Find the following indefinite integral

[jodecy]

Homework Statement

find the indefinite integral of ∫x√(1-x^2) dx

Homework Equations

The Attempt at a Solution

∫x√(1-x^2) dx

let x = sinθ
dx = cosθ dθ

now sin^2θ + cos^2θ = 1
=> cosθ = √1-sin^2θ ( for the form √(1-x^2))

∫x√(1-x^2) dx => ∫sinθ (√1-sin^2θ) cosθ dθ
=> ∫sinθ cos^2θ dθ

integrating :- let u = cosθ
du = -sinθ dθ

=> ∫ sinθ cos^2θ dθ = ∫ -u^2 du
=> -1/3 u^3 + c
=> -1/3 cos θ + c


is this correct?

 

[SammyS]

Homework Statement

find the indefinite integral of ∫x√(1-x^2) dx

Homework Equations

The Attempt at a Solution

∫x√(1-x^2) dx

let x = sinθ
dx = cosθ dθ

now sin^2θ + cos^2θ = 1
=> cosθ = √1-sin^2θ ( for the form √(1-x^2))

∫x√(1-x^2) dx => ∫sinθ (√1-sin^2θ) cosθ dθ
=> ∫sinθ cos^2θ dθ

integrating :- let u = cosθ
du = -sinθ dθ

=> ∫ sinθ cos^2θ dθ = ∫ -u^2 du
=> -1/3 u^3 + c
=> -1/3 cos³ θ + c


is this correct?

Hello jodecy. Welcome to PF !

At the very least, you need to give the final result in terms of the variable, x .

A more direct substitution is to let u = 1-x² .

 

[jodecy]

ohhhh so i need to state θ = arcsinx

therefore my answer in terms of x would be -1/3 cos^3(arcsinx )?

when i come back i'll try subst with u = 1-x^2

 

[Robert1986]

Well, I wouldn't even mess with trig substitution, as SammyS said; that's just overkill and too much work for this problem.

But, it is also not "correct" in the sense that it is really hard to understand what $(-1/3)\cos^3(\arccos(x))$. So, here is a better way.

Draw a right triangle. Pick one of the non-right angles to be $\theta$. Now, you now that $\sin (\theta)=x$, and you know that sine is opposite over hypotenuse. Now, label the opposite side with an $x$ and the hypotenuse with a $1$. Now, do you see how to find the cosine of $\theta$ out of that?

 

[jodecy]

ok so i worked it out using the subst of u = 1 - x^2 and got the answer to be
-1/3 (1-x^2) ^ 3/2 + c. however the original method i used was that presented to me by my lecturer so I'm wondering ...