Posted by Zooey on Tuesday, September 27, 2011 at 7:24pm.
recall that tan' = sec^2
If we let v = pi/7 x, then dv = pi/7 dx
and we have Int[tan^3(v) sec^2(v)] 7/pi dv
Now, let u = tan(v). du = sec^2(v) dv Our integral now becomes
7/pi Int[u^3 du] = 7/pi * 1/4 u^4 + C
= 7/4pi tan^4(pi/7 x) + C
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If x = 3tan(u) then 9+x^2 = 9 + 9tan^2(u) = 9 sec^2(u)
dx = 3sec^2(u) du
The integral now becomes
Int[3tan(u) * 3sec*u)* 3sec^2(u)] du = 9Int[sec(u) tan(u)*sec^2(u)]
Now, if v = sec(u), dv = sec(u) tan(u) du, and the integral now becomes
9 Int[v^2 dv] = 1/3 v^3 = 1/33 sec^3(u) = 1/3 sqrt(9+x^2)^(3/2) + C
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This could also have been obtained by letting u=9+x^2, so du = 2xdx, making the integral
Int[u^1/2 * du/2] = 1/3(9+x^2)^(3/2) + C
--------------------------------
Let u = sin(x), so du = cos(x) dx and the integral becomes
Int[u^3 du] = 1/4 u^4 = 1/4 sin^4(x) + C
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