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Consider the following function. x6 1 + x2 dx (a) Determine an appropriate trigonometric substitution. Use x = sin(θ), where − π 2 ≤ θ ≤ π 2 , since the integrand contains the expression 1 + x2 . Use x = tan(θ), where − π 2 < θ < π 2 , since the integrand contains the expression 1 + x2 . Use x = sec(θ), where 0 ≤ θ < π 2 or π ≤ θ < 3π 2 , since the integrand contains the expression 1 + x2 . (b) Apply the substitution to transform the integral into a trigonometric integral. Do not evaluate the integral. x6 1 + x2 dx = dθ

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Answer:

Step-by-step explanation:

In the first part, we are given the function:

[tex]\int \dfrac{x^6}{\sqrt{1+x}^2} \ dx[/tex]

Suppose we make x = tan θ

Then dx = sec² θ.dθ

[tex]= \int \dfrac{tan^6 (\theta)}{\sqrt{1+ tan ^2 \theta }}* dx[/tex]

[tex]= \int \dfrac{tan^6 (\theta)}{\sqrt{1+ tan ^2 \theta }}* sec ^2 (\theta) * d\theta[/tex]

Since; sec² θ - tan² θ = 1

sec² θ = 1+ tan² θ

[tex]sec \ \theta = \sqrt{1 + tan^2 \ \theta}[/tex]

∴

[tex]= \int \dfrac{tan^6 (\theta)}{sec \ \theta}* sec ^2 (\theta) * d\theta[/tex]

[tex]= \int tan^6 (\theta)* sec (\theta) * d\theta[/tex]

Thus; In the first part, Use x = tan θ, where  [tex]- \dfrac{\pi}{2} < \theta <\dfrac{\pi}{2}[/tex], since the integrand comprise the expression [tex]\sqrt{1+x^2}[/tex]

From the second part by using substitution method;

[tex]\int \dfrac{x^6}{\sqrt{1+x^2}} \ dx = \int \mathbf{tan^6(\theta) * sec ( \theta) } \ d \theta[/tex]

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