Here's a quick look at some of the examples of functions which we will learn how to integrate, applying some of the standard formulas and principals of integration . Most of these elementary examples of integration involve trigonometric and exponential functions, polynomials etc.

(i) e^{x} sin e^{x} *(Solved by substituting t = **e*^{x)}

(ii) cos^{3} x sin x *(Solved by substituting t = cos x, so dt/dx = -sin x)*

(iii) x^{5}/(1+x^{12}) *(Solved by substituting t = x*^{5}^{)}

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(iv) sin^{4} x cos x *(Solved by substituting t = sin x, so dt/dx = cos x)*

*(v) sin3 x, cos x cos 2x (Solved by using compound and multiple angle formulas from trigonometry)*

*(vi) *(x + 1)(x + log x)^{2} / 2x

(vii) xe^{x} / (x + 1)^{2}

(viii) e^{x} (1 − sin x) / (1 − cos x)

(ix) e^{3x} sin 4x

(x) e^{4x} cos 2x cos 4x

We will also pick up a few simple but illustrative examples of computing area(s) under curves using definite integration, between limits.

We also sketch the curves to help students get a better idea of what they're actually doing.

**Problem: Find the area of the region bounded by the parabola y**^{2} = 4x and the line y = 4x.

*We first find the points of intersection which are (0, 0) and (1/4,1). We then proceed to evaluate the integral of (equation of parabola - line) within the limits x=0 and x=1/4. This will give us the required area. We also sketch the curves involved: the parabola and the line, to help users get a better understanding via visualizations. *

**Problem: Find the area bounded by the curve y =1/x**^{2} , the axis of x and the ordinate x = 1.

*We solve this problem using definite integrals as well as concepts from limits. Let A be the point where the ordinate intersects the x-axis, i.e. OA = 1. We take a variable ordinate MP where OM = t and consider the area of the finite region GAMP G. If this area tends to a finite limit as the ordinate MP recedes to infinity, then this limit is said to be the area of the infinite region under consideration. We then solve another similar problem, where the curve is replace by 1/sqrt(x)*