How to solve this sum?
Attachments:
Answers
Answered by
2
∫[sin(x+π/6)÷sinx] /dx
∫ [sin x cos π/6+cosx sin π/6]/sin x dx
=∫cos π/6 + sin π/6 cotx dx
∫ √3/2+(1/2)cotx dx
[√3x/2 +(1/2)[log (sin x)] ] limits π/2 to π/4
[√3π/4+(1/2)[log(sin π/2)]-[√3π/8+(1/2)[log(sib π/4)]
[ √3π/4+(1/2)log(1)] - [√3π/8+(1/2)log(1/√2)]
√3/8+0-(1/2)log(1/√2)
∫ [sin x cos π/6+cosx sin π/6]/sin x dx
=∫cos π/6 + sin π/6 cotx dx
∫ √3/2+(1/2)cotx dx
[√3x/2 +(1/2)[log (sin x)] ] limits π/2 to π/4
[√3π/4+(1/2)[log(sin π/2)]-[√3π/8+(1/2)[log(sib π/4)]
[ √3π/4+(1/2)log(1)] - [√3π/8+(1/2)log(1/√2)]
√3/8+0-(1/2)log(1/√2)
kvnmurty:
There are mistakes in the answer. . Chk please
Answered by
1
It is simple if the numerator is expanded and each term is integrated separately.
Answer = sqrt3 × pi/8 + Ln sqrt2.
See the enclosed picture. .
Answer = sqrt3 × pi/8 + Ln sqrt2.
See the enclosed picture. .
Attachments:
:-)
Similar questions