Calculus Help?????????????

Question

Help with any or all of these problems would be gratefully received and incredibly appreciated. 

1. ∫ x cos (3x^2) dx 

2. ∫ (x dx) / (1 + 4x^2)

3. ∫ (dx) / (1 + 4x^2)

Answer 1

1.The expression is .

let .

Dervative  with respect to x.

.

The above values are substitute in .

Apply formula :.

Substitute the value of .

1.The expression is .

Answer 2

2) The expression  is  ∫ (x dx) / (1 + 4x²).

 Let  x² = t

Dervative on each side with respect to x.

2x dx  = dt => x dx  = dt /2

The above values are substitute in  ∫ (x dx) / (1 + 4x²)

= ∫ (dt/2) / (1 + 4t)

 = ∫ 1/ 2(1+4t) dt

Dervative on each side with respect to t.

= 1/2 ∫ 1/(1+4t) dt

= 1/2 log (1+4t).d/dt (1+4t)

= 1/2 log (1+4t).4(1)

= 2 log (1+4x²)

Comments

The expression is ∫ (x dx) / (1 + 4x²).

Let 1 + 4x² = t ⇒ 8x dx = dt ⇒ x dx = (1/8)dt.

= ∫ (1/8)dt / t

= (1/8) ∫ dt / t

= (1/8) log(t) + c.

= (1/8) log(1 + 4x²) + c.

Answer 3

3) The expression is  ∫ (dx) / (1 + 4x²).

=∫ (dx) / (1 +( 2x)²)

Apply  formula  ∫ 1/x²+ a² dx  = 1/a Tan^-1 (x/a)

=∫ 1/ 1²+(2x)²dx   =  1/a Tan^-1(1/2x)

 

 

Comments

The expression is ∫ dx / (1 + 4x²).

= ∫ dx / [ 1 + (2x)² ]

Let 2x = t ⇒ 2 dx = dt ⇒ dx = (1/2)dt.

= ∫ (1/2)dt / (1 + t²)

= (1/2) ∫ dt / (1 + t²)

= (1/2) tan^{- 1}(t) + c.

= (1/2) tan^{- 1}(2x) + C.