For trigonometric substitution to solve the above integral, fill in the blanks below using the picture of the triangle given.

Question

I am given the integral ∫(3x²)/sqrt(49+4x²) dx. I am also given a right triangle, with side B going along the horizon, side A is attached to the right end of side B, and it's perpendicular to side B, and side C is the hypotenuse. Between B and C is angle Q. Using the integral I am to fill in the following blanks.

side A=2x

side B =

side C=

(2x/7) =

(2/7) dx = dQ

(sqrt(49+4x²))/7 =

Then I have to evaluate the integral.

Answer 1

I = ∫3x²/√(49+4x²)

let x = (7/2)tan(u)

If x = (7/2) tan(u) then tan(u) = 2x/7 Which makes A=2x, B= 7 and C=√(49+4x²)

dx = (7/2)sec²(u) du

√(49+4x²) = 7√(1+tan²(u)) = 7sec(u)

3x²=(3/4)49tan²(u)

(3/4)49∫tan²(u)(7/2)sec²(u)/(7sec(u))du = (147/8)∫tan²(u)sec(u)du

Hold onto 147/8 for a while

∫tan²(u)sec(u)du = ∫(sec²(u)-1)sec(u)du = ∫sec³(u)du - ∫sec(u)du

∫sec(u)du = ln(tan(u)+sec(u)) Remember this

∫sec³(u)du = sec(u)tan(u)/2 + ln(tan(u)+sec(u))/2

putting both together

sec(u)tan(u)/2 - ln(tan(u)+sec(u))/2

Using the triangle to put it back into terms of x

sec(u) = (√(49+4x²)/7

tan(u) = 2x/7

I = (147/8)[ (√(49+4x²))/7)(2x/7)/2 - (ln(2x/7 + (√(49+4x²))/7))/2] + C

Which can be simplified

side A=2x

side B = 7

side C= √(49+4x²)

(2x/7) = tan(Q)

(2/7) dx =sec²(Q) dQ

(√(49+4x²))/7 = sec(Q)