$\int_a^b f(x) d x=F(b)-F(a)$

$F^{\prime}(x)=f(x)$

Remark:- $\quad(F(x)+c)^{\prime}=f(x)$

$\int_a^b f(x) d x =F(x)+\left.c\right|_a ^b$

$=F(b)+c-(F(a)+c)$

$=F(b)-F(a)$

$\underline{{\text{Ex.}}} \quad \int_0^1 \frac{1}{1+x^2} d x$

$=\left.\tan ^{-1} x\right|_{x=0} ^{x=1}$

$=\left.(\tan ^{-1} 1\right)-(\tan ^{-1} 0)$

$[ \quad \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2} ]$

$\frac{\pi}{4}-0=\frac{\pi}{4}$

$\int_a^b f(x) d x=F(b)-F(a)$

$F^{\prime}(x)=f(x)$

It is not easy to compute anti derivative of $f(x)$. always.

Then what to do?

$\underline{{\text{Ex:}}}$

$I=\int_0^{\pi / 4} {2 \tan x\left(\sec ^2 x d x) .\right.}$

${\tan x=t}$

$\sec ^2 x d x=d t$

$x=0, tan0=0=1 \\ x=\frac{\pi}{4}, \tan \frac{\pi}{4}=1=t$

$I=\int_{t=0}^1 2 t d t$

$=\left.2 \frac{t^2}{2}\right|_0 ^1$

$=1-0=1$

$I =\int_a^b f(g(x)) \quad {g^{\prime}(x) d x}$

$\quad g(x)=u$

$\biggl[{x=a, u=g(a)} \ \text{and} \ x=b, u=g(b)\biggl]$

$g^{\prime}(x) d x=d u$

$I =\int_{g(a)}^{g(b)} f(u)du$

$\quad \boxed{f(x)=F^{\prime}(x)}$

$\underline{{\text{Ex:}}}$

$I=\int_1^4 \frac{3 \sqrt{x}}{\left(1+x^{3 / 2}\right)^2} d x$

$\frac{1}{1+x^{3 / 2}}=u$

$\biggl[ x=1, \quad u=\frac{1}{2} \\ \quad x=4, u=\frac{1}{1+8}=\frac{1}{9} \biggl]$

$\frac{-1 \frac{3}{2} x^{1 / 2}}{\left(1+x^{1 / 2}\right)^2} d x=d u$

$[ \quad \frac{3 \sqrt{x} d x}{\left(1+x^{3 / 2}\right)^2}=-2 d u ]$

$I=\int_{1 / 2}^{1 / 9}-2 d u=-\left.2 u\right|_{1 / 2} ^{1 / 9}$

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

$=-\frac{2}{5}+1=7 / 9$

$\underline{{\text{Ex:}}}$

$\int_0^2 \frac{x^3 d x}{\sqrt{x^4+9}}$

$\sqrt{\left(x^4+9\right)}=t$

$\frac{1}{2}(\left(x^4+9\right)^{-1 / 2} 4 x^3 d x=d t$

$\frac{ x^3 d x}{\sqrt{x^4+9}}=\frac{d t}{2}$

$x=0 \quad t=\sqrt{9}=3$

$x=2 \quad t=\sqrt{25}=5$

$=\int_3^5 \frac{d t}{2}=\frac{1}{2}(5-3)=1$

$\underline{{\text{Problem.}}}$

$I=\int_{-1}^{-1 / 2} 4 t^{-2} \sin ^2\left(1+\frac{t}{t}\right) d t$

$1+\frac{1}{t}=u ;-t^{-2} d t=d u, t^{-2} d t=-d u$

$\quad t=-1, u=0 ; \quad t=-\frac{1}{2}, u=-1$

$I= -\int_0^{-1} 4 \sin ^2 u d u=-4 \int_0^{-1} \frac{1-\cos 2 u}{2} d u$

$= -2 \int_0^{-1}(1-\cos 2 u) d u=-2\left[u-\frac{\sin 2 u}{2}\right]_0^{-1}$

$=-2\left[-1-0-\frac{\sin (-2)}{2}+0\right]=2-\sin 2$

$\underline{\text{Property 1:-}}$

$\int_a^b f(x) d x=\int_a^b f(t) d t$

$x=t \quad d x=dt$

$x=a \rightarrow t=a$

$x=b \rightarrow t=b$

$\underline{\text{Property 2:}}$

$\int_a^b f(x) d x=-\int_b^a f(x) d x$

in particular $\int_a^a f(x) d x=0$

$\int_a^b f(x) d x=F(b)-F(a)=-[F(a)-F(b)]$

$=-\int_b^a f(x)dx$

$\underline{\text{Property \ 3:-}}$

a < c < b

$\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x$

$F^{\prime}(x)=f(x)$

$\int_a^b f(x) d x=F(b)-F(a)=F(b)-F(c)+F(c)-F(b)$

$\int_a^c f(x) d x=F(c)-F(a)$

$\int_c^b f(x) d x=F(b)-F(c)$

$=\int_c^b f(x)dx + \int_a^c f(x) dx$

$\underline{\text{Property 4.}}$

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

$\int_a^b f(x) d x \quad x=a+b-t , dx = -dt$

$\biggl[ x = a , t = b \\ \quad x = b, t = a \biggl]$

$=\int_{t=b}^{t=a} f(a+b-t)(-d t)$

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

$\underline{\text{Property 5.}}$

$\quad \int_0^a f(x) d x=\int_0^a f(a-x) d x$

$\int_0^a f(x) d x \quad x=a-t \quad x=0, t=a$

$[ \quad dx = -dt \quad x=a, t=0 ]$

$=\int_{t=a}^{t=0} f(a-t)-d t=\int_0^a f(a-t) d t$

$=\int_0^a f(a-x) d x$

$\underline{\text{Property 6.}}$

$\int_0^{2a} f(x) dx = \int_0^a f(x) dx+ \int_a^{2a} f(x) dx \rightarrow (Property 3)$

$\int_0^{2 a} f(x) d x \quad x=2 a-t ,d x=-d t$

$=\int_0^0 f(2 a-t) (-dt)$

$x=a, \rightarrow t=a$

$x=2 a \rightarrow t=0$

$=\int_0^a f(2 a-k) d t =\int_0^a f(2 a-x) d x$

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

$\int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_0^a f(2 a-x) d x$

$\underline{\text{Property 7.}}$

$\int_0^{2 a} f(x) d x=\int_0^a f(x) d x+\int_0^a f(2 a-2) d x \rightarrow 1$

$\text { If }f(2 a-x)=f(x) \text { then }$

$\left.\int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\right]$

$\text { if } f(2 a-x)=-f(x)$

$\left.\int_0^{2 a} f(x) d x=0\right)$

$\underline{\text{Property 8.}}$

$\int_{-a}^a f(x) d x=\left\{\begin{array}{l}2 \int_0^a f(x) d x \ if f(a) \text { is even }\end{array}\right.$

$\int_{-a}^a f(x) d x=0$ if $f(x)$ is odd

even function $f(-x)=f(x)$

odd function $f(-x)=-f(x)$.

$\int_{-a}^0 f(x) d x =\int_a^0 f(-t)(-d t) \quad x=-t , d x=-d t$

$=\int_0^a f(-t) d t \quad x=-a , t=a ,x=0 , t=0$

$\int_{-a}^a f(x) d x =\int_0^a f(-t) d t+\int_0^a f(t) d t$

$\int_{-a}^a f(x) d x=\int_0^a f(-x) d x+\int_0^a f(x) d x$

If $f(x)$ is even, i.e., $f(-x)=f(x)$

$\int_{-a}^a f(x) d x=2 \int_{-a}^a f(x) d x$

$\text { If } f(-x)=-f(x)$

$\int_{-a}^a f(x) d x=0$

$\int_{-a}^a f(x) d x=\int_0^a f(-x) d x+\int_0^a f(x) d x$

If $f(x)$ is even, i.e., $f(-x)=f(x)$

$\int_{-a}^a f(x) d x=2 \int_{-a}^a f(x) d x$

$\text { If } f(-x)=-f(x)$

$\int_{-a}^a f(x) d x=0$

$= \int_0^2-(x-2) d x+\int_0^4(x-2) d x$

$= -\frac{x^2}{2}+\left.2 x\right|_0 ^2+\frac{x^2}{2}-\left.2 x\right|_2 ^4$

$= -\frac{4}{2}+4-0+\frac{16}{2}-8-\frac{4}{2}+4$

= -2 + 4 - 0 + 8 - 8 - 2 + 4 = 4

$I =\int_0^{\pi / 2} \frac{4 \sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \int_0^a f(x) d x$

$I =\int_0^{\pi / 2} \frac{4 \sqrt{\sin \left(\frac{\pi}{2}-x\right)} d x}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}=\int_0^a f(a-x) d x$

$I =\int_0^{\pi / 2} \frac{4 \sqrt{\cos x} d x}{\sqrt{\cos x}+\sqrt{\sin x}}$

$\text { (1) }+(2)$

$2 I =\int_0^{\pi / 2} 4 d x=4 \times \frac{\pi}{2}$

$I = {\pi}$

$I=2 \int_0^{\pi / 2} \log \cos x d x \cdot-(1) \int_0^\pi f(x) d x$

$I=2 \int_0^{\pi / 2} f(\log \cos (\frac{\pi}{2}-x)) d x$

$I=2 \int_0^{\pi / 2} \log \sin x d x-2$

(1) + (2)

$2 I =2 \int_0^{\pi / 2}(\log \sin x+\log \cos x) d x$

$I =\int_0^{\pi / 2} \log \left(\frac{2 \sin x \cos x}{2}\right) d x$

$I =\int_0^{\pi / 2} \log \frac{\sin 2 x}{1} d x-\int_0^{\pi / 2} \log 2 d x$

2 x=k

$I=\int_0^\pi \log \sin t\left(\frac{1}{2} d t\right)-\frac{\pi}{2} \log 2$

$=\frac{1}{2} \int_0^\pi \log \sin t d t-\frac{\pi}{2} \log 2$

$\int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x \quad f(2 a-x)=f(x)$

$I=\frac{1}{2} \times 2 \int_0^{\pi / 2} \log \sin (\pi-t) d t-\frac{\pi}{2} \log 2$

$I=\int_0^{\pi / 2} \log \sin t d t-\frac{\pi}{2} \operatorname{log}$

$I =\int_0^1 x(1-x)^n d x . \quad \int_0^a f(x) d x$

$I =\int_0^1(1-x)(1-(1-x))^n d x=\int_0^a f(a-x) d x$

$=\int_0^1(1-x) x^n d x=\int_0^1 x^n-x^{n+1} d x$

$=\frac{x^{n+1}}{n+1}-\left.\frac{x^{n+2}}{n+2}\right|_0 ^1$

$=\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}$

$I =\int_0^1 x(1-x)^n d x . \quad \int_0^a f(x) d x$

$I =\int_0^1(1-x)(1-(1-x))^n d x=\int_0^a f(a-x) d x$

$=\int_0^1(1-x) x^n d x=\int_0^1 x^n-x^{n+1} d x$

$=\frac{x^{n+1}}{n+1}-\left.\frac{x^{n+2}}{n+2}\right|_0 ^1$

$=\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}$

$I =\int_0^\pi \frac{xd x}{1+\sin x} =\int_0^\pi \frac{\pi-x}{1+\sin(\pi-x)} d x$

$[{\int_0^a f(x) d x}] = [{\int_0^a f(a-x) dx }]$

$=\int_0^\pi \frac{(\pi-x) d x}{1+\sin x} ,\quad 2 I=\int_0^\pi \frac{\pi d x}{1+\sin x}$

$2 I=\pi \int_0^\pi \frac{d x}{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2}=\pi \int_0^\pi \frac{\sec ^2 \frac{x}{2} d x}{\left(1+\tan \frac{x}{2}\right)^2}$

Let $\tan \frac{x}{2}=t \frac{1}{2} \sec ^2 \frac{x}{2} d x=d t \quad{2 I=2 \pi}$

$2 I=\pi \int_0^{\infty} \frac{2 d t}{(1+t)^2}=\left.\pi \frac{(-2)}{1+t}\right|_0 ^{\infty}=-2 \pi(0-1)$