Posted by **Zooey** on Tuesday, September 27, 2011 at 7:24pm.

1. Find the indefinite integral.

Indefinite integral tan^3(pix/7)sec^2(pix/7)dx

2. Find the indefinite integral by making the substitution x=3tan(theta).

Indefinite integral x*sqrt(9+x^2)dx

3. Find the indefinite integral.

Indefinite integral cos(x)sin^3(x)dx

- Calc BC -
**Steve**, Wednesday, September 28, 2011 at 5:06am
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

----------------------

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

------

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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