(int_{}^{}x^{2}dx/sqrt({25-9x^{2}})) Solve Th

vjca

From: US
Posts: 10
Votes: 10
(int_{}^{}x^{2}dx/sqrt({25-9x^{2}})) Solve this integral using the best trigonometric substution

 

Answer No.1

Anonymous

From: -
Posts: -
Votes: -
Put 9/25x^2=tan^theta and then solve 1/54 (25 sin^(-1)((3 x)/5)-3 x sqrt(25-9 x^2)) + c

Answer No.2

Anonymous

From: -
Posts: -
Votes: -
1/54 (25 sin^(-1)((3 x)/5)-3 x sqrt(25-9 x^2)) + c

Answer No.3

Anonymous

From: -
Posts: -
Votes: -
Make x=5/3 sin(theta) to make the integral sin^2(theta)/sqrt(25-25sin^2(theta)). =integral of sin^2(theta)/(5sqrt(1-sin^2(theta))) =integral of sin^2(theta)/(5cos(theta)) =integral of tan(theta)sin(theta)/5 Integrate by parts: u=tan(theta) dv=sin(theta)/5 du=sec^2(theta) v=-cos(theta)/5 =-tan(theta)cos(theta)/5 - integral of -sec^2(theta)cos(theta)/5 =(cos(theta)+cos(theta)tan^2(theta))/5 =-tan(theta)cos(theta)/5 - sin(theta)- integral of (cos(theta)tan^2(theta))/5 =(sin^2(theta)/5cos(theta) intsin^2(theta)/(5cos(theta))=(-tan(theta)cos(theta)/5)-sin(theta)-(int sin^2(theta)/5cos(theta) 2(intsin^2(theta)/5cos(theta))=(-tan(theta)cos(theta)/5)-sin(theta) int sin^2(theta)/5cos(theta)=((-tan(theta)cos(theta)/5)-sin(theta))/2

Your Comment/Solution

Post As: Anonymous [Change]
Enter