Notes On Integration by Substitution - CBSE Class 12 Maths
While solving integrals, where the integrand is a function of a function. âˆ«f(g(x)) g'(x) dx Let g(x) = t On differentiating both sides with respect to x, we get g'(x)dx = dt âˆ«f(t) dt Example: âˆ« ((1+x)ex)/(cos2(xex)) dx xex = t x .ex dx + ex . 1 dx = dt â‡’ x ex dx + ex  dx = dt â‡’ ex(x + 1)dx = dt âˆ« ((1+x)ex)/(cos2(xex)) dx = âˆ« dt/cos2t = âˆ« sec2t dt = tan t + C = tan( xex) + C Hence, âˆ« ((1+x)ex)/(cos2(xex)) dx = tan(xex) + C Integrals of some trigonometric functions i) âˆ« tan x dx = âˆ« sin x/cos x dx Let cos x = t On differentiating both sides, we get - sinx dx = dt = âˆ« 1/t (-dt) = - log|t| + C = - log|cos x| + C = log|1/cos x| + C âˆ« tan x dx = log|sec x| + C ii) âˆ« cot x dx = âˆ« cosx/sinx dx Let sinx = t On differentiating both sides, we get cosx dx = dt = âˆ«1/t dt = log|t| + C = log|sinx| + C âˆ« cotx dx = log|sinx| + C iii) âˆ« secx dx = âˆ« (secx (sec x + tan x)/(secx + tan x)) ...(i) sec x + tan x = t sec x tan x dx + sec x dx = dt â‡’ sec x (tan x + sec x) dx = dt = âˆ« 1/t dt = log|t| + C âˆ« sec x dx = log|sec x + tan x| + C iv âˆ« cosec x dx = âˆ« (cosec x (cosec x - cot x))/(cosec x - cot x) dx = âˆ« (cosec2 x - cosec x. cot x)/(cosec x - cot x) dx ...(i) cosec x - cot x = t (- cosec x. cot x + cosec2 x) dx = dt âˆ´ âˆ« (cosec2 x - cosec x. cot x)/(cosec x - cot x) dx = âˆ«1/t dt  = log|t| + C = log|cosec x - cot x| + C   Integral   Solution   âˆ« tan x dx   log|sec x| + C   âˆ« cot x dx   log|sin x| + C   âˆ« sec x dx   log|sec x + tan x | + C   âˆ« cosec dx   log|cosec x -cot x| + C

#### Summary

While solving integrals, where the integrand is a function of a function. âˆ«f(g(x)) g'(x) dx Let g(x) = t On differentiating both sides with respect to x, we get g'(x)dx = dt âˆ«f(t) dt Example: âˆ« ((1+x)ex)/(cos2(xex)) dx xex = t x .ex dx + ex . 1 dx = dt â‡’ x ex dx + ex  dx = dt â‡’ ex(x + 1)dx = dt âˆ« ((1+x)ex)/(cos2(xex)) dx = âˆ« dt/cos2t = âˆ« sec2t dt = tan t + C = tan( xex) + C Hence, âˆ« ((1+x)ex)/(cos2(xex)) dx = tan(xex) + C Integrals of some trigonometric functions i) âˆ« tan x dx = âˆ« sin x/cos x dx Let cos x = t On differentiating both sides, we get - sinx dx = dt = âˆ« 1/t (-dt) = - log|t| + C = - log|cos x| + C = log|1/cos x| + C âˆ« tan x dx = log|sec x| + C ii) âˆ« cot x dx = âˆ« cosx/sinx dx Let sinx = t On differentiating both sides, we get cosx dx = dt = âˆ«1/t dt = log|t| + C = log|sinx| + C âˆ« cotx dx = log|sinx| + C iii) âˆ« secx dx = âˆ« (secx (sec x + tan x)/(secx + tan x)) ...(i) sec x + tan x = t sec x tan x dx + sec x dx = dt â‡’ sec x (tan x + sec x) dx = dt = âˆ« 1/t dt = log|t| + C âˆ« sec x dx = log|sec x + tan x| + C iv âˆ« cosec x dx = âˆ« (cosec x (cosec x - cot x))/(cosec x - cot x) dx = âˆ« (cosec2 x - cosec x. cot x)/(cosec x - cot x) dx ...(i) cosec x - cot x = t (- cosec x. cot x + cosec2 x) dx = dt âˆ´ âˆ« (cosec2 x - cosec x. cot x)/(cosec x - cot x) dx = âˆ«1/t dt  = log|t| + C = log|cosec x - cot x| + C   Integral   Solution   âˆ« tan x dx   log|sec x| + C   âˆ« cot x dx   log|sin x| + C   âˆ« sec x dx   log|sec x + tan x | + C   âˆ« cosec dx   log|cosec x -cot x| + C

Previous
Next
âž¤