What is $\int \frac{{(arcsin x)}^{9}}{\sqrt{1 - x^{2}} d x}$ $int ((arcsinx)^9) / (sqrt(1-x^2) dx$ ?

Question

What is $\int \frac{{(arcsin x)}^{9}}{\sqrt{1 - x^{2}} d x}$ $int ((arcsinx)^9) / (sqrt(1-x^2) dx$ ?

1 Answer

Impact of this question

2632 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-29T02:07:16

$\int \frac{{(arcsin x)}^{9}}{\sqrt{1 - x^{2}}} d x = \frac{{(arcsin x)}^{10}}{10} + c$ $color(blue)[int ((arcsinx)^9) / (sqrt(1-x^2)] dx=(arcsinx)^10/10+c]$

Explanation:

$\int \frac{{(arcsin x)}^{9}}{\sqrt{1 - x^{2}}} d x$ $int ((arcsinx)^9) / (sqrt(1-x^2)] dx$

lets suppose:

$u = arcsin x$ $u=arcsinx$

$d u = \frac{1}{\sqrt{1 - x^{2}}} \cdot d x$ $du=1/sqrt(1-x^2)*dx$

$d x = \sqrt{1 - x^{2}} \cdot d u$ $dx=sqrt(1-x^2)*du$

$\int \frac{{(arcsin x)}^{9}}{\sqrt{1 - x^{2}} d x} = \int \frac{{(u)}^{9} \cdot \sqrt{1 - x^{2}}}{\sqrt{1 - x^{2}}} \cdot d u$ $int ((arcsinx)^9) / (sqrt(1-x^2) dx]=int ((u)^9*sqrt(1-x^2))/ (sqrt(1-x^2)]*du$

$\int u^{9} \cdot d u = \frac{u^{10}}{10} = \frac{{(arcsin x)}^{10}}{10} + c$ $intu^9*du=u^10/10=(arcsinx)^10/10+c$

Science

Math

Humanities

... and beyond

What is $\int \frac{{(arcsin x)}^{9}}{\sqrt{1 - x^{2}} d x}$ $int ((arcsinx)^9) / (sqrt(1-x^2) dx$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is int ((arcsinx)^9) / (sqrt(1-x^2) dx∫(arcsinx)9√1−x2dxint ((arcsinx)^9) / (sqrt(1-x^2) dx?

1 Answer

Explanation:

Related questions

Impact of this question

What is $\int \frac{{(arcsin x)}^{9}}{\sqrt{1 - x^{2}} d x}$ $int ((arcsinx)^9) / (sqrt(1-x^2) dx$ ?