trigonnomic substition

[mark]

my answer doesnt match up with the books, i was wondering if anyone could chech
my work to see were i went wrong

the problem is

intergral from 0 to 1 ((5x^3) sqrt(1-x^2) dx)=

x = sin theta
dx = cos theta dtheta

intergral from 0 to 1 (5(sin^3 theta)cos theta sqrt(1- sin^2 theta) d_theta)=

5 times the intergral from 0 to 1 ((sin^3 theta)cos theta sqrt(cos^2 theta))=

5 times the intergral from 0 to 1 ((sin^3 theta)cos theta * cos theta )=

5 times the intergral from 0 to 1 ((sin theta sin^2 theta cos^2 theta)=

5 times the intergral from 0 to 1 ((sint theta (1 - cos^2 theta)cos^2 theta)=

u = cos theta
du = -sin theta d_theta

5 times the intergral from 0 to 1 ((1- u^2)u^2 du)=

5 times the intergral from 0 to 1 ((u^2 - u^4)du)=

(5/3)(cos^3 theta)- (cos^5 theta) from 0 to 1 =

cos theta = (sqrt(1-x^2))

(5/3)((sqrt(1-x^2))^3 - (sqrt(1-x^2))^5 from 0 to 1

i get zero but the book says 2/3

 

[pka]

\(\displaystyle \L
\begin{array}{l}
\int\limits_0^1 {5x^3 \sqrt {1 - x^2 } dx} \\
u = 1 - x^2 ,\quad x^2 = 1 - u,\quad du = - 2xdx,\quad \begin{array}{c}
x\ 01 \\
\hline
u\ 10 \\
\end{array} \\
\frac{5}{{ - 2}}\int\limits_0^1 {x^2 \sqrt {1 - x^2 } \left( { - 2xdx} \right)} \\

\frac{5}{{ - 2}}\int\limits_1^0 {\left( {1 - u} \right)\sqrt u \left( {du} \right)} = \frac{{ - 5}}{2}\int\limits_1^0 {\left( {u^{\frac{1}{2}} - u^{\frac{3}{2}} } \right)du} \\
\end{array}\)

 

[mark]

(0^1/2 - 0^3/2) - (1^1/2 - 1^3/2) = 0

it still equals zero, i dont get how the book got 2/3

 

[pka]

\(\displaystyle \L
\frac{{ - 5}}{2}\int\limits_1^0 {\left( {u^{\frac{1}{2}} - u^{\frac{3}{2}} } \right)du} = \left. {\frac{{ - 5}}{2}\left( {\frac{2}{3}u^{\frac{3}{2}} - \frac{2}{5}u^{\frac{5}{2}} } \right)} \right|_1^0 = \frac{2}{3}\)

 

[galactus]

Let \(\displaystyle x=sin({\theta})\) and \(\displaystyle dx=cos({\theta})d{\theta}\)

You have:

\(\displaystyle \L\\5\int_{0}^{\frac{{\pi}}{2}}sin^{3}({\theta})\sqrt{1-sin^{2}({\theta})}cos({\theta})d{\theta}\)

\(\displaystyle 1-sin^{2}({\theta})=cos^{2}({\theta})\)

\(\displaystyle \L\\5\int_{0}^{\frac{{\pi}}{2}}sin^{3}({\theta})cos^{2}({\theta})d{\theta}\)

Break off \(\displaystyle sin^{2}({\theta})=1-cos^{2}({\theta})\)

\(\displaystyle \L\\5\int_{0}^{\frac{{\pi}}{2}}(1-cos^{2}({\theta}))sin({\theta})cos^{2}({\theta})d{\theta}\)

Now, let \(\displaystyle u=cos({\theta}), -du=sin({\theta})d{\theta}\)

\(\displaystyle cos(\frac{{\pi}}{2})=0, cos(0)=1\)

\(\displaystyle \L\\-5\int_{1}^{0}(u^{2}-u^{4})du\)

\(\displaystyle =\L\\5\int_{0}^{1}(u^{2}-u^{4})du\)