BeMath-1-x-5-x-x-8-1-dx-2-1-2-1-sin-1-x-x-3-dx-

Question Number 108208 by bemath last updated on 15/Aug/20

\frac{B e M a t h}{⊟}

(1) \int \frac{x^{5} - x}{x^{8} + 1} d x

(2) \underset{\frac{1}{\sqrt{2}}}{\int^{1}} \frac{\sin^{- 1} (x)}{x^{3}} d x

Answered by john santu last updated on 15/Aug/20

\frac{★ JS ♠}{≎}

J = \int \frac{x (x^{4} - 1)}{x^{8} + 1} d x

l e t z = x^{2} \Rightarrow J = \frac{1}{2} \int \frac{z^{2} - 1}{z^{4} + 1} d z

d e c o m p o s i t i o n

z^{4} + 1 = {(z^{2} + 1)}^{2} - 2 z^{2}

= (z^{2} + z \sqrt{2} + 1) (z^{2} - z \sqrt{2} + 1)

\frac{z^{2} - 1}{z^{4} + 1} = \frac{1}{2} (\frac{z \sqrt{2} - 1}{z^{2} - z \sqrt{2} + 1} - \frac{z \sqrt{2} + 1}{z^{2} + z \sqrt{2} + 1})

J = \frac{1}{4} \int (\frac{z \sqrt{2} - 1}{z^{2} - z \sqrt{2} + 1} - \frac{z \sqrt{2} + 1}{z^{2} + z \sqrt{2} + 1}) d z

J = \frac{1}{4 \sqrt{2}} \ln (\frac{x^{4} - x^{2} \sqrt{2} + 1}{x^{4} + x^{2} \sqrt{2} + 1}) + c

Commented by bobhans last updated on 15/Aug/20

\frac{C ooll}{great}

Answered by bobhans last updated on 15/Aug/20

(2) \frac{bobhans}{° ∙ \equiv ∙ °}

I = \underset{1 / \sqrt{2}}{\int^{1}} \frac{\sin^{- 1} (x)}{x^{3}} dx

by part {\begin{cases} u = \sin^{- 1} (x) \to du = \frac{dx}{\sqrt{1 - x^{2}}} \\ v = - \frac{1}{{2 x}^{2}} \end{cases}

I = {[- \frac{\sin^{- 1} (x)}{{2 x}^{2}}]}_{\frac{1}{\sqrt{2}}}^{1} + \underset{1 / \sqrt{2}}{\int^{1}} \frac{dx}{x^{2} \sqrt{1 - x^{2}}}

I = 0 + \frac{1}{2} \underset{π / 4}{\int^{π / 2}} \frac{\cos h dh}{\sin^{2} h \sqrt{1 - \sin^{2} h}}

I = \frac{1}{2} \underset{π / 4}{\int^{π / 2}} \csc^{2} h dh = - \frac{1}{2} {[\cot h]}_{\frac{π}{4}}^{\frac{π}{2}}

I = \frac{1}{2}

Answered by mathmax by abdo last updated on 15/Aug/20

2) I = \int_{\frac{1}{\sqrt{2}}}^{1} \frac{arcsinx}{x^{3}} dx by parts u^{^{'}} = x^{- 3} and v = arcsinx

\Rightarrow I = {[- \frac{1}{2} x^{- 2} arcsinx]}_{\frac{1}{\sqrt{2}}}^{1} + \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{1} x^{- 2} \frac{dx}{\sqrt{1 - x^{2}}}

= \frac{π}{4} - \frac{π}{4} + \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{1} \frac{dx}{x^{2} \sqrt{1 - x^{2}}} = \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{1} \frac{dx}{x^{2} \sqrt{1 - x^{2}}}

=_{x = sint} \frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{cost dt}{\sin^{2} t cost} = \frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{2 dt}{1 - \cos (2 t)}

=_{2 t = u} \int_{\frac{π}{2}}^{π} \frac{du}{2 {1 - cosu}} =_{\tan (\frac{u}{2}) = z} \frac{1}{2} \int_{1}^{+ \infty} \frac{2 dz}{(1 + z^{2}) (1 - \frac{1 - z^{2}}{1 + z^{2}})}

= \int_{1}^{+ \infty} \frac{dz}{1 + z^{2} - 1 + z^{2}} = \frac{1}{2} \int_{1}^{+ \infty} \frac{dz}{z^{2}} = \frac{1}{2} {[- \frac{1}{z}]}_{1}^{+ \infty} = \frac{1}{2} \Rightarrow

I = \frac{1}{2}

Answered by mathmax by abdo last updated on 15/Aug/20

1) complex method let decompose F (x) = \frac{x^{5} - x}{x^{8} + 1}

x^{8} + 1 = 0 \Rightarrow x^{8} = e^{i (2 k + 1) π} \Rightarrow z_{k} = e^{\frac{i (2 k + 1) π}{8}} and k \in [[0, 7]]

\Rightarrow F (x) = \frac{x^{5} - x}{\prod_{k = 0}^{7} (x - z_{k})} = \sum_{k = 0}^{7} \frac{a_{k}}{x - z_{k}}

a_{k} = \frac{z_{k}^{5} - z_{k}}{{8 z}_{k}^{7}} = \frac{z_{k}^{6} - z_{k}^{2}}{- 8} \Rightarrow \int F (x) dx = - \frac{1}{8} \int \sum_{k = 0}^{7} \frac{z_{k}^{6} - z_{k}^{2}}{x - z_{k}} dx

= - \frac{1}{8} \sum_{k = 0}^{7} (z_{k}^{6} - z_{k}^{2}) \ln (x - z_{k}) + C