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```        1.∫(Sin4x/Cos2x).dx=?
2.∫{1+sin(x/2)}1/2 dx=?
Please give elaborate answers for these at the earliest...```
8 years ago

## Answers : (2)

```							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
```
8 years ago
```							dear chilukuri,
1.∫(Sin4x/Cos2x).dx
Possible intermediate steps:integral sec(2 x) sin(4 x) dxSimplify the integrand sin(4 x) sec(2 x) to get 4 sin(x) cos(x):= integral 4 sin(x) cos(x) dxFactor out constants:= 4 integral sin(x) cos(x) dxFor the integrand sin(x) cos(x), substitute u = cos(x) and du = -sin(x) dx:= -4 integral u duThe integral of u is u^2/2:= -2 u^2+constantSubstitute 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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8 years ago
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