∫[-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
1) find the integral from 1to -1 if(5sinx-2tanx+3x^5)dx
2) find the integral of x^3/(x^4+1)dx
2 answers
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
let u = x^4+1, and so du = 4x^3 dx. So, your integral becomes
∫ 1/4 du/u
which should look familiar
3 answers