INTEGRATION BY SUBSTITUTION WORKSHEET

Integrate each of the following with respect to x :

Question 1 :

Question 2 :

x/√1 + x²

Question 3 :

x²/1 + x⁶ 

Question 4 :

(e^x - e^-x)/(e^x + e^-x)

Question 5 :

(10x⁹ + 10^x log_e10)/(10^x + x¹⁰)

Question 6 :

(sin √x)/√x

Answers

1. Answer :

In the function, the power of 'e' is x³. The derivative is x³ is 3x² and it is being as a part of the function and we will be able to write the 3x² along with dx using multiplication.

Now substitute a new variable for x³ for which we have the derivative 3x² being a part of the given function.

Let u = x³.

u = x³

Differentiate with respect to x.

du/dx = 3x²

Multiply both sides by dx.

du = 3x²dx

Substitute x³ = u and 3x²dx = du in (1)

= ∫e^udu

= e^u + C

Substitute u = x³.

2. Answer :

= ∫(x/√1 + x²)dx

= ∫(xdx)/√1 + x² ----(1)

Let t = 1 + x².

t = 1 + x²

Differentiate with respect to x.

dt/dx = 0 + 2x

dt/dx = 2x

Multiply both sides by dx.

dt = 2xdx

Divide both sides by 2.

dt/2 = xdx

Substitute 1 + x² = t and xdx = dt/2 in (1).

∫(xdx)/√1 + x²  = ∫(dt/2)/√t

= (1/2)∫dt/√t

= (1/2)∫dt/t^-1/2

  = (1/2)∫t^-1/2dt

= (1/2)t^{-1/2 + 1}/(-1/2 + 1) + C

= (1/2)t^1/2/(1/2) + C

= (1/2)t^1/2(2/1) + C

= t^1/2 + C

= √t + C

Substitute t = x² + 1.

=  √(1 + x²) + C

3. Answer :

= ∫[x²/(1 + x⁶)] dx

=  ∫(x²dx)/[1 + (x³)²] ----(1)

Let t = x³.

t = x³

Differentiate with respect to x.

dt/dx = 3x²

Multiply both sides by dx.

dt = 3x²dx

Divide both sides by 3.

dt/3 = x²dx

Substitute x³ = t and x²dx = dt/3 in (1)

∫(x²dx)/(1 + (x³)² = ∫(dt/3)/(1 + t²)

= (1/3)∫1/(1 + t²)dt

= (1/3)tan^-1t + c 

Substitute t = x³.

= (1/3)tan^-1(x³) + c

4. Answer :

= ∫[(e^x - e^-x)/(e^x + e^-x)]dx

= ∫(e^x - e^-x)dx/(e^x + e^-x) ----(1)

Let t = e^x + e^-x.

t = e^x + e^-x

Differentiate with respect to x.

dt/dx = e^x + e^-x(-1)

dt/dx = e^x - e^-x

Multiply both sides by dx.

dt = (e^x - e^-x)dx

Substitute e^x + e^-x = t and (e^x - e^-x)dx = dt in (1).

∫(e^x - e^-x)dx/(e^x + e^-x) = ∫dt/t

= ∫(1/t)dt

= lnt

Substitute t = e^x + e^-x.

= ln(e^x + e^-x) + c

5. Answer :

= ∫[(10x⁹ + 10^xln10)/(10^x + x¹⁰)]dx

= ∫(10x⁹ + 10^xln10)dx/(10^x + x¹⁰) ----(1)

Let u = 10^x + x¹⁰.

u = 10^x + x¹⁰

Differentiate with respect to x.

du/dx = 10^xln10 + 10x⁹

Multiply both sides by dx.

du = (10^xln10 + 10x⁹)dx

du = (10x⁹ + 10^xln10)dx

Substitute 10^x + x¹⁰ = u and (10x⁹ + 10^xln10)dx in (1).

= ∫du/u

= ∫(1/u)du

= lnu + c

Substitute u = 10^x + x¹⁰.

= ln(10^x + x¹⁰) + c

6. Answer :

= ∫[(sin√x)/√x]dx

= ∫(sin√x)(dx/√x) ----(1) 

Let u = √x.

u = √x

u = x^1/2

du/dx = (1/2)x^{1/2 - 1}

du/dx = (1/2)x^-1/2

du/dx = 1/(2x^1/2)

du/dx = 1/(2√x)

Multiply both sides by 2dx.

2du = dx/√x

Substitute √x = u and dx/√x = 2du in (1).

∫(sin√x)(dx/√x) = ∫(sinu)(2du)

= 2∫sinudu

= 2(-cosu) + c

= -2cosu + c

Substitute u = √x.

= -2cos√x + c

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