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.

  • 5
    $\begingroup$ 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 ... $\endgroup$
    – Matti P.
    Feb 6, 2019 at 12:02
  • $\begingroup$ 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. $\endgroup$
    – user150203
    Feb 7, 2019 at 7:22

2 Answers 2


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

  • $\begingroup$ What about $$\int_0^\pi \sin xdx =2$$ Why doesn't the substitution $\sin x=t$ work? $\endgroup$
    – Zacky
    Feb 6, 2019 at 12:38
  • 2
    $\begingroup$ @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 $\endgroup$
    – 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.


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