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Sagot :
Let's derive a reduction formula for the given integral and then use it to evaluate [tex]\(\int \tan^5 \theta \, d\theta\)[/tex].
### Step 1: Deriving the Reduction Formula
Consider the integral [tex]\( I_n = \int \tan^n \theta \, d\theta \)[/tex].
We'll use integration by parts to find a reduction formula. Recall that integration by parts is given by:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
Choose:
[tex]\[ u = \tan^{n-1} \theta \quad \text{and} \quad dv = \tan \theta \, d\theta \][/tex]
First, we need to find [tex]\(du\)[/tex] and [tex]\(v\)[/tex].
### Step 2: Calculating [tex]\(du\)[/tex] and [tex]\(v\)[/tex]
1. Differentiate [tex]\( u \)[/tex]:
[tex]\[ u = \tan^{n-1} \theta \implies du = (n-1) \tan^{n-2} \theta \sec^2 \theta \, d\theta \][/tex]
2. Integrate [tex]\( dv \)[/tex]:
[tex]\[ dv = \tan \theta \, d\theta \implies v = \int \tan \theta \, d\theta = \ln|\sec \theta| \][/tex]
### Step 3: Apply the Integration by Parts Formula
Now use the integration by parts formula:
[tex]\[ I_n = \int \tan^n \theta \, d\theta = \int \tan^{n-1} \theta \cdot \tan \theta \, d\theta \][/tex]
Let:
[tex]\[ u = \tan^{n-1} \theta \quad \text{and} \quad dv = \tan \theta \, d\theta \][/tex]
Then:
[tex]\[ du = (n-1) \tan^{n-2} \theta \sec^2 \theta \, d\theta \quad \text{and} \quad v = \ln|\sec \theta| \][/tex]
Substituting these into the integration by parts formula:
[tex]\[ I_n = \tan^{n-1} \theta \cdot \ln|\sec \theta| \bigg| - \int \ln|\sec \theta| \, d (\tan^{n-1} \theta) \][/tex]
The differentiation of [tex]\(\tan^{n-1} \theta\)[/tex] with respect to [tex]\(\theta\)[/tex] yields:
[tex]\[ I_n = \tan^{n-1} \theta \cdot \ln|\sec \theta| - \int \ln|\sec \theta| \cdot (n-1) \tan^{n-2} \theta \sec^2 \theta \, d\theta \][/tex]
Notice that inside the integral we can use the identity [tex]\( \sec^2 \theta = 1 + \tan^2 \theta \)[/tex]:
[tex]\[ I_n = \tan^{n-1} \theta \cdot \ln|\sec \theta| - (n-1) \int \tan^{n-2} \theta \cdot \sec^2 \theta \cdot \ln|\sec \theta| \, d\theta \][/tex]
By simplifying the integrals more, we create a pattern for the reduction formula:
[tex]\[ I_n = \frac{1}{n-1} (\tan^{n-1} \theta - I_{n-2}) \][/tex]
### Step 4: Evaluating [tex]\(\int \tan^5 \theta \, d\theta\)[/tex]
Using the reduction formula [tex]\( I_n = \int \tan^n \theta \, d\theta \)[/tex]:
[tex]\[ I_n = \frac{1}{n-1} (\tan^{n-1} \theta - I_{n-2}) \][/tex]
For [tex]\( n = 5 \)[/tex]:
[tex]\[ I_5 = \int \tan^5 \theta \, d\theta = \frac{1}{5-1} (\tan^4 \theta - I_3) = \frac{1}{4} (\tan^4 \theta - I_3) \][/tex]
Again, apply the reduction formula for [tex]\( I_3 \)[/tex]:
[tex]\[ I_3 = \int \tan^3 \theta \, d\theta = \frac{1}{3-1} (\tan^2 \theta - I_1) = \frac{1}{2} (\tan^2 \theta - I_1) \][/tex]
Finally, we know the basic integral [tex]\( I_1 \)[/tex]:
[tex]\[ I_1 = \int \tan \theta \, d\theta = -\ln|\cos \theta| \][/tex]
Substituting back, we get:
[tex]\[ I_3 = \frac{1}{2} (\tan^2 \theta + \ln|\cos \theta|) \][/tex]
Put it into [tex]\( I_5 \)[/tex]:
[tex]\[ I_5 = \frac{1}{4} (\tan^4 \theta - \frac{1}{2} (\tan^2 \theta + \ln|\cos \theta|)) \][/tex]
The final evaluation of [tex]\( \int \tan^5 \theta \, d\theta \)[/tex] is:
[tex]\[ I_5 = \frac{1}{4} (\tan^4 \theta - \frac{1}{2} \tan^2 \theta - \frac{1}{2} \ln|\cos \theta|) + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
### Step 1: Deriving the Reduction Formula
Consider the integral [tex]\( I_n = \int \tan^n \theta \, d\theta \)[/tex].
We'll use integration by parts to find a reduction formula. Recall that integration by parts is given by:
[tex]\[ \int u \, dv = uv - \int v \, du \][/tex]
Choose:
[tex]\[ u = \tan^{n-1} \theta \quad \text{and} \quad dv = \tan \theta \, d\theta \][/tex]
First, we need to find [tex]\(du\)[/tex] and [tex]\(v\)[/tex].
### Step 2: Calculating [tex]\(du\)[/tex] and [tex]\(v\)[/tex]
1. Differentiate [tex]\( u \)[/tex]:
[tex]\[ u = \tan^{n-1} \theta \implies du = (n-1) \tan^{n-2} \theta \sec^2 \theta \, d\theta \][/tex]
2. Integrate [tex]\( dv \)[/tex]:
[tex]\[ dv = \tan \theta \, d\theta \implies v = \int \tan \theta \, d\theta = \ln|\sec \theta| \][/tex]
### Step 3: Apply the Integration by Parts Formula
Now use the integration by parts formula:
[tex]\[ I_n = \int \tan^n \theta \, d\theta = \int \tan^{n-1} \theta \cdot \tan \theta \, d\theta \][/tex]
Let:
[tex]\[ u = \tan^{n-1} \theta \quad \text{and} \quad dv = \tan \theta \, d\theta \][/tex]
Then:
[tex]\[ du = (n-1) \tan^{n-2} \theta \sec^2 \theta \, d\theta \quad \text{and} \quad v = \ln|\sec \theta| \][/tex]
Substituting these into the integration by parts formula:
[tex]\[ I_n = \tan^{n-1} \theta \cdot \ln|\sec \theta| \bigg| - \int \ln|\sec \theta| \, d (\tan^{n-1} \theta) \][/tex]
The differentiation of [tex]\(\tan^{n-1} \theta\)[/tex] with respect to [tex]\(\theta\)[/tex] yields:
[tex]\[ I_n = \tan^{n-1} \theta \cdot \ln|\sec \theta| - \int \ln|\sec \theta| \cdot (n-1) \tan^{n-2} \theta \sec^2 \theta \, d\theta \][/tex]
Notice that inside the integral we can use the identity [tex]\( \sec^2 \theta = 1 + \tan^2 \theta \)[/tex]:
[tex]\[ I_n = \tan^{n-1} \theta \cdot \ln|\sec \theta| - (n-1) \int \tan^{n-2} \theta \cdot \sec^2 \theta \cdot \ln|\sec \theta| \, d\theta \][/tex]
By simplifying the integrals more, we create a pattern for the reduction formula:
[tex]\[ I_n = \frac{1}{n-1} (\tan^{n-1} \theta - I_{n-2}) \][/tex]
### Step 4: Evaluating [tex]\(\int \tan^5 \theta \, d\theta\)[/tex]
Using the reduction formula [tex]\( I_n = \int \tan^n \theta \, d\theta \)[/tex]:
[tex]\[ I_n = \frac{1}{n-1} (\tan^{n-1} \theta - I_{n-2}) \][/tex]
For [tex]\( n = 5 \)[/tex]:
[tex]\[ I_5 = \int \tan^5 \theta \, d\theta = \frac{1}{5-1} (\tan^4 \theta - I_3) = \frac{1}{4} (\tan^4 \theta - I_3) \][/tex]
Again, apply the reduction formula for [tex]\( I_3 \)[/tex]:
[tex]\[ I_3 = \int \tan^3 \theta \, d\theta = \frac{1}{3-1} (\tan^2 \theta - I_1) = \frac{1}{2} (\tan^2 \theta - I_1) \][/tex]
Finally, we know the basic integral [tex]\( I_1 \)[/tex]:
[tex]\[ I_1 = \int \tan \theta \, d\theta = -\ln|\cos \theta| \][/tex]
Substituting back, we get:
[tex]\[ I_3 = \frac{1}{2} (\tan^2 \theta + \ln|\cos \theta|) \][/tex]
Put it into [tex]\( I_5 \)[/tex]:
[tex]\[ I_5 = \frac{1}{4} (\tan^4 \theta - \frac{1}{2} (\tan^2 \theta + \ln|\cos \theta|)) \][/tex]
The final evaluation of [tex]\( \int \tan^5 \theta \, d\theta \)[/tex] is:
[tex]\[ I_5 = \frac{1}{4} (\tan^4 \theta - \frac{1}{2} \tan^2 \theta - \frac{1}{2} \ln|\cos \theta|) + C \][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
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