1x3 25 1x2 dx For trigonometric substitution to solve the a

1x3/ 25 + 1x2 dx For trigonometric substitution to solve the above integral, fill in the blanks below using the picture of the triangle given. side A = 1 x side B = side C = [(1*)/5] = [1/5] dx = [( {25 + 1 x2})/5] =

Solution

x=3/7sint dx=3/7cost integral(3/7)^2*(sint)^2*(3/7cost)dt*3cost if A=7x B=sqrt(9-49x^2) C=3 7x/3=sinQ 7/3dx=cosQdQ sqrt(9-49x^2)/3=cosQ