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# Rationalization by Trigonometric Substitution JEE Notes | EduRev

## JEE : Rationalization by Trigonometric Substitution JEE Notes | EduRev

The document Rationalization by Trigonometric Substitution JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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D. RATIONALIZATION BY TRIGONOMETRIC SUBSTITUTION

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].

SOME STANDARD INTEGRALS

Ex.26 Evaluate  where a > 0

Sol.

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.27 Integrate

Sol.

Ex.28 Integrate (3x + 1) / (2x2 – 2x + 3).

Sol. Here (d/dx) (2x2 – 2x + 3) = 4x – 2.

Ex.29 Integrate

Sol.

Ex.30 Evaluate   dx.

Sol.

Ex.31 Integrate

Sol.

Ex.32 Integrate x2/(x4 + x2 + 1)

Sol.

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

Ex.33 Evaluate

Sol.

Ex.34 Evaluate

Sol.

Ex.35 Evaluate

Sol.

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.36 Evaluate I =

Sol.

Substituting ax2 = b sin2 θ ⇒

dividing Nr and Dr by cos2θ. we get

Ex.37 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

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## Mathematics (Maths) Class 12

209 videos|222 docs|124 tests

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