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Integration by Substitution

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Integration by Substitution - Lesson Summary

While solving integrals, where the integrand is a function of a function.

∫f(g(x)) g'(x) dx

Let g(x) = t

On differentiating both sides with respect to x, we get

g'(x)dx = dt

∫f(t) dt

Example:

∫ ((1+x)ex)/(cos2(xex)) dx

xex = t

x .ex dx + ex . 1 dx = dt

⇒ x ex dx + ex  dx = dt

⇒ e x(x + 1)dx = dt

∫ ((1+x)ex)/(cos2(xex)) dx = ∫ dt/cos2t

= ∫ sec2t dt

= tan t + C

= tan( xex) + C

Hence, ∫ ((1+x)ex)/(cos2(xex)) dx = tan(xex) + C

Integrals of some trigonometric functions

i) ∫ tan x dx = ∫ sin x/cos x dx

Let cos x = t

On differentiating both sides, we get

- sinx dx = dt

= ∫ 1/t (-dt)

= - log|t| + C

= - log|cos x| + C

= log|1/cos x| + C

∫ tan x dx = log|sec x| + C


ii) ∫ cot x dx

= ∫ cosx/sinx dx

Let sinx = t

On differentiating both sides, we get

cosx dx = dt

= ∫1/t dt

= log|t| + C

= log|sinx| + C

∫ cotx dx = log|sinx| + C


iii) ∫ secx dx

= ∫ (secx (sec x + tan x)/(secx + tan x)) ...(i)

sec x + tan x = t

sec x tan x dx + sec x dx = dt

⇒ sec x (tan x + sec x) dx = dt

= ∫ 1/t dt

= log|t| + C

∫ sec x dx = log|sec x + tan x| + C


iv ∫ cosec x dx

= ∫ (cosec x (cosec x - cot x))/(cosec x - cot x) dx

= ∫ (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

∴ ∫ (cosec2 x - cosec x. cot x)/(cosec x - cot x) dx = ∫1/t dt

 = log|t| + C

= log|cosec x - cot x| + C

  Integral

  Solution

  ∫ tan x dx

  log|sec x| + C

  ∫ cot x dx

  log|sin x| + C

  ∫ sec x dx

  log|sec x + tan x | + C

  ∫ cosec dx

  log|cosec x -cot x| + C

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