Integration and Definite Integrals

The integral of sin(x) dx is -cos(x) + C. This is because the derivative of cos(x) is -sin(x), so the derivative of -cos(x) is sin(x). Positive cos(x) is the derivative of sin(x), not its integral.

Trapezodial rule is $\int_{x_0}^{x_0 + nh}dx =\dfrac{h}{2}[(y_0 + y_n) + 2(y_1 + y_2 + .... + y_{n-1})]$ $\int_{1}^{2}xe^xdx ]$ = 1000e

$\int^{\pi/4}_0 \dfrac{(1-\tan x)}{(1+\tan x)}dx = \int^{\pi/4}_0 \dfrac{(\cos x-\sin x)}{(\cos x+\sin x)}dx$ Let, cos x + sin x = t $\therefore$(- sinx + cos x) dx = dt

$\int^{\sqrt 2}_0 \dfrac{dt}{t}dx = [\log t] = \log \sqrt 2 = \dfrac{1}{2} \log 2$

$\int_0^{\pi/2} \dfrac{e^{ix}}{e^{-ix}}dx = \int_0^{\pi/2} e^{2ix} dx$ $\left( \dfrac{e^{2ix}}{2i}\right) = \dfrac{1}{2i}[e^{ix}-1] = \dfrac{1}{2i}[\cos \pi + i \sin \pi -1] =\dfrac{1}{2i}[-1 + 0 -1] = \dfrac{-2}{2i} = \dfrac{-1}{i} \times \dfrac{i}{i} = \dfrac{-i}{-1} = i $

$I = \int\limits_0^\pi sin^3 \theta d \theta \\ = \int\limits_0^\pi \bigg(\frac{3sin\theta - sin 3 \theta}{4}\bigg)d\theta \hspace{2cm} sin 3 \theta = 3 sin \theta - 4 sin^3 \theta \\ = \bigg[\frac{-3}{4}cos \theta\bigg]_0^\pi = \bigg[\frac{\omega s 3 \theta}{12}\bigg]_0^\pi = \bigg[ \frac{3}{4}+ \frac{3}{4}\bigg]-\bigg[ \frac{1}{12}+ \frac{1}{12}\bigg] = \frac{4}{3}$

Using the property integral from 0 to a of f(x) dx = integral from 0 to a of f(a-x) dx, let I be the integral. Then I = integral from 0 to a of sqrt(a-x)/(sqrt(a-x) + sqrt(x)) dx. Adding the two expressions for I gives 2I = integral from 0 to a of 1 dx = a, so I = a/2.

The integral of (sin t)/t from 0 to infinity is a standard Dirichlet integral, which evaluates to pi/2.

Correct options is (2).

$\int\limits_{-a}^a F(x) = 0 \\ F(x) = \int\limits_0^2\frac{(x-1)^2 sin (x-1)}{(x-1)^2 + cos(x-1)}dx=0$

Correct option is (2).