Definite Integration

Let \(x=a+b-u\).

$dx=-du .$

When \(x=a, u=b\);

when \(x=b, u=a\).

${\int }_a^{b}f\left(x\right)dx={\int }_{u=b}^{u=a}f\left(a+b-u\right)\left(-du\right)$

$\Rightarrow {\int }_a^{b}f\left(x\right)dx={\int }_{u=a}^{u=b}f\left(a+b-u\right)du$

Rename \(u\) to \(x\) :

\(\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\).

Let $f\left(x\right)=x\left|x\right|$

$f(-x)=-x\left|x\right|=-f\left(x\right)$

So, $f\left(x\right)$ is an odd function.

We know that if $f\left(x\right)$ is an odd function then ${\int }_{-a}^{a}f\left(x\right)dx=0$.

Thus, ${\int }_{-1}^{1}x\left|x\right|\mathrm{d}x=0$

Let, $I={\int }_0^{\pi /2}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$

Using ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx$

$\Rightarrow I={\int }_0^{\pi /2}\frac{\cos x-\sin x}{1+\sin x\cos x}dx$

$\Rightarrow I=-{\int }_0^{\pi /2}\frac{\sin x-\cos x}{1+\sin x\cos x}dx=-I$

$\Rightarrow 2I=0\Rightarrow I=0$

$I={\int }_0^{\pi /2}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$

Put $x\mapsto \frac{\pi }{2}-x$

$\Rightarrow I={\int }_0^{\pi /2}\frac{\cos x-\sin x}{1+\sin x\cos x}dx$

$\Rightarrow I=-{\int }_0^{\pi /2}\frac{\sin x-\cos x}{1+\sin x\cos x}dx=-I$

$2I=0\Rightarrow I=0$

Let $f\left(x\right)=x\left|x\right|$

$f(-x)=-x\left|x\right|=-f\left(x\right)$

So, $f\left(x\right)$ is an odd function.

We know that if $f\left(x\right)$ is an odd function then ${\int }_{-a}^{a}f\left(x\right)dx=0$.

Thus, ${\int }_{-1}^{1}x\left|x\right|\mathrm{d}x=0$

Let \(x=a+b-u\).

$d x=-d u .$

When \(x=a, u=b\); when \(x=b, u=a\).

${\int }_a^{b}f\left(x\right)dx={\int }_{u=b}^{u=a}f(a+b-u)(-du)$

$\Rightarrow {\int }_a^{b}f\left(x\right)dx={\int }_{u=a}^{u=b}f(a+b-u)du$

Rename \(u\) to \(x\) : \(\int_a^b f(x) d x=\int_a^b f(a+b-x) d x\).

\(f(x)=[x]^2-[x]-6=([x]+2)([x]-3)\)

\(f(x)=0 \Rightarrow[x]=3\) or \([x]=-2\)

Infinitely many solutions.

\(f(x)>0 \Rightarrow[x] \in(-\infty,-2) \cup(3, \infty)\)

\(\Rightarrow x \in(-\infty,-2) \cup[4, \infty)\)

\(f(x)<0 \Rightarrow[x] \in(-2,3)\)

\(\Rightarrow x \in[-1,3)\)

\(\int_0^2 f(x) d x=\int_0^1(0-0-6) d x+\int_1^2(1-1-6) d x\)

\(=-6-6 =-12\)

\( I=\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{d x}{1+\sqrt[3]{\tan 2 x}} \)

\(I=\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{\cos ^{1 / 3} 2 x}{\sin ^{1 / 3} 2 x+\cos ^{1 / 3} 2 x} d x \ldots (1)\)

Using king rule :

\(I=\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} \frac{\sin ^{1 / 3} 2 x}{\cos ^{1 / 3} 2 x+\sin ^{1 / 3} 2 x} \ldots (2)\)

adding (1) and (2), we get

\(2 I=\int_{\frac{\pi}{24}}^{\frac{5 \pi}{24}} 1 d x \)

\(2 I=\frac{5 \pi}{24}-\frac{\pi}{24}=\frac{\pi}{6} \)

\(I=\frac{\pi}{12}\)

$I={\int }_{1/4}^{3/4}\cos \left(2co{t}^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx$

${\int }_{1/4}^{3/4}\cos \left(2\left({\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right)\right)dx$

${\int }_{1/4}^{3/4}\frac{1-{\tan }^2\left({\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right)}{1+{\tan }^2\left({\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right)}dx$

$={\int }_{1/4}^{3/4}\frac{1-\left(\frac{1+x}{1-x}\right)}{1+\left(\frac{1+x}{1-x}\right)}dx{\int }_{1/4}^{3/4}\frac{-2x}{2}dx$

$={\int }_{1/4}^{3/4}(-x)dx=-{\left(\frac{x^2}{2}\right)}_{1/4}^{3/4}$

$=-\frac{1}{2}\left[\frac{9}{16}-\frac{1}{16}\right]$

$=-\frac{1}{4}$

$I={\int }_{1/4}^{3/4}\cos \left(2co{t}^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx$

${\int }_{1/4}^{3/4}\cos \left(2\left({\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right)\right)dx$

${\int }_{1/4}^{3/4}\frac{1-{\tan }^2\left({\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right)}{1+{\tan }^2\left({\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right)}dx$

$={\int }_{1/4}^{3/4}\frac{1-\left(\frac{1+x}{1-x}\right)}{1+\left(\frac{1+x}{1-x}\right)}dx{\int }_{1/4}^{3/4}\frac{-2x}{2}dx$

$={\int }_{1/4}^{3/4}(-x)dx=-{\left(\frac{x^2}{2}\right)}_{1/4}^{3/4}$

$=-\frac{1}{2}\left[\frac{9}{16}-\frac{1}{16}\right]$

$=-\frac{1}{4}$

Let \(I=\int_0^{2 a} f(x) d x\)

\(I=\int_0^a f(x) d x+\int_a^{2 a} f(x) d x\)

In \(\int_a^{2 a} f(x) d x\), put \(x=2 a-t, d x=-d t\)

\(\int_a^{2 a} f(x) d x=\int_a^0 f(2 a-t)(-d t)=\int_0^a f(2 a-t) d t\)

Given \(f(2 a-t)=f(t)\)

\(\Rightarrow \int_a^{2 a} f(x) d x=\int_0^a f(t) d t\)

\(\Rightarrow I=\int_0^a f(x) d x+\int_0^a f(x) d x=2 \int_0^a f(x) d x\)

Let \(I=\int_0^{2 a} f(x) d x\)

Using the property of breaking the limits :

\(I=\int_0^a f(x) d x+\int_a^{2 a} f(x) d x\)

In \(\int_a^{2 a} f(x) d x\), put \(x=2 a-t, d x=-d t\)

\(\int_a^{2 a} f(x) d x=\int_a^0 f(2 a-t)(-d t)=\int_0^a f(2 a-t) d t\)

Given \(f(2 a-t)=f(t)\)

\(\Rightarrow \int_a^{2 a} f(x) d x=\int_0^a f(t) d t\)

Now, using the property of changing the variable :

\(\Rightarrow I=\int_0^a f(x) d x+\int_0^a f(x) d x=2 \int_0^a f(x) d x\)