Help me find integrals of the following?

Question

(1+cosecx)^1/2
1/(sinx)^4+(cosx)^4
1/(sinx)^6+(cosx)^6

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colin, the method u r sayin is true for
1/(sinx)^4 + 1/(cosx)^4 and not for
1/(sinx)^4 + (cosx)^4

actually it is 1/[(sinx)^4 + (cosx)^4]

Answer 1

These integrals are all elementary but plenty tough.
Let me get you started on the second one. If I have
time, I'll come back and finish it. I haven't been able
to crack the others yet.
Write the integrand as
1/(sin^ 4 x + cos^ 4 x).
Complete the square in the denominator to
rewrite this as
1/[sin^ 4 x + 2cos^ 2 x sin^ 2 x + cos^ 4 x - 2 sin^ 2 x
cos^ 2 x]
= 1/ [(sin^ 2 x + cos^ 2 x)^ 2 - 2 sin^ 2 x cos^ 2 x]
= 1/[ 1- 1/2*(sin^ 2 2x}]
Now you can let u = 2x, x = u/2 dx = 1/2* du
To get
1/2*Int[ du/ 1 - 0.5 sin^ 2 u]
Now use the standard half angle substitution
z = tan(u/2) to evaluate the resulting integral.

Answer 2

Sin integrates to -cos
cos integrates to sin
cosec integrates to sec

1/(Sinx)^4 = (Sinx)^-4

same rule applies to cos

Standard integration applies (Raise power by 1, divide by new power)

(1 + cosec x )^1/2
(sinx)^-4 + (cosx)^4
(sinx)^-6 + (cosx)^6

CLUE: Int (Cosx)^-8 = ((Sinx)^-7)/7 + c

CLUE 2: Substitution in Q1


While we are on it, which idiot decided to make COsec 1/sin and sec 1/COs?