Q.$\int_{0}^{1} \frac{4\tan^{-1}x}{1 + x^2}\,dx =$

Substitute $t=\log x$: the integral becomes $\int_0^1 t^2\,dt=\tfrac13$.

Let $t=\log x$, so $dt=\tfrac{dx}{x}$. Limits: $x=1\to t=0$, $x=e\to t=1$. Then

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Substitute $t=\tan x$ (so $dt=\sec^2 x\,dx$): the integral becomes $\int_0^1 e^t\,dt=e-1$.

Let $t=\tan x$. Then $dt=\sec^2 x\,dx=\dfrac{dx}{\cos^2 x}$. Limits: $x=0\to t=0$, $x=\tfrac{\pi}{4}\to t=1$. So

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Substitute $t=\sqrt{x}$; the integral becomes $2\int_0^1 e^t\,dt=2(e-1)$.

Let $t=\sqrt{x}$, so $dt=\dfrac{dx}{2\sqrt{x}}$, i.e. $\dfrac{dx}{\sqrt{x}}=2\,dt$. Limits: $x=0\to t=0$, $x=1\to t=1$. Then

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$\int_0^a\sqrt{a^2-x^2}\,dx$ is a quarter-circle area $=\tfrac{\pi a^2}{4}$.

Using the standard formula $\int\sqrt{a^2-x^2}\,dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}$, evaluate from $0$ to $a$:

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Substitute $u=\tan^{-1}x$; the integral becomes $4\int_0^{\pi/4}u\,du=\frac{\pi^2}{8}$.

Let $u=\tan^{-1}x$, so $du=\dfrac{dx}{1+x^2}$. When $x=0,\ u=0$; when $x=1,\ u=\frac{\pi}{4}$. Then …

With $u=a^2-x^2$, $\int_0^a \frac{x\,dx}{2\sqrt{a^2-x^2}} = \frac12\big[-\sqrt{a^2-x^2}\big]_0^a = \frac{a}{2}$.

Let $u = a^2 - x^2$, so $du = -2x\,dx$, i.e. $x\,dx = -\tfrac12 du$. Also $\int \frac{x\,dx}{\sqrt{a^2-x^2}} = -\sqrt{a^2-x^2}$. Therefore

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A definite integral's value does not depend on the name of the integration variable, so writing $\phi(x)$ in place of $x$ within the same limits does not change the value.

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