Q.Evaluate $\int_{0}^{1} \dfrac{\log(1+x)}{1+x^2}\, dx$.

Show $f(\pi/2-x)=-f(x)$, which forces the definite integral to equal its own negative, hence zero.

Given $f(x)=\dfrac{\sin x-\cos x}{1+\sin x\cos x}$ and the property $\displaystyle\int_0^a f(x)dx=\int_0^a f(a-x)dx$ with $a=\dfrac{\pi}{2}$.

$f\left(\dfrac{\pi}{2}-x\right)=\dfrac{\sin\left(\frac{\pi}{2}-x\right)-\cos\left(\frac{\pi}{2}-x\right)}{1+\sin\left(\frac{\pi}{2}-x\right)\cos\left(\frac{\pi}{2}-x\right)}=\dfrac{\cos x-\sin x}{1+\cos x\sin x}=-f(x)$

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Use $g(\pi/4-x) = \log 2 - g(x)$ (from the tangent-subtraction formula) together with the given property to solve for the integral.

Given $g(x)=\log(1+\tan x)$, evaluate $I=\displaystyle\int_0^{\pi/4}g(x)\,dx$ using $\displaystyle\int_0^a f(x)dx=\int_0^a f(a-x)dx$ with $a=\dfrac{\pi}{4}$.

$\tan\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-\tan x}{1+\tan x}$

$1+\tan\left(\dfrac{\pi}{4}-x\right)=1+\dfrac{1-\tan x}{1+\tan x}=\dfrac{(1+\tan x)+(1-\tan x)}{1+\tan x}=\dfrac{2}{1+\tan x}$

$g\left(\dfrac{\pi}{4}-x\right)=\log\dfrac{2}{1+\tan x}=\log 2-\log(1+\tan x)=\log 2-g(x)$

By the given property:

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Substitute $x=a+b-t$ into the left side; the limits swap and the integral reduces to the right side.

Let $I=\displaystyle\int_a^b f(x)\,dx$.

Put $x = a+b-t$, so $dx=-dt$.

When $x=a$, $t=b$; when $x=b$, $t=a$.

$I = \displaystyle\int_{b}^{a} f(a+b-t)\,(-dt) = \int_a^b f(a+b-t)\,dt$

Since the definite integral's value does not depend on the name of the dummy variable, we may replace $t$ by $x$:

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Use the property $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ to show the integral equals its own complement, then solve.

Let $I=\displaystyle\int_0^{\pi/2}\dfrac{\sin^4x}{\sin^4x+\cos^4x}\,dx$ ... (1)

Using $\displaystyle\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ with $a=\dfrac{\pi}{2}$, and $\sin\left(\dfrac{\pi}{2}-x\right)=\cos x$, $\cos\left(\dfrac{\pi}{2}-x\right)=\sin x$:

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Use the property $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ to show the integral equals its own complement, then solve for it.

Let $I=\displaystyle\int_0^{\pi/2}\dfrac{\sin^3x}{\sin^3x+\cos^3x}dx$.

Using $x\to \dfrac\pi2-x$ (so $\sin x\to\cos x$, $\cos x\to\sin x$):

$I=\displaystyle\int_0^{\pi/2}\dfrac{\cos^3x}{\cos^3x+\sin^3x}dx$

Adding both expressions for $I$:

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