integral

• Dec 27th 2011, 03:03 AM
Ozymandias
integral
Find the value of $\displaystyle \int \frac{1}{(1+x^2)^{3 \over 2}}dx$

I have been trying to solve this question since many years but I can't reach anywhere...

I have tried substituting $\displaystyle x^2$ ,$\displaystyle x^2+1$ and $\displaystyle 1/(x^2+1)$ but I didnot get the answer.
• Dec 27th 2011, 03:10 AM
AgentSmith
Re: integral
Substitute $\displaystyle x= \tan {\lambda}$.

\displaystyle \begin{align*} \int \frac{1}{(1+x^2)^{\frac{3}{2}}} \ dx&= \int \frac{\sec^2{\lambda}}{(\sec^2{\lambda})^{\frac{3} {2}}}d\lambda \\ &=\int \frac{1}{\sec{\lambda}} \ d\lambda \\ &= \int \cos{\lambda} \ d\lambda \\ &= \sin{\lambda}+C \\ &= \sin[\tan^{-1}{(x)}]+C\end{align*}

Do you understand now?
• Dec 27th 2011, 03:11 AM
Ozymandias
Re: integral
Quote:

Originally Posted by AgentSmith
Substitute $\displaystyle x= \tan {\lambda}$.

\displaystyle \begin{align*} \int \frac{1}{(1+x^2)^{\frac{3}{2}}} &= \int \frac{\sec^2{\lambda}}{(\sec^2{\lambda})^{\frac{3} {2}}}d\lambda \\ &=\int \frac{1}{\sec{\lambda}}d\lambda \\ &= \int \cos{\lambda} d\lambda \\ &= \sin{\lambda}+C \\ &= \sin[\tan^{-1}{(x)}]+C\end{align*}

Do you understand now?

I most certainly do not understand!
The answer given in the book is $\displaystyle \frac{x}{\sqrt{x^2+1}}+C$ ...
• Dec 27th 2011, 03:16 AM
AgentSmith
Re: integral
Quote:

Originally Posted by Ozymandias
I most certainly do not understand!
The answer given in the book is $\displaystyle \frac{x}{\sqrt{x^2+1}}+C$ ...

That is the same thing...

$\displaystyle \sin(\tan^{-1}{x})=\sin{\left [\sin^{-1}{\left(\frac{x}{\sqrt{x^2+1}}}\right) \right]}=\frac{x}{\sqrt{x^2+1}}$
• Dec 27th 2011, 03:20 AM
Ozymandias
Re: integral
okay! thank you for your explanation
• Dec 27th 2011, 05:09 AM
Prove It
Re: integral
The reason being...

\displaystyle \displaystyle \begin{align*} \sin^2{\theta} + \cos^2{\theta} &\equiv 1 \\ \frac{\sin^2{\theta}}{\sin^2{\theta}} + \frac{\cos^2{\theta}}{\sin^2{\theta}} &\equiv \frac{1}{\sin^2{\theta}} \\ 1 + \frac{1}{\tan^2{\theta}} &\equiv \frac{1}{\sin^2{\theta}} \\ \frac{\tan^2{\theta} + 1}{\tan^2{\theta}} &\equiv \frac{1}{\sin^2{\theta}} \\ \frac{\tan^2{\theta}}{\tan^2{\theta} + 1} &\equiv \sin^2{\theta} \\ \sin{\theta} &\equiv \pm \frac{\tan{\theta}}{\sqrt{\tan^2{\theta} + 1}} \end{align*}

Therefore

\displaystyle \displaystyle \begin{align*} \sin{\left(\tan^{-1}{x}\right)} &= \pm \frac{\tan{\left(\tan^{-1}{x}\right)}}{\sqrt{ \tan^2{\left(\tan^{-1}{x}\right) + 1}}} \\ &= \pm \frac{x}{\sqrt{x^2 + 1}} \end{align*}