Prove that ∫{{{ sec }^2}theta tan θ }dθ =1/2{{ tan }^2}theta +{C_1}=1/2{{ se
Question: Prove that
\[\int{{{\sec }^{2}}\theta \tan \theta }d\theta =\frac{1}{2}{{\tan }^{2}}\theta +{{C}_{1}}=\frac{1}{2}{{\sec }^{2}}\theta +{{C}_{2}}\]
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Solution: The solution consists of 1 page
Deliverables: Word Document
Deliverables: Word Document
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