How do you integrate #int x^3/sqrt(64+x^2)# by trigonometric substitution?
4 Answers
#I = 64/3(x^2 + 64)^(3/2) - 64sqrt(x^2 + 64) + C#
Explanation:
Use the substitution
#I = int (8tantheta)^3/sqrt(64 + (8tantheta)^2) * 8sec^2theta d theta#
#I = int (512tan^3theta)/sqrt(64 + 64tan^2theta) * 8sec^2theta d theta#
#I = int(512tan^3theta)/sqrt(64(1 + tan^2theta)) * 8sec^2theta d theta#
Now use
#I = int(512tan^3theta)/sqrt(64sec^2theta) * 8sec^2theta d theta#
#I = int(512tan^3theta)/(8sectheta) * 8sec^2theta d theta#
#I = int512sec thetatan^3theta d theta#
#I =int512secthetatantheta(tan^2theta) d theta#
We now use
#I = int512secthetatantheta(sec^2theta - 1) d theta#
Now let
#I = int512secthetatantheta * (u^2- 1) (du)/(secthetatantheta)#
#I= int(512u^2 - 512)du#
#I = 512/3u^3 - 512u + C#
#I = 512/3sec^3theta - 512sectheta#
From our initial substitution, we know that
#I = 64/3(x^2 + 64)^(3/2) - 64sqrt(x^2 + 64) + C#
Hopefully this helps!
Explanation:
I know you specified trigonometric substitution but I don't see why you'd use on in this case as a more obvious one stands out to me, because we have an
Someone's already submitted the trigonometric substitution method so thought I may as well share.
#I = int (x^3)/(sqrt{64+x^2})"d"x# .
Let
Then,
#I = 1/2 int x^2/(sqrt{64+x^2}) "d"u# ,
#= 1/2 int u(64+u)^(-1/2) "d"u# .
At this point, I used integration by parts. Let
#I = 1/2 ( [2u(64+u)^(1/2)]-2int(64+u)^(1/2) "d"u)#
#I = u(64+u)^(1/2) - 2/3(64+u)^(3/2) + C#
Then, I factored out
#I = 1/3sqrt{64+x^2}(3x^2-2(64+x^2)) + C#
#= color(blue)(1/3sqrt{64+x^2}(x^2-128) + C)# .
Explanation:
Here is another way to solve the Problem, without using
substitution.
N.B.: The Final Integrals were obtained using the Rule,
Enjoy Maths.!
After using
After using
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