Obtain the antiderivatives of tge following indefinite integrals, given the additional conditions.?

Question

(a) y=∫x^3 (3x^2-3)dx, where y(1)=-1
(b) y=∫3csc^2 x-5secxtanxdx, where y(pi/4)=1.

Answer 1

(a) 

y(x) = ∫x³ (3x²-3)dx

y(x) = ∫(3x^5 - 3x³) dx

y(x) =3 ∫(x^5 - x³) dx

y(x) =3 [ (x^6)/6 - (x^4)/4)  +c]               [ ∫xⁿ =  (xⁿ⁺¹)/n+1 ] 

Given an initial condition y(1) = -1 .

y(1) =3 [ (1^6)/6 - (1^4)/4)  +c] 

-1 = 3 [ 1/6 - 1/4  +c] 

-1 = 3 [ -1/12 +c] 

-1/12 +c = -1/3 

c =  -1/3  +1/12

c = -3/4 .

y(x) =3 [ (x^6)/6 - (x^4)/4)  -3/4 ]

So y(x) =3 [ (x^6)/6 - (x^4)/4)  -3/4 ] .

Answer 2

(b) 

 y(x) = ∫(3csc² x - 5secx tanx ) dx 

 y(x) = 3∫csc² x dx  - 5 ∫secx tanx dx 

 y(x) = -3cot x  - 5 secx + c                    [  ∫csc² x dx = -cot x  and ∫secx tanx dx = sec x + c ]

Given an initial condition y(π/4) = 1 .

y(π/4) = -3cot π/4  - 5 secπ/4 + c   

1 = -3(1)  - 5√2  + c   

1 = -3  - 5√2  + c  

c = 4 + 5√2 

So  y(x) = -3cot x  - 5 secx + 4 + 5√2