Ask a New Question
Menu

Integration of (sinx + cosx)^2

from 0 to pi/2.
I get pi/2 + 1 as my final answer... wondering if that's correct?

1 answer

Hmmm. Isn't (sin + cos)^2=1+2cosxsinx ?

int(1+2cosx sinx)dx= x +sin^2 x

over the limit= PI/2+1 -0-0= wow, your answer.
Similar Questions
  1. Simplify #3:[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
    1. answers icon 1 answer
  2. tanx+secx=2cosx(sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 =
    1. answers icon 0 answers
  3. I need to prove that the following is true. Thanks(cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds
    1. answers icon 0 answers
  4. prove the identity(sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2 sinX^6= sinx^2 ^3 = (1-cosX^2)^3 = (1-2CosX^2 + cos^4) (1-cosX^2)
    1. answers icon 0 answers
more similar questions