Answers
Answered by
5
_____________________________
Attachments:
![](https://hi-static.z-dn.net/files/dab/63cab84a3e452b5881e6b67750d65538.jpg)
![](https://hi-static.z-dn.net/files/d97/1b59afdd6c2607c2b12b33b63b182d49.jpg)
Answered by
0
Answer:
Using the trigonometric identities
cos x = sin (x+π/2) and sin x + sin y = 2 sin((x+y)/2) cos((x-y)/2) gives
1/(sin x + cos x) = 1/(sin x + sin(x+π/2)) = 1/(2sin(x+π/4)cos(-π/4)) = csc(x+π/4)/√2
We can integrate using the formula ∫ csc x dx = ln|tan(x/2)| + C to get
∫ dx/(sinx+cos x) = (1/√2) * ∫ [csc(x+π/4)] dx = (1/√2) ln|tan(x/2 + π/8)| + C
Using the trigonometric identities
cos x = sin (x+π/2) and sin x + sin y = 2 sin((x+y)/2) cos((x-y)/2) gives
1/(sin x + cos x) = 1/(sin x + sin(x+π/2)) = 1/(2sin(x+π/4)cos(-π/4)) = csc(x+π/4)/√2
We can integrate using the formula ∫ csc x dx = ln|tan(x/2)| + C to get
∫ dx/(sinx+cos x) = (1/√2) * ∫ [csc(x+π/4)] dx = (1/√2) ln|tan(x/2 + π/8)| + C
Similar questions