Reduction Formula
Last Updated :
26 Dec, 2023
Integration involving higher-order terms is difficult to handle and solve. So, to simplify the solving process of higher-order terms and get rid of the lengthy-expression solving process of higher-order degree terms – Integration processes can be simplified by using Reduction Formulas.
The reduction formula comes to the rescue to simplify the higher-order terms. Integration of higher-order terms consisting of logarithmic, algebraic, and trigonometric functions are simplified by reduction formulas. n the Reduction formula, higher-order degree terms are given a degree n. Reduction formulas with degree n are derived from the integration base formulas. All rules of integration apply to these reduction formulas as well.
Reduction Formula for different expressions are listed below:
Reduction Formulas for Logarithmic expressions
∫ lognx dx = xlognx -n∫logn-1x dx
∫xnlogmx dx = xn+1logmx/ n+1 – m/n+1 .∫xnlogm-1x dx
Reduction Formulas for Algebraic expressions
∫ xn/mxn+k dx = x/m – y/k∫ 1/mxn+k dx
Reduction Formulas for Trigonometric expressions
∫ sinnx dx = -1/n sinn-1x. cosx + n-1/n∫sinn-2x dx
∫ cosnx dx = 1/n cosn-1x.sinx + n-1/n∫cosn-2x dx
∫ tannx dx = 1/n-1 tann-1x – ∫tann-2x dx
∫ sinnx.cosmx dx = sinn+1x. cosm-1x / n+m +. m-1/n+m∫ sinnx.cosm-2x dx
Reduction Formulas for Exponential expressions
∫ xnemx dx = 1/m. xnemx – n/m ∫xn-1emx dx
Reduction Formulas for Reduction Formulas for Inverse Trigonometric expressions
∫ xn arc sinx dx = (xn+1/n+1) arc sinx – (1/n+1)∫(xn+1/(1-x2)1/2) dx
∫ xn arc cosx dx = (xn+1/n+1) arc cosx + (1/n+1)∫(xn+1/(1-x2)1/2) dx
∫ xn arc tanx dx = (xn+1/n+1) arc tanx – (1/n+1)∫(xn+1/(1+x2)1/2) dx
Sample Problems
Problem1: Simplify ∫ x2.log2x dx
Solution:
Using formula ∫xnlogmx dx = xn+1logmx/ n+1 – m/n+1 .∫xnlogm-1x dx
n=2, m=2
∫ x2.log2x dx = x3log2x/3 – 2/3.∫x2logx dx
= x3log2x/3 – 2/3.∫x2logx dx
= x3log2x/3 – 2/3. (x3.logx/3 – 1/3. ∫x2 dx)
= x3log2x/3 – 2/3. (x3.logx/3 – 1/3. x3/3)
= x3log2x/3 – 2/9. x3.logx – 2/27. x3
Problem2: Simplify ∫ tan5x dx
Solution:
Using formula ∫ tannx dx = 1/n-1 tann-1x – ∫tann-2x dx
∫ tan5x dx = 1/4 tan4x – ∫tan3x dx
= 1/4 tan4x – ∫tan3x dx
= 1/4 tan4x – ( 1/2tan2x – ∫ tanx dx)
= 1/4 tan4x – 1/2tan2x + 1/2. ln secx
Problem3: Simplify ∫ xe3x dx
Solution:
Using formula ∫ xnemx dx = 1/m. xnemx – n/m ∫xn-1emx dx
= 1/3.xe3x – n/m ∫e3x dx
= 1/3.xe3x – n/m . 3. e3x dx
Problem4: Simplify ∫ log2x dx
Solution:
Using ∫ lognx dx = xlognx -n∫logn-1x dx
∫ log2x dx = 2log2x -2∫logx dx
= 2log2x -2∫logx dx
= 2log2x -2xlogx
Problem5: Simplify ∫ tan2x dx
Solution:
Using ∫ tannx dx = 1/n-1 tann-1x – ∫tann-2x dx
n=2
∫ tan2x dx = tanx – ∫tan0x dx
∫ tan2x dx = tanx – x
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