Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need a help solving this integration equation,$$\int {y \over \sqrt{1+y^2}} dy$$

I am quite weak with maths. I would really appreciate if someone would guide me solving this.

share|improve this question
3  
Try computing the derivative of $f(x) = \sqrt{1+x^2}$. –  copper.hat Jan 8 '13 at 3:27

3 Answers 3

up vote 5 down vote accepted

Let $u=1+y^2$.

Then:

$du=2y\,dy\implies {1\over 2}\,du=y\,dy$

And:

$$\displaystyle{\int{y\over \sqrt{1+y^2}}\,dy={1\over 2}\int {du\over \sqrt u}= {1\over 2}\int u^{-1/2}\,du={1\over 2}{u^{1/2}\over 1/2}+C=\sqrt{1+y^2}+C.}$$

share|improve this answer
    
kudos! I wish i could think this stuff –  Kishan Thobhani Jan 8 '13 at 3:40

First check that

$$\int\frac{f'(x)\,dx}{\sqrt{f(x)}}=2\sqrt{f(x)}+C\,\,,\,\,C=\text{ a constant}$$

and now just observe that

$$y=\frac{1}{2}2y=\frac{1}{2}\frac{d}{dy}(1+y^2)=\frac{1}{2}(1+y^2)'$$

share|improve this answer

You may also substitute $y=\tan(\theta)$, and $dy=\sec^2(\theta)d\theta$. $$\int \frac{y}{\sqrt{1+y^2}} dy=\int \frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}}\sec^2(\theta)d\theta=\int \tan(\theta)\sec(\theta)d\theta=\sec(\theta)$$ $$\sec(\theta)=\sqrt{1+\tan^2(\theta)}=\sqrt{1+y^2}$$ So the answer is $\sqrt{1+y^2}+C$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.