1) find the integral from 1to -1 if(5sinx-2tanx+3x^5)dx
2) find the integral of x^3/(x^4+1)dx
∫[-1,1] 5sinx - 2tanx + 3x^5 dx
If you look carefully, all three terms are odd functions, so the whole integrand is odd. So, the integral from -1 to 1 is zero.
However, if you want to do the integration, note that since tanx = sinx/cosx = -d(cosx)/cosx, we have
-5cosx + 2log cosx + 1/2 x^6
for ∫x^3/(x^4+1) dx
let u = x^4+1, and so du = 4x^3 dx. So, your integral becomes
∫ 1/4 du/u
which should look familiar
To find the integral of a function, you can use the fundamental theorem of calculus and anti-derivatives. I'll explain the process for both of your questions:
1) Find the integral from 1 to -1 of (5sinx - 2tanx + 3x^5) dx:
To find the integral of this function, you'll need to evaluate the antiderivative of each term separately and then subtract the antiderivative evaluated at -1 from the antiderivative evaluated at 1.
Let's break down each term:
- The antiderivative of sinx is -cosx.
- The antiderivative of tanx is -ln|cosx|.
- The antiderivative of x^5 is (1/6)x^6.
Now, let's find the antiderivative of each term:
- The antiderivative of 5sinx is -5cosx.
- The antiderivative of 2tanx is -2ln|cosx|.
- The antiderivative of 3x^5 is (1/6)x^6.
Now, substitute the limits of integration into each term:
- Evaluate -5cosx at -1 and 1. You get -5cos(1) - (-5cos(-1)).
- Evaluate -2ln|cosx| at -1 and 1. You get -2ln|cos(1)| - (-2ln|cos(-1)|).
- Evaluate (1/6)x^6 at -1 and 1. You get (1/6)(1^6) - (1/6)(-1^6).
Simplify the results and subtract:
-5cos(1) + 5cos(-1) - 2ln|cos(1)| + 2ln|cos(-1)| + (1/6)(1^6) - (1/6)(-1^6)
This will give you the value of the integral from 1 to -1 of (5sinx - 2tanx + 3x^5) dx.
2) Find the integral of x^3/(x^4+1) dx:
To find the integral of this function, you can use a substitution method. Let u = x^4 + 1, then du = 4x^3 dx. Rearrange the equation to solve for dx.
dx = du/(4x^3)
Now substitute the values back into the integral:
∫(x^3/(x^4+1)) dx = ∫(1/(4u)) du
Now integrate ∫(1/(4u)) du:
= (1/4)ln|u| + C
Substitute back the value of u:
= (1/4)ln|x^4+1| + C
This will give you the value of the integral of x^3/(x^4+1) dx.