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]
