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What's the integral of #int tanx(secx)^(3/2) dx#?

Answer 1

Charles Arocha

#inttanx(secx)^(3/2)dx = 2/3(secx)^(3/2) + C#

For this integral, we will use #u# substitution. Let #u = secx# then #du = secxtanxdx#

So we have #inttanx(secx)^(3/2)dx = int(secx)^(1/2)*secxtanxdx=intu^(1/2)du#

Applying the formula #intx^ndx = x^(n+1)/(n+1) + C# we get

#intu^(1/2)du = u^(3/2)/(3/2)+C = 2/3u^(3/2) + C#

Finally, we substitute back in for #u#:

#2/3u^(3/2) + C = 2/3(secx)^(3/2) + C#

Thus

#inttanx(secx)^(3/2)dx = 2/3(secx)^(3/2) + C#

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Answer 2

William Abdalla

The integral of ( \tan(x) \left(\sec(x)\right)^{\frac{3}{2}} ) with respect to ( x ) is ( \frac{2}{\sqrt{\sec(x)}} + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.