DIFFERENTIATION BY USING TRIGONOMETRIC SUBSTITUTION

Trigonometric Identities

Let us look at some examples to understand how to differentiate a function using trigonometric substitution. 

Example 1 :

Differentiate the following

tan^-1[√(1 - cos x)/(1+cosx)]

Solution :

Let y  =  tan^-1[√(1 - cos x)/(1+cosx)]

Trigonometric formula for (1-cos x)/(1+cosx)  =  tan²(x/2)

By applying trigonometric formula, we may find derivatives easily. 

 y  =  tan-1[√tan²(x/2)]

y = x/2

Differentiating with respect to "x"

dy/dx  =  1/2

Example 2 :

Differentiate the following

tan^-1[6x/1-9x²]

Solution :

Let y  =  tan^-1[6x/1-9x²]

y  =  tan^-1[2(3x) /1-(3x)²]

Let 3x  =  tan θ

y  =  tan^-1[2 tan θ /1-tan²θ]

2 tan θ /1-tan²θ  =  tan 2θ

y  =  tan^-1[tan 2θ]

y = 2θ

dy/dx  =  2 (dθ/dx)  ------(1)

3x  =  tan θ

3(1)  =  sec²θ(dθ/dx)

3/sec²θ  =  (dθ/dx)

Applying dθ/dx  =  3/sec²θ in (1)

dy/dx  =  2 (3/sec²θ) 

   =  2(3)/(1  + tan²θ)

   =  6/(1 + (3x)²)

   =  6/(1 + 9x²)

Example 3 :

Differentiate the following

cos (2tan^-1[√(1-x)/(1+x)])

Solution :

Let y  =  cos (2tan^-1[√(1-x)/(1+x)])

let x = cos θ

Differentiating with respect to x, we get

dx/dθ  =  -sin θ

 y  =  cos (2tan^-1[√(1-cosθ)/(1+cosθ)])

 y  =  cos (2tan^-1[√tan²(θ/2])

 y  =  cos (2tan^-1[tan(θ/2])

 y  =  cos θ

Differentiating with respect to θ.

dy/dx  =  -sin θ (dθ/dx)    -----(1)

Now let us apply the value of dx/dθ  =  -sin θ in (1)

dy/dx  =  -sin θ (-1/sinθ) 

dy/dx  =  1

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