INTEGRATION BY PARTS TRIGONOMETRIC SUBSTITUTION

Integration by parts method is generally used to find the integral when the integrand is a product of two different types of functions or a single logarithmic function or a single inverse trigonometric function or a function which is not integrable directly.

Formula :

∫udv  =  uv - ∫vdu

By differentiating 'u' we get 'du', by integrating dv we get v.

Practice Questions

Question 1 :

Integrate the following with respect to x:

 x sin-1x /√(1-x2)

Solution :

  =  ∫ x sin-1x /√(1-x2dx

t = sin-1x  ===>  x  =  sin t

dt  =  1/√(1-x2dx

  =  ∫ x t dt

  =  ∫ t sin t dt

u = t            dv  =  ∫ sin t dt

du  = dt          v  =  - cos t

∫ udv  =  uv - ∫vdu

  =  t(-cos t) - ∫ (-cos t) dt

  =  - t cos t + ∫ cos t dt

  =  - t cos t + sin t + c

 =  - t √(1 - sin2t) + sin t + c

 =  - sin-1x √(1 - x2) + x + c

Question 2 :

Integrate the following with respect to x:

 x5 e^x2 

Solution :

  =  ∫ x5 e^x2  dx

  =  ∫ (x2)2 x e^x2 dx

Let x2  =  t

2x dx  =  dt

x dx  =  dt/2

 (x2)2 x e^x2 dx  =   t2 et (dt/2)

  =  (1/2)  t2 et dt

u  =  t2     dv  =  et dt

u'  =  2t     v  =  et 

u''  =  2      v1  =  et

∫ udv  =  uv - u'v1 + u''v2 - u'''v3 + ...................

  =  (1/2) [t2(et) - 2t(et) + 2 (et)]

  =  (1/2) [e(t2 - 2t + 2)] + c

By applying the value of "t", we get

  =  (1/2) [e^x2 ((x2)- 2(x2) + 2)] + c

  =  (1/2)e^x2 (x4- 2x2 + 2) + c

Question 3 :

Integrate the following with respect to x:

 tan-1 (8x /1 - 16x2)

Solution :

  =  ∫  tan-1 (8x /1 - 16x2) dx 

  =  ∫ tan-1 (2(4x) /1 - (4x)2) dx

Let 4x  =  tan θ

4 dx  =  sec2θ dθ

  =  ∫ tan-1 (2 tan θ /1 - tan2θ) dx

=  ∫ tan-1 (tan 2θ) dx

=  ∫ 2θ (sec2θ/4) dθ

=  (1/2) ∫ θ sec2θ dθ

u = θ            dv  =  sec2θ

du =  dθ          v  = tan θ

∫ udv  =  uv - ∫vdu

  =  θtan θ - tan θ dθ

  =  θ tan θ - log sec θ  + c

  =   4x tan-1 (4x) - log √(1+(4x)2) + c

  =   4x tan-1 (4x) - log √(1+16x2) + c

Question 4 :

Integrate the following with respect to x:

 sin-1 (2x /1 + x2)

Solution :

  =  ∫sin-1 (2x /1 + x2) dx

Let x  =  tan θ

dx  =  sec2θ dθ

  =  ∫ sin-1 (2 tan θ /1 + (tan θ)2) dx

  =  ∫ sin-1 (2 tan θ /(1 + tan2θ)) dx 

  =  ∫ sin-1 (sin 2θ) dx

  =  ∫ 2θ dx

  =  ∫ 2θ sec2θ dθ

u = θ             dv  =  sec2θ dθ

du  =  1           v  =  tan θ

∫ udv  =  uv - ∫vdu

  =  2 [θtan θ - tan θ dθ]

  =  2 [θ tan θ - log sec θ + c]

  =  2 [x tan-1 x - log  √(1+tan2θ) + c

  =  2 [x tan-1 x - log  √(1+x2) + c

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Angular Speed and Linear Speed

    Dec 07, 22 05:15 AM

    Angular Speed and Linear Speed - Concepts - Formulas - Examples

    Read More

  2. Linear Speed Formula

    Dec 07, 22 05:13 AM

    Linear Speed Formula and Examples

    Read More

  3. Angular Speed and Linear Speed Worksheet

    Dec 07, 22 05:08 AM

    Angular Speed and Linear Speed Worksheet

    Read More