# Use substitution to solve the definite integral

• chwala
chwala
Gold Member
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.

Last edited:
Use $$\cos^2 \theta \equiv \tfrac12(1 + \cos 2\theta)$$ twice.

MatinSAR and chwala
pasmith said:
Use $$\cos^2 \theta \equiv \tfrac12(1 + \cos 2\theta)$$ twice.
that is ##\cos^4θ = \dfrac{1}{2}(1+\cos 4θ) ?##

chwala said:
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.

MatinSAR and chwala
Orodruin said:
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!

No one seems to have caught this:
chwala said:
##\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##!

chwala
Mark44 said:
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##.

PhDeezNutz and chwala
Mark44 said:
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.

Orodruin said:
Probably because it (the first expression) is a typo. It is missing the ##\cos\theta## from the ##dx##.
Correct. Cheers guys.

chwala said:
Correct.
Did you doubt it?

SammyS and chwala

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