(Solved) Prove that ∫ sec ^2theta tan θ dθ =1/2 tan ^2theta +C_1=1/2 sec ^2theta
Question: Prove that
\[\int{{{\sec }^{2}}\theta \tan \theta }d\theta =\frac{1}{2}{{\tan }^{2}}\theta +{{C}_{1}}=\frac{1}{2}{{\sec }^{2}}\theta +{{C}_{2}}\]
Deliverable: Word Document 
Solution: Compute the area enclosed by the graph y=x(1-x^2)^1/2
![[Solved] Compute the area of the region](/images/solutions/MC-solution-library-35623.jpg "[Solved] Compute the area of the region R bounded above by the")
[Solved] Compute the area of the region R bounded above by the
![[See Steps] Compute the area of the](/images/solutions/MC-solution-library-35624.jpg "[See Steps] Compute the area of the region R between the graphs")
[See Steps] Compute the area of the region R between the graphs
(Steps Shown) Sketch the region bounded by the given curves and
![[Solution Library] The area between the graph](/images/solutions/MC-solution-library-35626.jpg "[Solution Library] The area between the graph of the positive")
[Solution Library] The area between the graph of the positive
Chi-Square Test for Goodness of Fit - MathCracker.com