Consider the integral dx
If we change the variable from x to θ by the substitution x = a sinθ, then the identity
1 – sin2θ = cos2θ
allows us to get rid of the roots sign because
Notice the difference between the substitution u = a2 – x2 (in which the new variable is a function of the old one) are the substitution x = a sin θ (the old variable is a function of the new one).
In general, we can make a substitution of the form x = g(t) by using the Substitution Rule in reverse. To make our calculations simpler, we assume that g has an inverse function; that is, g is one-to-one.
This kind of substitution is called inverse substitution.
We can make the inverse substitution x = a sin θ provided that it defines a one-to-one function. This can be accomplished by restricting θ to lie in the interval [–π/2, π/2].
Ex.1 Evaluate where a > 0
We let x = a secθ, where 0 < θ < π/2 or π < θ < 3π/2. Then dx = a sec θ tan θ dθ and
Ex.3 Integrate (3x + 1) / (2x2 – 2x + 3).
Sol. Here (d/dx) (2x2 – 2x + 3) = 4x – 2.
Ex.5 Evaluate dx.
Ex.7 Integrate x2/(x4 + x2 + 1)
dividing the numerator and the denominator both by x2.
Now the denominator x2 + 1 + 1/x2 can be written either as The diff. coeff. of x-1/x is 1 + 1/x2 and that of x + 1/x is 1-1/x2. So we write
In the first integral put x – 1/x = t so that and in the second integral put
Put x = αcos2θ + β sin2θ so that dx = 2 (β- α) sinθ cos θ dθ
Also(x – α) = (β - α) sin2θ, and (β - x) = (β - α) cos2θ
Making these substitutions,in the given integral =
But x = αcos2θ + βsin2 θ; 2x = α (1 + cos 2θ) + β(1 - cos 2θ)
i.e., (β - α) cos 2θ = (α + β - 2x) or cos 2θ = (α + β - 2x) / (β - α)
Ex.11 Evaluate I =
Substituting ax2 = b sin2 θ ⇒
dividing Nr and Dr by cos2θ. we get
Ex.12 Integrate 1/(1 + 3 sin2x).
Sol. Dividing Nr. and Dr. by cos2 x, we have
Now putting 2 tan x = 5 so that 2 sec2x dx = dt, we have