (-1 and 1) to evaluate the definite integral. This will give you the final value of the integral.