# Integrating $\int_{-1} ^{1}\frac{1}{1+x^2}dx$ with the substitution $x^2=t$ gives an incorrect value of $0$. What went wrong?

Consider this integral.

$$\int_{-1} ^{1}\dfrac{1}{1+x^2}dx$$

Its easy to solve as $$\tan ^{-1} x$$ is the anti derivative of $$\dfrac{1}{1+x^2}dx$$ . Therefore,

$$\int_{-1} ^{1}\dfrac{1}{1+x^2}dx \implies \left[\tan^{-1} x \right]_{-1} ^{1}$$

$$\implies \dfrac{\pi}{2}$$

But if I do this

Let $$x^2=t$$ so $$dx=\dfrac{dt}{2\sqrt{t}}$$. When $$x=-1,t=1$$ and when $$x=1,t=1$$. Therefore,

$$\int_{-1} ^{1}\dfrac{1}{1+x^2}dx \implies \int_{1} ^{1}\dfrac{1}{2\sqrt t (1+t)}dt$$

Since the upper and lower limits are same therefore the expression reduces $$0$$. I know it's incorrect, but I cannot figure out my mistake.

• The mistake is that you chose $x^2 = t$. Of course $x= -1$ and $x=1$ map to the same value of $t$, thus the problem ... Feb 6, 2019 at 12:02
• Slighly extending @MattiP.'s comment - the derivative of $x^2$ changes over the region of integration which is another reason why your substitution isn't working.
– user150203
Feb 7, 2019 at 7:22

Thanks @Swapnil and @Matti P for pointing that out.

Actually I can use $$x^2=t$$ but I'll have to split the integration. When $$x=-1,t=1$$ and when x<0 $$dx=\dfrac{dt}{-2\sqrt{t}}$$ and when $$x=1,t=1$$ and when x>0 $$dx=\dfrac{dt}{2\sqrt{t}}$$.

$$\int_{-1} ^{1}\dfrac{1}{1+x^2}dx \implies \int_{1} ^{0}\dfrac{1}{-2\sqrt t (1+t)}dt + \int_{0} ^{1}\dfrac{1}{2\sqrt t (1+t)}dt$$

$$\implies 2\int_{0} ^{1}\dfrac{1}{2\sqrt t (1+t)}dt$$

This after certain substitution it will turn out to be $$\dfrac{\pi}{2}$$ as well

• What about $$\int_0^\pi \sin xdx =2$$ Why doesn't the substitution $\sin x=t$ work? Feb 6, 2019 at 12:38
• @Zacky it works but you should be careful on when and where to apply substitution. When $x \in (0,\pi /2)$ $\sin x$ and $\cos x$ both are positive, so if $\sin x = t$ , $\cos x = \sqrt{1-t^2}$ and when $x \in (\pi /2,\pi )$ $\sin x$ is +ve but $\cos x$ is -ve , therefore if $\sin x = t$ , $\cos x = -\sqrt{1-t^2}$ Now split the limit and use it. $$\int _{0} ^{1} \dfrac{t\ dt}{\sqrt{1-t^2} } + \int _{1} ^{0} \dfrac{t\ dt}{-\sqrt{1-t^2} }$$ Solve it to get answer as $2$. I don't know why you asked this, as I have written exactly the same thing in my answer
– user585765
Feb 6, 2019 at 14:06

$$x^2=t$$ does not mean $$dx=\dfrac{dt}{2\sqrt{t}}$$.

$$x^2=t \implies x=\pm\sqrt{t}$$

So $$dx = \pm \dfrac{dt}{2\sqrt{t}}$$ and to the best of my knowledge, you cannot use this for integration.

EDIT: OK, as suggested by the OP (Loop Back) in his answer, yes we can determine whether $$x^2=t \implies x=\sqrt{t}$$ or $$x=-\sqrt{t}$$ by splitting the integral and using appropriate value.