Question: solve: integral of (sqrt((1/4t)+(1/9sqrt(t))) from 1 to 16.
The expression is: √(1/4 t+1/9∙√t)
Excluding the limits of integration, for the moment,
I=∫√(1/4 t+1/9∙√t) dt
I=1/6 ∫√(9t+4∙√t) dt
Let u=√t, then u^2=t and 2u du=dt
I=1/6 ∫√(9u^2+4u)∙2u du
Complete the square
Multiply the quadratic expression by 4 times the coefft of the u^2-term
I=1/36 ∫√(324u^2+144u)∙2u du
I=1/36 ∫√(324u^2+144u+16-16)∙2u du
I=1/36 ∫√((18u+4)^2-16)∙2u du
Let 4w=18u+4, then w=9/2 u+1, u=2/9(w-1), and 2 dw=9 du. This gives us,
I=1/36 ∫√((4w)^2-16)∙2u∙2dw/9
I=4/9^2 ∫√(w^2-1)∙u dw
I=8/9^3 ∫√(w^2-1)∙(w-1) dw
I=8/9^3 ∫w√(w^2-1) dw - 8/9^3 ∫√(w^2-1) dw
I=8/9^3*I_1 - 8/9^3*I_2
I_1=∫w(w^2-1)^(1/2) dw = 1/3∙(w^2-1)^(3/2)
I_1=1/3∙(w^2-1)^(3/2)
I_2=∫√(w^2-1) dw=∫(w^2-1)/√(w^2-1) dw=∫w^2/√(w^2-1) dw - ∫1/√(w^2-1) dw
i.e. I_2=I_3 + I_4
I_3=∫w^2 (w^2-1)^(-1/2) dw=w∙(w^2-1)^(1/2) - ∫√(w^2-1) dw
I_3=w∙√(w^2-1) - I_2
I_4=∫1/√(w^2-1) dw
Let v^2=w^2-1, then w^2=1+v^2 and w dw=v dv
I_4=∫1/v (v dv)/w = ∫dv/w = ∫dv/√(1+v^2 )
I_4=asinh(v)
I_4=asinh(√(w^2-1))
I_2=I_3 - I_4
=w√(w^2-1) - I_2 - asinh(√(w^2-1))
2I_2=w√(w^2-1) - asinh(√(w^2-1))
I_2=1/2 {w√(w^2-1) - asinh(√(w^2-1)) }
I=8/9^3 {I_1 - I_2}
I=8/9^3 {1/3∙(w^2-1)^(3/2) - 1/2 {w√(w^2-1) - asinh(√(w^2-1)) } }
I=8/9^3 {√(w^2-1) [1/3 (w^2-1) - 1/2 w]+1/2*asinh(√(w^2-1)) }
Where w=1/2(9u+2), and u=√t.
From w=1/2(9u+2), we can get w^2-1= 9/4 (9t+4√t) and √(w^2-1)=3/2 √(9t+4√t)
Substitution gives us,
I=8/9^3 {3/2 √(9t+4√t) [3/4 (9t+4√t) - 1/2∙1/2 (9√t+2) ] + 1/2* asinh(3/2√(9t+4√t) ) }
I=8/9^3 {3/8√(9t+4√t) [3(9t+4√t) - (9√t+2) ]+1/2*asinh(3/2√(9t+4√t) ) }
I=8/9^3 {3/8√(9t+4√t) [27t+3√t-2] + 1/2* asinh(3/2√(9t+4√t) ) }
Evaluating this between the limits of t=1 to t=16, we get,
[t=1]: l1 = 0.4285657238
[t=16]: l2 = 23.02780286
Value = l2 – l1 = 23.02780286 – 0.4285657238 = 22.5992371362
Answer: 22.6 (approx)