# 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

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