I'm having trouble with trig substitution. This is what I've done so far, but I'm not sure if I did everything right. This is the integral: $$\int \frac{x^2}{(1+x^2)^\frac{3}{2}}$$

and my substitution is: $x= \tan\Theta$

$$\int \frac{\tan\Theta ^2}{(\sqrt{\sec\Theta ^2})^3}\sec\Theta ^2d\Theta $$

$$\int \frac{\sec\Theta ^2-1}{\sec\Theta }$$

$$\int \sec\Theta -\int \cos\Theta $$

$$(\ln(\sec\Theta +\tan\Theta ))-(\sin\Theta)$$

$$\ln(\frac{1}{\sqrt{1-x^2}}+ \frac{x}{\sqrt{1-x^2}}) - x + C$$



The problem here is simply when "back substituting".

Having set $x =\tan \theta \implies (\sec\theta = \sqrt{1 + x^2} \,\text{ and }\, \sin \theta = \dfrac{x}{\sqrt{1+x^2}})$

This gives us the final answer: $$\ln \left(\sqrt{x^2+1}+x\right)-{{x}\over{\sqrt{x^2+1}}}+C$$


Okay up to the last step. Now $$ \tan(\theta) = x \\ \sec(\theta) = \sqrt{1+x^2}\\ \sin(\theta) = {\frac{x}{\sqrt{x^2+1}}} $$

So your answer is $$\log \left(\sqrt{x^2+1}+x\right)-{{x}\over{\sqrt{x^2+1}}}+C$$

Differentiate the answer and square it to get $${\frac{x^4}{\left(x^2+1\right)^3}}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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