INTEGRATION BY PARTS WITH INVERSE TRIGONOMETRIC FUNCTIONS

Example 1 :

Integrate tan^-1 x

Solution :

∫ tan^-1 x dx

∫ u dv  =  u v - ∫ v du

u  =  tan^-1 x            dv = dx

du  =  1/(1 + x²)         v = x

=  ∫ tan^-1 x dx

=  (tan^-1 x) x - ∫ x [1/(1 + x²)] dx

1 + x²  =  t

2xdx  =  dt

xdx  =  dt/2 

=  (tan^-1 x) x - ∫ (dt/2)  (1/t) dx

=  (tan^-1 x) x - ∫ x [1/(1+x²)] dx

=  (tan^-1 x)x - ∫ x/(1+x²) dx

t  =  1+x²

dt  =  2xdx

x dx  =  dt/2

=  (tan^-1 x) x - (1/2)∫ dt/t

=  x (tan^-1 x) - (1/2) log t + C

=  x tan^-1 x- (1/2) log (1+x²) + C

Example 2 :

(sin^-1x) (e^sin^-1x)/√(1 - x²)

Solution :

∫ (sin^-1x) (e^sin^-1x)/√(1 - x²)

let t  =  sin^-1x

dt  =  1/√(1 - x²) dx

∫(sin^-1x) (e^sin^-1x)/√(1 - x²)dx  =  ∫te^t dt

u  =  t        dv  =  e ^t

du  =  dt     v  =  e^t

∫ u dv  =  uv - ∫ vdu

=  t e^t - ∫ e^t dt

=  t e^t - e^t + C

=  e^t(t - 1) + C

=  e^sin^-1x (sin^-1x - 1) + C

Example 3 :

Integrate tan^-1 [(3x-x³)/(1-3x²)]

Solution :

x  =  tan θ

=  tan^-1[(3tanθ - (tanθ)³)/(1-3(tan θ)²)]

=  tan^-1[(3tanθ - tan³θ)/(1-3tan²θ)]

=  tan^-1(tan 3 θ)

=  3θ

y  =  3∫tan^-1 x dx

u  =  tan^-1 x           dv  =  dx

du  =  1/(1+x²)           v  =  x

∫ u dv  =  uv - ∫v du

=  3{(tan^-1x) x -  ∫ x [1/(1+x²)] dx}

=  3{(tan^-1x) x -  ∫x/(1+x²) dx}

t  =  1+x²

dt  =  2x dx

x dx  =  dt/2

=  3{(tan^-1x) x -  ∫ (dt/2)(1/t)}

=  3{(tan^-1x) x -  (1/2)∫ 1/t dt}

=  3xtan^-1x -  (3/2) log(1+x²) + 3C

=  3xtan^-1x -  (3/2) log (1+x²) + C

Example 4 :

Integrate x sin^-1 (x²)

Solution :

Let x²  =  t

2x dx  =  dt

xdx  =  dt/2

=  ∫x sin^-1 (x²) dx

=  (1/2) ∫ sin^-1(t) dt

now we are going to apply the substitution method

u  =  sin^-1(t)             dv = dt

du  =  1/√(1-t²)         v = t 

∫ u dv  =  uv - ∫v du

=  (1/2) [tsin^-1(t) -∫t (1/√(1-t²))dt]

=  (1/2) [tsin^-1(t) - ∫ t/√(1-t²)dt] ------ (1)

Let a = 1-t²

da  =  -2t dt

-da/2  =  t dt

∫ t/√(1-t²) dt  =  ∫ (-da/2)/√a

=  - (1/2)∫ (1/√a) da

= - (1/2) [√a/(1/2)]

= - (1/2) [2√a]

= - √(1-t²) + C

By applying =  (1/2) [x²(sin^-1 x²) - (- √(1-t²))]+ C

=  (1/2) [x²(sin^-1 x²) + √(1-(x²)²)]+ C

=  (1/2) [x²(sin^-1 x²) + √(1-x⁴)]+ C

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