1.∫(Sin⁴x/Cos²x).dx=?

2.∫{1+sin(x/2)}^1/2 dx=?

Please give elaborate answers for these at the earliest...

I = (sinx/2 + 1)^1/2 dx

(sinx/2+1) = (sinx/4)² + (cosx/4)² + 2cosx/4sinx/4   =  (sinx/4 + cosx/4)²

after putting this

I ={ sinx/4 + cosx/4 }dx

I= -4cosx/4 + 4sinx/4 + c

please approve my ans if u like it

dear chilukuri,

1.∫(Sin⁴x/Cos²x).dx

Possible intermediate steps:
integral sec(2 x) sin(4 x) dx
Simplify the integrand sin(4 x) sec(2 x) to get 4 sin(x) cos(x):
= integral 4 sin(x) cos(x) dx
Factor out constants:
= 4 integral sin(x) cos(x) dx
For the integrand sin(x) cos(x), substitute u = cos(x) and du = -sin(x) dx:
= -4 integral u du
The integral of u is u^2/2:
= -2 u^2+constant
Substitute back for u = cos(x):
= -2 cos^2(x)+constant

2.∫{1+sin(x/2)}^1/2 dx

(4*(-Cos[x/4] + Sin[x/4])*Sqrt[1 + Sin[x/2]])/ (Cos[x/4] + Sin[x/4])

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