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Extra resources for Analysis of the conservation law for matter and gravitational field introduced by Einstein

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3 x ⌠ 1 + 2 sec θ  3  ⌠ dx sec θ tan θ  dθ =   3 tan θ  2 ⌡ 2x − 4x − 7  2  ⌡ 1 3 sec θ + sec 2 θ dθ = ⌠  2 ⌡ 2 1 3 = ln sec θ + tan θ + tan θ + c 2 2 And here is the right triangle for this problem. aspx Calculus II Example 7 Evaluate the following integral. ∫e 4x 1 + e 2 x dx Solution This doesn’t look to be anything like the other problems in this section. However it is. To see this we first need to notice that, e2 x = ( e x ) 2 With this we can use the following substitution. = e x tan = θ e x dx sec 2 θ dθ Remember that to compute the differential all we do is differentiate both sides and then tack on dx or dθ onto the appropriate side.

1 A B = + ( x − 1)( x + 1) x − 1 x + 1 Setting numerator equal gives, 1= A ( x + 1) + B ( x − 1) Picking value of x gives us the following coefficients. aspx Calculus II Integrals Involving Roots In this section we’re going to look at an integration technique that can be useful for some integrals with roots in them. We’ve already seen some integrals with roots in them. Some can be done quickly with a simple Calculus I substitution and some can be done with trig substitutions. However, not all integrals with roots will allow us to use one of these methods.

It turns out that a trig substitution will work nicely on the second integral and it will be the same as we did when we had square roots in the problem. aspx Calculus II Okay, at this point we’ve got two options for the remaining integral. We can either use the ideas we learned in the section about integrals involving trig integrals or we could use the following formula. m = ∫ cos θ dθ 1 m −1 sin θ cos m −1 θ + cos m − 2 θ dθ m m ∫ Let’s use this formula to do the integral. 1 3 sin θ cos3 θ + ∫ cos 2 θ dθ 4 4 1 31 1  = sin θ cos3 θ +  sin θ cos θ + ∫ cos 0 θ dθ  4 42 2  1 3 3 = sin θ cos3 θ + sin θ cos θ + θ 4 8 8 4 = ∫ cos θ dθ cos 0 θ = 1!

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