Properties of Definite Integrals

Let $I = \int_0^{\pi/2} \cos^2 x dx \dots (1)$
Using $P_3: I = \int_0^{\pi/2} \cos^2(\frac{\pi}{2}-x) dx = \int_0^{\pi/2} \sin^2 x dx \dots (2)$
Adding (1) & (2): $2I = \int_0^{\pi/2} (\sin^2 x + \cos^2 x) dx = \int_0^{\pi/2} 1 dx = [x]_0^{\pi/2} = \frac{\pi}{2}$.
$2I = \frac{\pi}{2} \implies I = \frac{\pi}{4}$.

Let $I = \dots (1)$. Using $P_3$, replace $x \to \frac{\pi}{2}-x$.
$I = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx \dots (2)$
Adding: $2I = \int_0^{\pi/2} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} dx = \int_0^{\pi/2} 1 dx = \frac{\pi}{2}$.
$I = \frac{\pi}{4}$.

Same logic as Q2. The powers cancel out upon adding.
Result: $I = \frac{\pi}{4}$.

Same logic as Q2. $f(x) + f(\frac{\pi}{2}-x) = 1$.
Result: $I = \frac{\pi}{4}$.

$|x+2|$ changes sign at $x=-2$. Split integral at -2.
$I = \int_{-5}^{-2} -(x+2) dx + \int_{-2}^5 (x+2) dx$.
$= -[\frac{x^2}{2}+2x]_{-5}^{-2} + [\frac{x^2}{2}+2x]_{-2}^5$.
$= -[(2-4) – (\frac{25}{2}-10)] + [(\frac{25}{2}+10) – (2-4)]$.
$= -[-2 – 2.5] + [22.5 – (-2)] = 4.5 + 24.5 = 29$.

Split at $x=5$. $I = \int_2^5 -(x-5) dx + \int_5^8 (x-5) dx$.
$= -[\frac{x^2}{2}-5x]_2^5 + [\frac{x^2}{2}-5x]_5^8$.
$= -[(\frac{25}{2}-25) – (2-10)] + [(32-40) – (\frac{25}{2}-25)]$.
$= -[-12.5 + 8] + [-8 + 12.5] = 4.5 + 4.5 = 9$.

Use $P_3$: replace $x$ with $1-x$.
$I = \int_0^1 (1-x)(1-(1-x))^n dx = \int_0^1 (1-x)x^n dx = \int_0^1 (x^n – x^{n+1}) dx$.
$= [\frac{x^{n+1}}{n+1} – \frac{x^{n+2}}{n+2}]_0^1 = \frac{1}{n+1} – \frac{1}{n+2} = \frac{1}{(n+1)(n+2)}$.

Use $P_3$: replace $x$ with $\frac{\pi}{4}-x$.
$I = \int_0^{\pi/4} \log(1 + \tan(\frac{\pi}{4}-x)) dx = \int_0^{\pi/4} \log(1 + \frac{1-\tan x}{1+\tan x}) dx$.
$= \int_0^{\pi/4} \log(\frac{1+\tan x + 1-\tan x}{1+\tan x}) dx = \int_0^{\pi/4} \log(\frac{2}{1+\tan x}) dx$.
$I = \int_0^{\pi/4} (\log 2 – \log(1+\tan x)) dx = \int_0^{\pi/4} \log 2 dx – I$.
$2I = \frac{\pi}{4}\log 2 \implies I = \frac{\pi}{8}\log 2$.

$= \int_0^{\pi/2} (\log \sin^2 x – \log(2\sin x \cos x)) dx = \int_0^{\pi/2} \log(\frac{\sin^2 x}{2\sin x \cos x}) dx$.
$= \int_0^{\pi/2} \log(\frac{\tan x}{2}) dx = \int_0^{\pi/2} (\log\tan x – \log 2) dx$.
We know $\int_0^{\pi/2} \log\tan x dx = 0$.
Result: $-\frac{\pi}{2}\log 2 = \frac{\pi}{2}\log(1/2)$.

$\sin^2 x$ is an even function. $I = 2\int_0^{\pi/2} \sin^2 x dx$.
From Q1, we know $\int_0^{\pi/2} \sin^2 x dx = \frac{\pi}{4}$.
$I = 2(\frac{\pi}{4}) = \frac{\pi}{2}$.

$f(x) = \sin^7 x$. $f(-x) = (\sin(-x))^7 = (-\sin x)^7 = -\sin^7 x = -f(x)$.
Since $f$ is an odd function, integral is $0$.

Let $I = \dots (1)$. Replace $x \to \frac{\pi}{2}-x$.
$I = \int_0^{\pi/2} \frac{\cos x – \sin x}{1+\cos x \sin x} dx = – \int_0^{\pi/2} \frac{\sin x – \cos x}{1+\sin x \cos x} dx = -I$.
$2I = 0 \implies I = 0$.

Let $I = \int_0^\pi \log(1+\cos x) dx$.
Using $P_3$: $I = \int_0^\pi \log(1+\cos(\pi-x)) dx = \int_0^\pi \log(1-\cos x) dx$.
$2I = \int_0^\pi \log(1-\cos^2 x) dx = \int_0^\pi \log(\sin^2 x) dx = 2\int_0^\pi \log\sin x dx$.
$I = \int_0^\pi \log\sin x dx = 2\int_0^{\pi/2} \log\sin x dx$.
Standard Result: $\int_0^{\pi/2} \log\sin x dx = -\frac{\pi}{2}\log 2$.
$I = 2(-\frac{\pi}{2}\log 2) = -\pi \log 2$.

Using $P_3$, replace $x \to a-x$. Same logic as Q2.
$2I = \int_0^a 1 dx = a \implies I = a/2$.

Split at $x=1$. $I = \int_0^1 -(x-1) dx + \int_1^4 (x-1) dx$.
$= -[\frac{x^2}{2}-x]_0^1 + [\frac{x^2}{2}-x]_1^4$.
$= -[-1/2] + [(8-4) – (1/2-1)] = 0.5 + 4.5 = 5$.

Given $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$.
$I = \int_0^a f(a-x)g(a-x) dx = \int_0^a f(x)(4-g(x)) dx$.
$I = 4\int_0^a f(x) dx – \int_0^a f(x)g(x) dx$.
$I = 4\int_0^a f(x) dx – I \implies 2I = 4\int_0^a f(x) dx \implies I = 2\int_0^a f(x) dx$.

Let $f(x) = x^3 + x\cos x + \tan^5 x$. All are odd functions ($f(-x)=-f(x)$).
$\int_{-\pi/2}^{\pi/2} f(x) dx = 0$.
Remaining term: $\int_{-\pi/2}^{\pi/2} 1 dx = [x]_{-\pi/2}^{\pi/2} = \pi$.

Let $I = \dots (1)$. Replace $x \to \frac{\pi}{2}-x$.
$I = \int_0^{\pi/2} \log(\frac{4+3\cos x}{4+3\sin x}) dx = \int_0^{\pi/2} -\log(\frac{4+3\sin x}{4+3\cos x}) dx = -I$.
$2I = 0 \implies I = 0$.