I need help calculating two integrals 1) ∫124+1xdx 2) ∫02πxnsin⁡(x)dx

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I need help calculating two integrals  1) ∫124+1xdx  2) ∫02πxnsin⁡(x)dx

lovagwb · Accepted Answer

For the first problem:  ∫4+1xdx=∫4x+1xdx  Let u=x, x=u2, 2udu=dx  ∫4u2+1u2udu  Try some trigonometric substitution.

Lindsey Gamble · Accepted Answer

For the second problem:  Set  I(n)=∫xnsin⁡(x)dx  Let u=xn, dv=sin⁡x, du=nxn−1, v=−cos⁡x by integration be parts.  I(n)=xn×cos⁡x+n∫xn−1cos⁡xdx  Let u=xn, dv=cos⁡x, du=(n−1)xn−2, v=sin⁡x by integration be parts.  I(n)=xn×cos⁡x+n(xn−1sin⁡x−(n−1)∫xn−2sin⁡xdx)=xn×cos⁡x+nxn−1sin⁡x−n(n−1)I(n−2)  Then use mathematic induction you will get the formula.

karton · Accepted Answer

You can indeed work out the second integral by parts. In:=∫0π/2xnsin⁡xdx=−xncos⁡x|0π/2+n∫0π/2xn−1cos⁡xdx Repeat with the cosine integral, Jn:=∫0π/2xncos⁡xdx=xnsin⁡x|0π/2−n∫0π/2xn−1sin⁡xdx This gives you the recurrence relations In=nJn−1 Jn=(π2)n−nIn−1 So that In=n((π2)n−1−(n−1)In−2) When you decrease n, you will eventually reach n=1 or n=0, you need to explictly compute I1 and I0. It is an easy matter to establish I0=J0=1 then I1=1

I need help calculating two integrals 1) \int_1^2\sqrt{4+\frac{1}{x}}dx 2) \int_0^{\frac{2}{\pi}}x^n\sin(x)dx

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