December 3rd, 2007, 11:47 AM  #1 
Newbie Joined: Dec 2007 From: phoenix Posts: 2 Thanks: 0  Integral Problem
So ive been trying to tackle this problem for a while now and im sure there is something very simple that im missing. Part A. Evaluate the integral by multiplying the numerator by an appropriate expression. a. S(1/1+sin(x))dx than part b is use the double angle formula to evaluate the integral b. S(1/1+cos2x)dx Thanks alot ahead of time. p.s. if there is any confusion on the S its suppose to be the stretched out s not a variable. 
December 3rd, 2007, 01:51 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,438 Thanks: 562 
You need parentheses to calarify the expressions to be integrated. If the trig function is supposed to be part of the denominator, put () around the whole denominator.

December 3rd, 2007, 07:57 PM  #3 
Newbie Joined: Dec 2007 From: phoenix Posts: 2 Thanks: 0 
a. S((1)/(1+sin(x)))dx b. S((1)/(1+cos2x))dx 
December 4th, 2007, 10:18 AM  #4 
Senior Member Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus 
a. Multiply the numerator and denominator by (1sin(x)). The denominator becomes 1  sin(x)^2, which is cos(x)^2. Then you can expand the fraction to become 1/cos(x)^2  sin(x)/cos(x)^2. Now this is easy to integrate. b. Substitute the double angle formula for cos(2x), being 2cos(x)^2  1. You're left with 1/(2cos(x)^2), which should be easy. 

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