∫▒dx/(x√(2x^2+5x+1)) 2. ∫▒dx/(x√(x^2+4x-2)) 3. ∫▒dx/(sinx+13cosx-11)

∫▒dx/(x√(2x^2+5x+1)) 2. ∫▒dx/(x√(x^2+4x-2)) 3. ∫▒dx/(sinx+13cosx-11) 4. ∫▒〖(x^2-x+4)ln⁡(x+√(x^2-1)
1. 1 answer
2. asked by washington
3. 1,306 views
Prove the following:[1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)]
1. 3 answers
2. asked by Anonymous
3. 945 views
Write the equation of the periodic function based on the graph. (1 point) Responses y = −3cosx y = −3cosx y = 3cosx y =
1. 1 answer
2. 102 views
Mathematics - Trigonometric Identities - Reiny, Friday, November 9, 2007 at 10:30pm(sinx - 1 -cos^2x) (sinx + 1 - cos^2x) should
1. 3 answers
2. asked by Anonymous
3. 784 views
Write the equation of the periodic function based on the graph.(1 point) Responses y = −3cosx y = −3cosx y = 3cosx y = 3cosx
1. 1 answer
2. 83 views
Simplify sin x cos^2x-sinxHere's my book's explanation which I don't totally follow sin x cos^2x-sinx=sinx(cos^2x-1)
1. 1 answer
2. asked by Tara
3. 1,054 views
Hello! Can someone please check and see if I did this right? Thanks! :)Directions: Find the exact solutions of the equation in
1. 1 answer
2. asked by Maggie
3. 848 views
I need help solving for all solutions for this problem:cos 2x+ sin x= 0 I substituted cos 2x for cos^2x-sin^2x So it became
1. 1 answer
2. asked by Martha
3. 635 views
tanx+secx=2cosx(sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 =
1. 0 answers
2. asked by shan
3. 984 views
the problem is2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is
1. 3 answers
2. asked by alex
3. 869 views