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**D. RATIONALIZATION BY TRIGONOMETRIC SUBSTITUTION**

Consider the integral dx

If we change the variable from x to Î¸ by the substitution x = a sinÎ¸, then the identity 1 â€“ sin^{2}Î¸ = cos^{2}Î¸ allows us to get rid of the roots sign because

Notice the difference between the substitution u = a^{2} â€“ x^{2} (in which the new variable is a function of the old one) are the substitution x = a sin Î¸ (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 Î¸ provided that it defines a one-to-one function. This can be accomplished by restricting Î¸ to lie in the interval [â€“Ï€/2, Ï€/2].

**SOME STANDARD INTEGRALS**

**Ex.26 Evaluate **** where a > 0**

**Sol.**

We let x = a secÎ¸, where 0 < Î¸ < Ï€/2 or Ï€ < Î¸ < 3Ï€/2. Then dx = a sec Î¸ tan Î¸ dÎ¸ and

**Ex.27 Integrate **

**Sol.**

**Ex.28 Integrate (3x + 1) / (2x ^{2} â€“ 2x + 3).**

**Sol.** Here (d/dx) (2x^{2} â€“ 2x + 3) = 4x â€“ 2.

**Ex.29 Integrate **

**Sol.**

**Ex.30 Evaluate **** dx.**

**Sol.**

**Ex.31 Integrate **

**Sol.**

**Ex.32 Integrate x ^{2}/(x^{4} + x^{2} + 1)**

**Sol. **

dividing the numerator and the denominator both by x^{2}.

Now the denominator x^{2} + 1 + 1/x^{2} can be written either as The diff. coeff. of x-1/x is 1 + 1/x^{2} and that of x + 1/x is 1-1/x^{2}. So we write

In the first integral put x â€“ 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 = Î±cos2Î¸ + Î² sin2Î¸ so that dx = 2 (Î²- Î±) sinÎ¸ cos Î¸ dÎ¸

Also(x â€“ Î±) = (Î² - Î±) sin2Î¸, and (Î² - x) = (Î² - Î±) cos2Î¸

Making these substitutions,in the given integral =

But x = Î±cos2Î¸ + Î²sin2 Î¸; 2x = Î± (1 + cos 2Î¸) + Î²(1 - cos 2Î¸)

i.e., (Î² - Î±) cos 2Î¸ = (Î± + Î² - 2x) or cos 2Î¸ = (Î± + Î² - 2x) / (Î² - Î±)

**Ex.36 Evaluate I =**

**Sol.**

Substituting ax^{2} = b sin^{2} Î¸ â‡’

dividing N^{r} and D^{r} by cos2Î¸. we get

**Ex.37 Integrate 1/(1 + 3 sin ^{2}x).**

**Sol.** Dividing N^{r}. and D^{r}. by cos^{2} x, we have

Now putting 2 tan x = 5 so that 2 sec^{2}x dx = dt, we have

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