Question #e678e

2 Answers
Dec 5, 2017

int_0^1 x*sqrt(1-x^4)*dx=pi/8

Explanation:

int_0^1 x*sqrt(1-x^4)*dx

=1/2*int_0^1 sqrt[1-(x^2)^2]*2x*dx

After using sinu=x^2 and cosu*du=2x*dx substitution, this integral became

1/2*int_0^(pi/2) sqrt[1-(sinu)^2]*cosu*du

=1/2*int_0^(pi/2) sqrt[(cosu)^2]*cosu*du

=1/2*int_0^(pi/2) (cosu)^2*du

=1/4*int_0^(pi/2) (1+cos2u)*du

=[u/4+1/8*sin2u]_0^(pi/2)

=pi/8

Dec 5, 2017

pi/8

Explanation:

.

int_0^1xsqrt(1-x^4)dx

u=x^2
x=sqrtu
u^2=x^4
du=2xdx
dx=(du)/(2x)=(du)/(2sqrtu)
sqrt(1-x^4)=sqrt(1-u^2)

Now, we substitute the above into the problem integral to turn it into an integral in terms of u:

int(sqrtu)(sqrt(1-u^2))((du)/(2sqrtu))=

1/2intsqrt(1-u^2)du

Now we can use trigonometric substitution to solve this:

u=sintheta
du=costhetad(theta)

1/2intsqrt(1-sin^2theta)costhetad(theta)

1/2intsqrt(cos^2theta)costhetad(theta)

1/2intcosthetacosthetad(theta)=1/2intcos^2thetad(theta)

cos^2theta=(1+cos(2theta))/2

1/4int(1+cos(2theta))d(theta)=1/4intd(theta)+1/4intcos(2theta)d(theta)

1/4theta+1/8sin(2theta)

Now, we can substitute back for u:

u=sintheta, then theta=arcsinu
costheta=sqrt(1-six^2theta)=sqrt(1-u^2)
sin(2theta)=2sinthetacostheta=2usqrt(1-u^2)

1/4theta+1/8sin(2theta)=(arcsinu+usqrt(1-u^2))/4

Now, we can substitute back for x:

int_0^1xsqrt(1-x^4)dx=(arcsinx^2+x^2sqrt(1-x^4))/4

We will evaluate this from 0 to 1:

(arcsin1^2+1^2sqrt(1-1^4))/4-(arcsin0^2+0^2sqrt(1-0^4))/4=

(pi/2+0)/4-(0+0)/4=pi/8