Another simple trigonometric substitution gone wrong

  • #1

Homework Statement



[tex]\int_0^1 \! 7x\sqrt{x^2+4} dx[/tex]


Homework Equations





The Attempt at a Solution



Noticing that the radical is of the form [tex]x^2 + a^2[/tex], I know to use [tex]a*tan\theta[/tex].

[tex]x = 2tan\theta[/tex]

[tex]dx = 2sec^2\theta d\theta[/tex]​


Then I simplified the radical to put it in terms of a trig function.

[tex]\sqrt{x^2+4}[/tex]

[tex]\sqrt{4tan^2\theta+4}[/tex]

I know that this should actually be + 1 instead of - 1, but since it's something squared, I assume it will be positive in the end.

[tex]\sqrt{4(tan^2\theta-1}[/tex]

[tex]2sec\theta[/tex]​


Putting this information back into my original integral and pulling out the 7...

[tex]7 \int 2tan\theta*2sec\theta*2sec^2\theta d\theta dx[/tex]​


Pulling out all of those 2's...

[tex]56 \int tan\theta*sec\theta*sec^2\theta d\theta[/tex]​


In trying to come up with a way to proceed, I figured I'd try a u-substitution. If I substitute [tex]u = sec\theta[/tex], then [tex]du = sec\theta*tan\theta d\theta[/tex] which would take care of most of the integral.

[tex]56 \int u^2 du[/tex]

[tex]56 \frac{u^3}{3}[/tex]

[tex]\frac{56sec^3\theta}{3} + C[/tex]​

I'm a little bit at a loss about how to continue from here, and it makes me feel like I did something wrong in the steps above. I now have an expression in terms of theta instead of x. I know that sec = 1/cos, and I assume sec^3 = 1/cos^3, but that cube is throwing me off. Any hints would be great, thanks.
 
Last edited:

Answers and Replies

  • #2
34,976
6,729
Your trig substitution will work, but it's not the best way for this problem. I'll take a closer look in another post, but in the meantime, the best approach for this integral is an ordinary substitution.

u = x2 + 4, du = 2x dx
 

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