Evaluate the integral.
∫5sec^(4)x dx
I got 5(secx+tanx)^(5)+C. is it right.
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
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
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
To evaluate the integral ∫5sec^4(x) dx, you can use the power reduction formula for secant. The power reduction formula states:
∫sec^n(x) dx = (1/(n-1)) * tan(x) sec^(n-2)(x) + (n-2)/(n-1) * ∫sec^(n-2)(x) dx
Using this formula, you can rewrite the integral as:
∫5sec^4(x) dx = 5 * (1/(4-1)) * tan(x) sec^2(x) + (4-2)/(4-1) * ∫5sec^2(x) dx
Simplifying further:
∫5sec^4(x) dx = (5/3) * tan(x) sec^2(x) + (2/3) ∫5sec^2(x) dx
Now, to evaluate the integral of sec^2(x), you can use the basic integral formula:
∫sec^2(x) dx = tan(x) + C
Therefore:
∫5sec^4(x) dx = (5/3) * tan(x) sec^2(x) + (2/3) (5tan(x) + C)
Simplifying this expression:
∫5sec^4(x) dx = (5/3) * tan(x) sec^2(x) + (10/3)tan(x) + C
So, your answer is not correct. The correct answer is:
∫5sec^4(x) dx = (5/3) * tan(x) sec^2(x) + (10/3)tan(x) + C