Calculus > U Substitution

**Contents:**

- Overview and Basic Example
- U Substitution for Trigonometric Functions
- U Substitution for Definite Integrals
- U Substitution for Exponential Functions

## 1. Overview and Basic Example

Integration by substitution is one of the first techniques you use in integral calculus. All the technique is doing is taking a rather complicated integral and turning it — using algebra — into integrals you can recognize and easily integrate. U substitution requires strong algebra skills and knowledge of rules of differentiation. Why? Because you’ll need to be able to look at the integral and see where a little algebra might get the form into one you can easily integrate — and as integration is really reverse-differentiation, knowing your rules of differentiation will make the task much easier. For example, the following sample problem uses the integral 2x(x^{2}+3)^{70}. Recognizing that if you differentiate x^{2} +3, you get 2x, is the key to successful u substitution.

**Sample problem:** Integrate 2x(x^{2}+3)^{70} using integration by substitution.

Step 1: **Choose a term to substitute for u.** Pick a term that when you substitute u in, it makes it easily to integrate. In this example, replacing (x^{2}) with u makes the function look more familiar for integrating:

u=x^{2}+3

Step 2: **Take the derivative of the u function.** This particular function uses the power rule, so:

du = 2x dx

Step 3: **Rewrite the problem **using the “u” and “du” you derived in Steps 1 and 2:

∫2x(x^{2}+3)^{70}dx

= ∫2x(u)^{70}dx

which can be rewritten as:

∫(u)^{70}2x dx

substituting du from Step 2:

∫(u)^{70}du

Step 4: **Integrate the function,** using the power function rule for integration.

∫(u)^{70}du=^{u71}⁄_{71} + C

Step 5: **Resubstitute your original term** (from Step 1) in place of u:

^{u71}⁄_{71} + C = ^{(X2 + 3)71}⁄_{71} + C

**Tip:** When using integration by substitution, always look for terms that are derivatives of each other. In the above example, the derivative of x^{2} is 2x.

## 2. U Substitution for Trigonometric Functions

U substitution is one way you can find integrals for trigonometric functions. With u substitution, you substitute “u” to simplify the process of integration, re-substituting the original term at the end of the process. The “trick” to successfully using u substitution is to be very familiar with the rules of integration, because you are going to be picking terms to substitute that leave you with something that can easily be integrated using the general rules.

## U Substitution Trigonometric Functions: Example

**Sample problem #1:** Integrate ∫sin 3x dx.

Step 1: **Select a term for “u.”** Look for substitution that will result in a more familiar equation to integrate. Substituting u for 3x will leave an easier term to integrate (sin u), so:

u = 3x

Step 2: **Differentiate u:**

du = 3 dx

Or (rewriting using algebra — necessary because you need to replace “dx”, not 3 dx:

⅓ du = dx

Step 3: **Replace all forms of x** in the original equation:

Substituting for u: ∫ sin 3x dx = ∫ sin u dx

Substituting for dx: ∫ sin u dx = ∫ sin u ⅓ du

Step 4: **Rewrite, bringing the constant in front **of the integral symbol (so that you can easily integrate):

∫ sin u ⅓ du = ⅓ ∫ sin u du

Step 5: **Integrate **using the usual rules of integration:

⅓ ∫ sin u du = ⅓ (-cos u) + C = -⅓ cos u + C

Step 6: **Re-substitute** for u:

“u” is left in the equation, so: ⅓ cos u + C = ⅓ cos 3x + C

*That’s it!*

**Sample problem #2:** Integrate ∫ 5 sec 4x dx

Step 1: Pick a term to **substitute for u**:

u = 4x

Step 2: **Differentiate**, using the usual rules of differentiation.

du = 4 dx

¼ du = dx (using algebra to rewrite, as you need to substitute dx on its own, not 4x)

Step 3: **Substitute u and du **into the equation:

∫ 5 sec 4x tan 4x dx = 5 ∫ sec u tan u ¼ du = ^{5}⁄_{4} ∫ sec u tan u du

Step 4: **Integrate**, using the usual rules of integration. For this problem, integrate using the rule D(sec x) = sec x tan x:

^{5}⁄_{4}∫ sec u tan u du = ^{5}⁄_{4} sec u + C

Step 5: **Resubstitute for u**:

^{5}⁄_{4} sec u + C = ^{5}⁄_{4} sec 4x + C

**Tip:** If you don’t know the rules by heart, compare your function to the general rules of integration and look for familiar looking integrands *before *you attempt to substitute anything for u.

*That’s all there is to U Substitution for Trigonometric Functions!*

## U Substitution for Definite Integrals

In general, a definite integral is a good candidate for u substitution if the equation contains both a **function** and that function’s **derivative.** When evaluating definite integrals, figure out the indefinite integral first and then evaluate for the given limits of integration.

**Sample problem:** Evaluate:

Step 1: **Pick a term for u. **Choose sin x for this sample problem, because the derivative is cos x.

u = sin x.

Step 2: **Find the derivative of u**:

du = cos x dx

Step 3: **Substitute u and du **into the function:

Step 4: **Integrate the function **from Step 3:

Step 5: **Evaluate at the given limits**:

*That’s it!*

## 4. U Substitution for Exponential Functions

With u substitution, you algebraically simplify a function so that its antiderivative can be easily recognized. U substitution is just like it sounds — you substitute in the variable u to perform the integration, which simplifies the process. At the end of your calculations, you re-substitute in your original terms for the u.

**Sample question**: Find the integral for the exponential function ^{ex+1}⁄_{ex} using u substitution.

Step 1: **Rewrite your function using algebra** to get it in a form where you can easily find an integral:

∫^{ex+1}⁄_{ex} = ∫(^{ex}⁄_{ex}+^{1}⁄_{ex}) = ∫(1+e^{-x})dx

Step 2: **Split up the function** into separate parts:

∫(1+e^{-x})dx = ∫1dx + ∫e^{-x}dx

Step 3: **Pick u **and find the **derivative of u**. For this example, pick “-x” in e^{-x}:

u = -x

du =-1*dx

Step 4: **Find a way to get rid of the symbol dx** using your second substitution in Step 3. Using algebra:

du =-1*dx

so

-1du=dx

Step 5: **Substitute the “u”, and “du”** from Steps 3 and 4 into the equation.

∫1dx + ∫e^{u}(-1)du

Step 6: **Solve the integrals:**

∫1dx + ∫e^{u}(-1)du = x – e^{u} + C

Step 7: **Resubstitute your terms back into the function.** u=-x, so:

x – e^{u} + C = x – e^{-x} + C

*That’s it!*

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