Q.Evaluate: $\int_0^1 \dfrac{\log x}{\sqrt{1-x^2}}\,dx$

Apply the property $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$ with $a+b = 2+7 = 9$; adding the original and reflected integrals gives $2I = \int_2^7 1\,dx$.

Let $I = \int_2^7 \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{9-x}}\,dx.$

By the property $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$, replace $x$ with $2+7-x = 9-x$:

$I = \int_2^7 \frac{\sqrt{9-x}}{\sqrt{9-x}+\sqrt{x}}\,dx.$

Add the two expressions for $I$: …

The variable of integration in a definite integral is a dummy variable, so $\int_a^b f(x)\,dx = \int_a^b f(t)\,dt$.

A definite integral $\int_a^b f(x)\,dx$ depends only on the function $f$ and the limits $a, b$; its value is a fixed number. The symbol used for the variable of integration ($x$, $t$, $u$, …) is merely a placeholder that is "integrated out," so renamin …

By the Beta-function formula $\displaystyle\int_{0}^{1}x^m(1-x)^n\,dx=\dfrac{m!\,n!}{(m+n+1)!}$, the integral equals $\dfrac{1}{10100}$.

1. We need $\displaystyle I=\int_{0}^{1}x(1-x)^{99}\,dx$, which is of the standard form $\displaystyle\int_0^1 x^m(1-x)^n\,dx$ with $m=1$, $n=99$.
2. This is the Beta function $B(m+1,n+1)=B(2,100)$, and for non-negative integers, $B(m+1,n+1)=\dfrac{m!\,n!}{(m+n+1)!}$.
3. Substituting $m=1$, $n=99$: $I=\dfrac{1!\times 99!}{(1+99+1)!}=\dfrac{99!}{101!}$. …

$\int_b^a\varphi=-\int_a^b\varphi$, so the sum cancels to $0$.

A basic property of definite integrals is $\displaystyle\int_b^a\varphi(x)\,dx=-\int_a^b\varphi(x)\,dx$.

Therefore

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$f(x)=-f(-x)$ means $f$ is an odd function, and the definite integral of an odd function over a symmetric interval is always zero.

The condition $f(x)=-f(-x)$ (equivalently $f(-x)=-f(x)$) is exactly the definition of an odd function.

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Using $f(\pi/2-x)=-f(x)$, the integral $\displaystyle\int_0^{\pi/2}\dfrac{\tan x-\cot x}{1+\tan x\cot x}\,dx$ equals $0$.

1. Since $\tan x\cdot\cot x=1$ (wherever both are defined), the denominator $1+\tan x\cot x=1+1=2$ throughout the interval.
2. So the integrand simplifies to $f(x)=\dfrac{\tan x-\cot x}{2}$.
3. Apply the substitution $x\to \dfrac{\pi}{2}-x$: $\tan\left(\dfrac{\pi}{2}-x\right)=\cot x$ and $\cot\left(\dfrac{\pi}{2}-x\right)=\tan x$.
4. So $f\left(\dfrac{\pi}{2}-x\right)=\dfrac{\cot x-\tan x}{2}=-f(x)$. …

Apply the King's-rule substitution $x\to\frac\pi2-x$ on $[0,\pi/2]$; since $\sin$ and $\cos$ swap under this substitution, the integrand becomes exactly its own negative, forcing the definite integral to be $0$.

1. Let $I = \displaystyle\int_0^{\pi/2} \dfrac{\sin x-\cos x}{1+\sin x\cos x}\,dx$.
2. Use the standard property $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ with $a=\pi/2$: replace $x$ by $\dfrac\pi2-x$.
3. Since $\sin\left(\dfrac\pi2-x\right)=\cos x$ and $\cos\left(\dfrac\pi2-x\right)=\sin x$, and $\sin x\cos x$ is symmetric under this swap: $I = \int_0^{\pi/2} \frac{\cos x-\sin x}{1+\cos x\sin x}\,dx$ …

The stated splitting property requires the symmetry condition $f(2a-x)=f(x)$.

1. In general, $\int_0^{2a}f(x)\,dx=\int_0^a f(x)\,dx+\int_a^{2a}f(x)\,dx$.
2. In the second integral substitute $x=2a-t$, $dx=-dt$; limits $x=a\to t=a$, $x=2a\to t=0$: $\int_a^{2a}f(x)\,dx=\int_0^a f(2a-t)\,dt$.
3. So $\int_0^{2a}f(x)\,dx=\int_0^a\big[f(x)+f(2a-x)\big]\,dx$.
4. This equals $2\int_0^a f(x)\,dx$ exactly when $f(2a-x)=f(x)$ for all $x\in[0,a]$, i.e. the graph is symmetric about the vertical line $x=a$. …