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## A Tricky Integral

Let’s solve a beautiful and tricky integral problem.

The Problem:

Let $$I=\int e^x/(e^{4x}+e^{2x}+1) dx$$ $$J= \int e^{-x}/(e^{-4x}+e^{-2x}+1)dx$$. Find the value of (J-I).

Solution:

$$I=\int e^x/(e^{4x}+e^{2x}+1) dx$$

$$J= \int e^{-x}/(e^{-4x}+e^{-2x}+1)dx$$

Let (e^x)=(z)

$$J-I=\int\frac{e^x(e^{2x-1})}{e^{4x}+e^{2x}+1}dx=\int\frac{z^2-1}{z^4+z^2+1}dz$$

$$=\frac{1}{2}ln\frac{(e^x+e^-x-1)}{(e^x+e^-x+1)}+c$$ ( where c is a constant of integration)

Categories

## Dimensional Analysis

Let’s discuss a problem useful for Physics Olympiad based on Dimensional Analysis.

The Problem: Dimensional Analysis

Consider an expression F=Ax(sin^{-1} (Bt)) where F represents force, x represents distance and t represents time. Dimensionally the quantity AB represents

(A) energy

(B) surface tension

(C) intensity of light

(D) pressure

Solution:

The quantity Ax on RHS must have the dimensions of force since there is force on LHS. B must have dimension of reciprocal of time. Thenn, the product AB will have the dimensions of energy per unit area per unit time, the same as those of intensity of light.

Categories

## Dimensional Analysis

The distance travelled by an object is given by x=(at+bt2)/(c+a) where t is time and a,b,c are constants. The dimension of b and c respectively are:

• [L2T-3], [LT-1]
• [LT-2], [LT-1]
• [LT-1], [L2T1]
• [LT-1], [LT2]

Solution:

The dimension of x is L. Hence, the dimension of a must be LT-1.

Dimension of c and a are the same.

Dimension of b therefore is L2T-3.