The value of _- ^ 0.05em 2y (1+ sin ⁡y )1+ cos^2 ⁡y dy is:

Question

Accepted Answer

Correct answer: D. (d) _- ^ 0.05em 2y (1+ sin ⁡y )1+ cos^2 ⁡y dy = _- ^ 0.05em 2y1+ cos^2 ⁡y dy+ _- ^ 0.05em 2y sin ⁡y1+ cos^2 ⁡y dy ( Odd ) ( Even ) =0+2.2 _0^ 0.05emy ( sin ⁡y1+ cos^2 ⁡y ) dyI=4 _0^ 0.05em y sin ⁡y1+ cos^2 ⁡y dyI=4 _0^ 0.05em ( -y ) sin ⁡y1+ cos^2 ⁡y dy2I=4 _0^ 0.05em sin ⁡y1+ cos^2 ⁡y dyI=2 _0^ 0.05em sin ⁡y1+ cos^2 ⁡y dy =2 (- tan^-1 ⁡ ( cos ⁡y ))_0^ =-2 ((- 4)- ( 4 )) =-2 (- 2 4)= ^2

$The value of \int_{- π}^{π} \frac{2 y ( 1 + sin y )}{1 + cos ^{2} y} dy is:$

Solution

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