Integrating quadratic polynomials

Integrating quadratic polynomials involves transformation the quadratic polynomial into a trigonometric expression which we can easily integrate. A quadratic polynomial is an algebraic expression of the form ax²+bx+c. It is a polynomial where the highest power is 2. Polynomials are algebraic expressions that consist of variables and coefficients. 

consider the expression:

The best approach is ensuring that the quadratic polynomial is in the form of a perfect square.

The nearest perfect square to our quadratic polynomial is given by

(x+1)² =(x+1)(x+1)= (x²+2x + 1)

(x²+2x + 1)+1 =x²+2x+2

thus x²+2x+2 = (x+1)² +1

using the above to rewrite our integral we get:

then using trigonometric substitution, we have:

then differentiating x in terms of tanθ;

then we rewrite the integral:

From the above transformation, the quadratic expressions has been converted to a trigonometric expression.

tan²θ+1 = sec²θ

from the relation:

hence we have:

Example problems involving Integrating quadratic polynomials

1. Evaluate

we need to express the given quadratic expression into a perfect square.

rewriting the expression 9 + 16x – 4x² as – 4x² + 16x +9

dividing by 4 to make coefficient of x² to be 1;

-4(x²-4x-9/4)

we now complete the square by adding 1/2(-4)² being careful not to change the value of the expression.

As you can see, we have added -2 squared and subtracted it immediately to ensure the expression value remains the same.

opening the brackets:

rearranging we have:

our integral is then transformed as follow:

The trigonometric substitution to be used in this expression is 2(x-2)=5sinθ

differentiating x with respect to sinθ:

hence substituting dx:

our integral then becomes:

from the trigonometric identities:

1-sin²θ = cos²θ, hence:

the expression under the square bracket is now a perfect square and we can get the square root. The expression thus becomes:

dividing by 5 and cosθ , the expression becomes:

from our substitution factor 2(x-2) = 5sinθ;

θ is thus the sine inverse of the above expression.

our working thus results to:

Revision Exercise

Related Topics

• Trigonometric substitutions: concise approach
• Introduction to heat conduction
• Fundamental Theorem of Calculus Examples and Solutions
• Basic Integration rules

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