Evaluate the following integrals using substitution

1. ʃxsin(x²)dx

2. ʃsin(π)t dt

3. (cos√t) /√t dt

4. ʃ (a+bx²)/√(3ax+bx³)dx
u = 3ax+bx³

Hello,

∫ x sin(x²) dx =

rewrite it as:

∫ sin(x²) x dx =

let:

x² = u

differentiate both sides:

d(x²) = du

2x dx = du

x dx = (1/2) du

then, substituting:

∫ sin(x²) x dx = ∫ sin u (1/2) du =

(pulling the constant out)

(1/2) ∫ sin u du =

(1/2)(- cos u) + C =

- (1/2)cos u + C

substitute back for u, ending with:

∫ x sin(x²) dx = - (1/2)cos(x²) + C

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I'm not sure whether you mean ∫ [sin(π)] t dt or rather ∫ sin(πt) dt; let's see both cases:

∫ [sin(π)] t dt =

being sin(π) a mere constant, pull it out:

sin(π) ∫ t dt =

sin(π) [1/(1+1)] t^(1+1) + C =

{[sin(π)]/2} t² + C

---------------------------------

∫ sin(πt) dt =

let:

πt = u

t = (1/π)u

dt = (1/π) du

then, substituting:

∫ sin(πt) dt = ∫ sin u (1/π) du =

(pulling the constant out)

(1/π) ∫ sin u du =

(1/π)(- cos u) + C =

- (1/π)cos u + C

being u = πt, the answer is:

∫ sin(πt) dt = - (1/π)cos(πt) + C

============================

∫ [cos(√t) /√t] dt =

rewrite it as:

∫ cos(√t) (1/√t) dt =

let:

√t = u

differentiate both sides:

d[t^(1/2)] = du

(1/2) t^[(1/2) -1] dt = du

(1/2)t^(-1/2) dt = du

(1/2)(1/√t) dt = du

(1/√t) dt = 2 du

then, substituting:

∫ cos(√t) (1/√t) dt = ∫ cos u 2 du =

2 ∫ cos u du =

2sin u + C

being u = √t, the answer is:

∫ [cos(√t) /√t] dt = 2sin(√t) + C


=============================

∫ [(a + bx²) /√(3ax + bx³)] dx =

rewrite the integral as:

∫ (3ax + bx³)^(-1/2) (a + bx²) dx =

as suggested, let:

3ax + bx³ = u

differentiate both sides:

d(3ax + bx³) = du

(3a + 3bx²) dx = du

(factoring out 3)

3 (a + bx²) dx = du

(a + bx²) dx = (1/3) du

then, substituting:

∫ (3ax + bx³)^(-1/2) (a + bx²) dx = ∫ u^(-1/2) (1/3) du =

(1/3) ∫ u^(-1/2) du =

(1/3) {1/[(-1/2)+1]} u^[(-1/2)+1] + C =

(1/3)[1/(1/2)]u^(1/2) + C =

(1/3)2√u + C =

(2/3)√u + C

substitute back 3ax + bx³ for u, ending with:

∫ [(a + bx²) /√(3ax + bx³)] dx = (2/3)√(3ax + bx³) + C



I hope this helps..