Question

Evaluate the integral.

∫5sec^(4)x dx

I got 5(secx+tanx)^(5)+C. is it right.

Answer 1

You can always check an integral by differentiating your result.
If I do that to your answer I get:
25(secx + tanx)^4 ( secxtanx + sec^2 x)

I doubt very much if that can be reduced to a simple
5(secx)^4

Wolfram answer can be reduced to :
(5/3)tanx (2 + sec^2 x)

http://integrals.wolfram.com/index.jsp?expr=5%28sec%28x%29%5E4&random=false

Answer 2

Hmmm. Tony, your reasoning is wrong on a couple of levels.

If you make the substitution

u = sec x, you'd have (sort of)

∫5u^4 = u^5

and that would have been sec^5(x). But you forgot about dx. It is an "integral" part of the problem :-)

Also, you reasoned that since
∫secx dx = secx+tanx, what you had seemed logical. But that would have been like saying

∫5u^4 du = (∫u)^5

Nope.

If you try the substitution

u = sec(x) then you also have to remember the dx:
du = secx tanx dx, so
dx = du/(secx tanx)
= du/(u√(u^2-1))
and your integral then becomes

∫5u^/√(u^2-1) du

No joy there, is there?

Visit Reiny's link and click to see the step-by-step solution

Answer 3

Hmmm. I see Reiny's link doesn't go to wolframalpha.com, so the first steps are

∫sec^4(x) dx
= ∫sec^2(x) (tan^2(x)+1)
= ∫tan^2(x) sec^2(x) + sec^2(x) dx
Now just substitute
u = tan(x)
du = sec^2(x) dx
and you have
∫(u^2 + 1) du
and the rest is easy