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Evaluate : int0^(2pi)e^xcos(pi/4+x/2)dx...

Evaluate : `int_0^(2pi)e^xcos(pi/4+x/2)dx`

Text Solution

Verified by Experts

Let,`( x/2​+π/4​=θ)`
` x=2θ−π/2` and `dx=2dθ`
Thus, `2int_(pi/4)^(5pi/4)e^(2theta-pi/2)cos(theta)d theta`
=`2e^(-pi/2)I_1`
Where`I_1`=`2int_(pi/4)^(5pi/4)e^(2theta)cos(theta)d theta`
=`1​/4e^(2θ)[(2cosθ+sinθ)]_(π/4)^(5π/4)-(1/4​)I_1`
or `5I_1/4=1/4[e^(5pi/2)​(-2 ...
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