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

Example 5 :

∫ 1/(4 + x²) dx

Solution :

∫1/(4 + x²) dx

Integrating 1/(a² + x²) dx = (1/a) tan^-1(x/a) + C

Comparing 4 + x² with a² + x²

a = 2 and x = x

∫1/(4 + x²) dx = (1/2) tan^-1(x/2) + C

Example 6 :

∫ tan^-1(2t) /(1 + 4t²) dt

Solution :

∫ tan^-1(2t) /(1 + 4t²) dt

Let u = tan^-1(2t)

du = 1/(1 + 4t²) dt

By changing the given integral in terms of u and du.

∫ tan^-1(2t) /(1 + 4t²) dt = ∫ u du

= u²/2 + C

Applying the value of u, we get

= [tan^-1(2t)]²/2 + C

Example 7 :

∫ 1/x√(x² - 9) dx

Solution :

∫ 1/x√(x² - 9) dx

Comparing ∫ 1/x√(x² - 9) dx with ∫ 1/x√(x² - a²) dx, we get (1/a) sec^-1(x/a) + C

∫ 1/x√(x² - 9) dx = (1/3) sec^-1(x/3) + C

Example 8 :

∫ 1/x√(x² - 25) dx

Solution :

∫ 1/x√(x² - 25) dx

Comparing ∫ 1/x√(x² - 9) dx with ∫ 1/x√(x² - a²) dx, we get (1/a) sec^-1(x/a) + C

∫ 1/x√(x² - 25) dx = (1/5) sec^-1(x/5) + C

Example 9 :

∫ tan(sin^-1x) /√(1 - x²) dx

Solution :

∫ tan(sin^-1x) /√(1 - x²) dx

Let u = sin^-1x

Differentiate with respect to x

du = 1/√(1 - x²) dx

Replacing u and du in the given integral, we get

∫ tan(sin^-1x) /√(1 - x²) dx = ∫ tan u du

= ∫ (sin u / cos u) du

= - ∫ (-sin u / cos u) du

= - ln cos u + C

= ln (cos u)^-1 + C

= ln (1/cos u) + C

= ln (sec u) + C

Applying the value of u, we get

= ln (sec sin^-1x) + C

Example 10 :

∫ tan(cos^-1x) /√(1 - x²) dx

Solution :

∫ tan(cos^-1x) /√(1 - x²) dx

Let u = cos^-1x

Differentiate with respect to x

du = -1/√(1 - x²) dx

-du = 1/√(1 - x²) dx

Replacing u and du in the given integral, we get

∫ tan(cos^-1x) /√(1 - x²) dx = -∫ tan u du

= ∫ (-sin u / cos u) du

=  ln cos u + C

Applying the value of u, we get

= ln (cos cos^-1x) + C

= ln (x) + C

Example 11 :

∫ sin (tan^-1x) /(1 + t²) dt

Solution :

∫ tan(cos^-1x) /√(1 - x²) dx

Let u = cos^-1x

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