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Question Number 54972 by peter frank last updated on 15/Feb/19
∫0π∫04cosz∫016−y2ydxdydz
Answered by kaivan.ahmadi last updated on 15/Feb/19
∫0π∫04coszyx]016−y2dydz=∫0π∫04cosz(y16−y2)dydz=∫0π−13(16−y2)32]04coszdz=−13∫0π((16−16cos2z)32−1632)dz=−16323∫0π(sin3z−1)dz=−16323×(sin2zcosz+2cosz3−z)0π=16323×[(−23−π)−(23)]=−16323×(−43−π)=2569+643π
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