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We can easily find the integration of $\frac {1}{1+ cos x}$ and $\frac {1}{1 -cos x}$ using trigonometric identities and fundamental integration techniques, The formula of these integral is given as $\int \frac {1}{1+ cos x} \; dx = \tan \frac {x}{2} + C$ or $\int \frac {1}{1+ cos x} \; dx = -\cot x […]
integration of 1/1+cosx , 1/1 -cos x Read More »
The integration of tan cube x $\tan ^3 x$, can be found using integration substitution and trigonometry identities . The integral of $\tan ^3 x$ with respect to (x) is: \[\int \tan^3 x \, dx =\frac {1}{2} \tan^2 x – \ln |sec x|+ C\] Here, (C) represents the constant of integration, which is added because
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We already know about electric field and electric potential. We also know that electrostatic field is completely characterized by vector function E(r). The electric field depicts the force exerted on other electrically charged objects by the electrically charged particle the field is surrounding. Now a question arises why do we need introduction of electric potential when we already have electric field for the description of electric force between charges.
Use of concept of electric potential Read More »
Question: If you are riding on a train that speeds past other train moving in the same direction on adjacent train.It appears the other train is moving backward. Why?
Solution:
Your reference frame is that of the train you are riding. If you are traveling with a relatively constant velocity (not over
Kinematics Good conceptual problems Read More »
Since the car is turning on a curve, The car will be accelerating as direction of the car changes even the velocity is constant. This acceleration is centripetal acceleration And centripetal force will be providing it,
There are three cases with turning
1) Unbanked curve.
2) Banked curve with no friction
3) Banked curve with friction
Mechanics of Turning a car on the curve Read More »
For integration of $\log(\tan x)$, we generally consider the definite integral over the interval from 0 to $\pi/2$ To calculate the definite integral of $\log(\tan x)$ from (0) to $\pi/2$, we use a technique involving symmetry and the properties of logarithms. The integral is: \[\int_{0}^{\pi/2} \log(\tan x) \, dx\] Let $I=\int_{0}^{\pi/2} \log(\tan x) \, dx$
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For integration of $\log(\cos x)$, we generally consider the definite integral over the interval from 0 to $\pi/2$ To calculate the definite integral of $\log(\cos x)$ from (0) to $\pi/2$, we use a technique involving symmetry and the properties of logarithms. The integral is: \[\int_{0}^{\pi/2} \log(\cos x) \, dx\] Let $I=\int_{0}^{\pi/2} \log(\cos x) \, dx$
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The integral ($\int e^{ax} \sin(bx) \, dx$) is computed as: \[\int e^{ax} \sin(bx) \, dx= \frac{a e^{ax} \sin(bx)}{a^2 + b^2} – \frac{b e^{ax} \cos(bx)}{a^2 + b^2}+ C\] Proof of Integration To solve the integral $\int e^{ax} \sin(bx) \, dx$, we can use the method of integration by parts, which is based on the formula: $$\int
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