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September 30, 2014

September 30, 2014

Posted by **Shelley** on Monday, January 29, 2007 at 1:23am.

1. integral from 0 to pi/4 of (tanx^2)(secx^4)dx

It says u = tan x to substitute

So if I use u = tan x, then my du = secx^2

then I have integral of (u^2)

then (1/3)u^3.....to get (1/3)tanx^3 then solve as 0 to pi/4 .......is that right?

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2. integral from 0 to 2root3 of (x^3)/(root(16-x^2))

I am pretty sure this has something to do with trig substitution. is it something with tan-1 or cos(theta)?

I am not sure how to go about this one at all.

Thanks again! =)

. integral from 0 to pi/4 of (tanx^2)(secx^4)dx

It says u = tan x to substitute

So if I use u = tan x, then my du = secx^2

then I have integral of (u^2)

INT u^2 sec^2x du

how are you going to get rid of the sec^2 x?

second: draw the imaginary triangle, 4 is the hypotenuse, with angle theta , x is the opposite side, the adjacent side is sqrt (16-x^2)

then (1/3)u^3.....to get (1/3)tanx^3 then solve as 0 to pi/4 .......is that right?

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