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# Integration by parts

Integration by parts method is generally used to find the integral when the integrand is a product of two different types of functions or a single logarithmic function or a single inverse trigonometric function or a function which is not integrable directly.

Integration by parts

Integration by parts method is generally used to find the integral when the integrand is a product of two different types of functions or a single logarithmic function or a single inverse trigonometric function or a function which is not integrable directly. From the formula for derivative of product of two functions we obtain this useful method of integration.

If u and v are two differentiable functions then we have

d (uv ) = vdu+udv

udv = d (uv ) - vdu

Integrating

Ōł½udv = Ōł½d (uv ) - Ōł½vdu

Ōł½udv = uv - Ōł½vdu

Ōł½udv in terms of another integral Ōł½vdu and does not give a final expression for the integral Ōł½udv. It only partially solves the problem of integrating the product u dv . Hence the term ŌĆśPartial IntegrationŌĆÖ has been used in many European countries. The term ŌĆ£Integration by PartsŌĆØ is used in many other countries as well as in our own.

The success of this method depends on the proper choice of u

(i) If integrand contains any non integrable functions directly from the formula, like logx, tanŌłÆ1 x etc., we have to take these non integrable functions as u and other as dv.

(ii) If the integrand contains both the integrable function, and one of these is xn (where n is a positive integer) then take u = xn.

(iii) For other cases the choice of u is ours.

Example 11.33

Evaluate the following integrals

(i) Ōł½xexdx

(ii)  Ōł½ x cos x dx

(iii) Ōł½log x dx

(iv) Ōł½sinŌłÆ1x dx

Solution

(i) Let I = Ōł½ xe x dx.

Since x is an algebraic function and ex is an exponential function,

so take u = x then du = dx

dv = e x dx ŌćÆ v = ex

Applying Integration by parts, we get

Ōł½udv = uv ŌłÆ Ōł½vdu

ŌćÆ Ōł½ xex dx = xe x ŌłÆ Ōł½e x dx

That is,  Ōł½ xe x dx = xe x ŌłÆ e x + c.

(ii) Let I = Ōł½ x cos x dx

Since x is an algebraic function and cos x is a trigonometric function,

so take u = x then du = dx

dv = cos xdx ŌćÆ v = sin x

Applying Integration by parts, we get

Ōł½udv = uv ŌłÆ Ōł½vdu

ŌćÆ Ōł½ x cos xdx = x sin x ŌłÆ Ōł½sin x dx

ŌćÆ Ōł½ x cos xdx = x sin x + cos x + c

(iii) Let I = Ōł½log x dx

Take u = log x then du = 1/x dx

dv = dx ŌćÆ v = x

Applying Integration by parts, we get

Ōł½udv = uv ŌłÆ Ōł½vdu

ŌćÆ log x dx = x log x ŌłÆ Ōł½x (1/x) dx

ŌćÆ Ōł½log x dx = x log x ŌłÆ x + c

(iv) Let I = Ōł½sinŌłÆ1x dx

u = sinŌłÆ1 (x), dv = dx

Then du = 1/ 1ŌłÆ x2 , v = x  Tags : Equation, Solved Example Problems | Mathematics , 11th Mathematics : UNIT 11 : Integral Calculus
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11th Mathematics : UNIT 11 : Integral Calculus : Integration by parts | Equation, Solved Example Problems | Mathematics