I'm trying to solve∫−11x2(1−x2)92dx

Setting x=sin⁡(t), we obtainI=∫−π2π2sin2(t)cos9(t)cos⁡(t)dt=∫−π2π2sin2(t)cos10(t)dt=∫−π2π2cos10(t)dt−∫−π2π2cos12(t)dtFrom here, we have∫−π2π2cos2m(t)dt=π22m(2mn)Hence,I=π210(105)−π212(126)=21π1024

x=sin⁡y∫x2(1−x2)92dx=∫x2(1−x2)9dx=∫sin2y(cos2y)9cos⁡ydy=∫sin2y(cos⁡y)9cos⁡ydy=∫sin2ycos10ydy

Setting x=sin⁡(t), we obtainI=∫−π2π2sin2(t)cos9(t)cos⁡(t)dt=∫−π2π2sin2(t)cos10(t)dt=∫−π2π2cos10(t)dt−∫−π2π2cos12(t)dtFrom here, we have∫−π2π2cos2m(t)dt=π22m(2mn)Hence,I=π210(105)−π212(126)=21π1024

x=sin⁡y∫x2(1−x2)92dx=∫x2(1−x2)9dx=∫sin2y(cos2y)9cos⁡ydy=∫sin2y(cos⁡y)9cos⁡ydy=∫sin2ycos10ydy

"I'm trying to solve∫−11x2(1−x2)92dx" was answered by our community

Markelovnsl

Answered question

2022-04-08

I'm trying to solve

\int_{- 1}^{1} x^{2} {(1 - x^{2})}^{\frac{9}{2}} d x

Answer & Explanation

maggionmoo

Beginner2022-04-09Added 16 answers

Setting

x = \sin (t)

, we obtain

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{2} (t) \cos^{9} (t) \cos (t) d t = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{2} (t) \cos^{10} (t) d t

= \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{10} (t) d t - \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{12} (t) d t

From here, we have

\int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2 m} (t) d t = \frac{π}{2^{2 m}} (\begin{matrix} 2 m \\ n \end{matrix})

Hence,

I = \frac{π}{2^{10}} (\begin{matrix} 10 \\ 5 \end{matrix}) - \frac{π}{2^{12}} (\begin{matrix} 12 \\ 6 \end{matrix}) = \frac{21 π}{1024}

rhyclelal80j6

Beginner2022-04-10Added 13 answers

x = \sin y

\int x^{2} {(1 - x^{2})}^{\frac{9}{2}} d x = \int x^{2} {(\sqrt{1 - x^{2}})}^{9} d x

= \int \sin^{2} y {({\sqrt{\cos}}^{2} y)}^{9} \cos y d y

= \int \sin^{2} y {(\cos y)}^{9} \cos y d y

= \int \sin^{2} y \cos^{10} y d y

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