Use substitution to solve the definite integral

[chwala]

Homework Statement

see attached

Relevant Equations

integration

I have $1-x^2 = 1- \sin^2 θ = \cos^2 θ$ and $dx =cos θ dθ$

$\int_0^{0.5} (1-x^2)^{1.5} dx = \int_0^{\frac{π}{6}} [cos ^2θ]^\frac{3}{2} dθ = \int_0^{\frac{π}{6}} [cos ^4θ] dθ$

Suggestions on next step.

 

[pasmith]

Use <br /> \cos^2 \theta \equiv \tfrac12(1 + \cos 2\theta) twice.

 

[chwala]

Use <br /> \cos^2 \theta \equiv \tfrac12(1 + \cos 2\theta) twice.

that is $\cos^4θ = \dfrac{1}{2}(1+\cos 4θ) ?$

 

[Orodruin]

that is $\cos^4θ = \dfrac{1}{2}(1+\cos 4θ) ?$

No. Write it $\cos^4\theta = (\cos^2\theta)^2$, apply the identity and expand the square.

 

[chwala]

No. Write it $\cos^4\theta = (\cos^2\theta)^2$, apply the identity and expand the square.

Thanks, i was confusing... the correct identity is

$\cos^{2} 2θ = \dfrac{1}{2}(1+\cos 4θ)$

anyway i realized that I will still need it in doing my work ;

$\int_0^{\frac{π}{6}} [cos ^4θ] dθ= \int_0^{\frac{π}{6}} \left[\dfrac{1}{4} + \dfrac{\cos 2θ}{2} + \dfrac{\cos^{2} 2θ}{4}\right] dθ$

$\int_0^{\frac{π}{6}} \left[\dfrac{1}{4} + \dfrac{\cos 2θ}{2} + \dfrac{\cos^{2} 2θ}{4}\right] dθ= \int_0^{\frac{π}{6}} \left[\dfrac{1}{4} + \dfrac{\cos 2θ}{2} + \dfrac{1}{8} +\dfrac{\cos 4θ}{8}\right] dθ$

$=\left[\dfrac{3θ}{8}+ \dfrac{\sin 2θ}{4} + \dfrac{\sin 4θ}{32}\right]_0^{\frac{π}{6}}$

This will lead me in the right direction. Cheers!

 

[Mark44]

No one seems to have caught this:

$\int_0^{\frac{π}{6}} [cos ^2θ]^\frac{3}{2} dθ = \int_0^{\frac{π}{6}} [cos ^4θ] dθ$

Hold on there! $(u^2)^{3/2} \ne u^4$!

 

[Orodruin]

No one seems to have caught this:

Hold on there! $(u^2)^{3/2} \ne u^4$!

Probably because it (the first expression) is a typo. It is missing the $\cos\theta$ from the $dx$.

 

[chwala]

No one seems to have caught this:

Hold on there! $(u^2)^{3/2} \ne u^4$!

Substitution for $dx$ using change of variables... It should be fine.

Let me check that again.

 

[chwala]

Probably because it (the first expression) is a typo. It is missing the $\cos\theta$ from the $dx$.

Correct. Cheers guys.

 

[Orodruin]

Correct.

Did you doubt it?