MHB How to Solve the Integral of 1/sqrt(7-x^2) – A Step-by-Step Guide

[karush]

242.ws3.2
$\displaystyle
I=\int \frac{dx}{\sqrt{7-x^2}}
=\arcsin\left(\frac{\sqrt{7}x}{7}\right)+C$
I know this is a standard Integral but was given to solve it
$x=\sqrt{7}\sin\left({u}\right) \therefore dx=\sqrt{7}\cos\left({u}\right) \, du$
I proceeded but

 

[kaliprasad]

242.ws3.2
$\displaystyle
I=\int \frac{dx}{\sqrt{7-x^2}}
=\arcsin\left(\frac{\sqrt{7}x}{7}\right)+C$
I know this is a standard Integral but was given to solve it
$x=\sqrt{7}\sin\left({u}\right) \therefore dx=\sqrt{7}\cos\left({u}\right) \, du$
I proceeded but

putting $x=\sqrt{7}\sin\left({u}\right) \therefore dx=\sqrt{7}\cos\left({u}\right) \, du$
we have
$\sqrt{7-x^2} = \sqrt{7}\cos (u)$
so integal becomes
$I=\int \frac{dx}{\sqrt{7-x^2}}= \int \frac{\sqrt{7}\cos\left({u}\right)}{\sqrt{7}\cos\left({u}\right)} \, du= \int du = u +C = \arcsin \frac{x}{\sqrt{7}} + C $

 

[karush]

really, that's what I did but didn't think it was right...

 

[HallsofIvy]

Then you should have told us that you had a solution and shown how you got it in your first post.

 

[karush]

$I=\int \frac{dx}{\sqrt{7-x^2}}= \int \frac{\sqrt{7}\cos\left({u}\right)}{\sqrt{7}\cos\left({u}\right)} \, du$
When I got to here I thought it was wrong.
But it was correct