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dx/((sinx)^6 + (cosx)^6)

dx/((sinx)^6 + (cosx)^6)

Grade:

1 Answers

Mohit Chhabra
18 Points
6 years ago
dx/sinx^6+(1-sinx^2)^3
dx/sinx^6+1-sinx^6-3sinx^2+3sinx^4
dx/3sinx^4-3sinx^2+1
dx/-3sinx^2(1-sinx^2)+1
dx/-3sinx^2cosx^2+1
4dx/4-3sin2x
now sin2x=2tanx/1+tanx^2
secx^2 dx/tanx^2-1.5tanx+1
put tanx=t
secx^2dx = dt [eq 1]
using eq 1
dt/t-1.5t+1
add and subtract 9/16 in denominator
dt/t-1.5t+1+9/16-9/16
dt/{t-3/4}^2+7/16
now use dx/(x^2+a^2) formula
{4/7^(1/2)}arctan[t-3/4]/[7^(1/2)/4]
now put the value of t=tanx
pls suggest if the answer helpful..

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