Solved Examples on Methods of Integration

## Method of Substitution

Example 1: Find

Sol:
Here we notice that cos x dx is the differential of sin x, and also of 1 + sin x. Thus, if we put u = 1 + sin x, then du = cos x dx and

Example 2:  Find

Sol: Since 4x2 = (2x)2 we put u = 2x, so that du = 2dx, dx = 1/2 du , and

Example 3: Find

Sol: Here the fact that the x in the numerator is essentially the derivative of the expression 9 - 4x2 inside the radical suggests the substitution u = 9 - 4x2. Then du = - 8x dx, and

## Trigonometric Integrals

Example 4:

Sol:

Example 5: The half-angle formula for the cosine enables us to write
Sol:

If we wish to express this result in terms of the variable x (instead of 2x), we use the double-angle formula sin 2x = 2 sin x cos x and write

Example 6: By using both of the half-angle formulas we get
Sol:

Example 7: ∫tan3 x sec5x dx =  ∫ tan2 x sec4 x sec x tan x dx
Sol:

## Trigonometric Substitutions

Example 8: Find

Sol:
This integ ral is o f th e first type, so we write
x= a sin θ, dx = a cos θ dθ,

Then

= —a ln (csc θ + cot θ) + a cos θ.     ...(7)
This completes the integration, and we now must write the answer in terms of the original variable x. We do this quickly and easily by drawing a right triangle whose sides are labeled in the simplest way that is consistent with the equation x = a sin θ or sin θ = x/a. This figure tellsus at once that

so from (7) we have

Example 9: Find

Sol:

x = a tan θ, dx = a sec2 θ dθ,

This yields

= ln (sec θ + tan θ)   ...(8)
The substitution equation x = a tan 6 or tan 6 = x/a is pictured in Fig, and from this figure we obtain

We therefore continue the calculation in (8) by writing

Students will notice that since

the constant -In a has been grouped together with the constant of integration c' and the quantity -In a + c' is then rewritten as c. Usually we don’t bother to make notational distinctions between one constant of integration and another, because all are completely arbitrary; but we do so here in the hope of clarifying the transition from (9) to (10).

## Completing the Square

Example 10: Find

Sol:

Since the coefficient of the term x2 under the radical is negative, we place the terms containing x in parentheses preceded by a minus sign, leaving space for completing the square,
3 + 2x — x2 = 3 — (x2 — 2x + ) = 4 — (x2 — 2x + 1)
= 4 — (x - 1)2 = a2 — u2,
where u = x - 1 and a = 2.
Since x = u + 1, we have dx = du and x + 2 = u + 3, and therefore

Q11: Find

Sol:

We complete the square on the terms containing x, and write
x2 + 2x + 10 = (x2 + 2x + ) + 10 = (x2 + 2x + 1) + 9
= (x + 1)2 + 9 = u2 + a2 ,
where u = x + 1 and a = 3. We now have = dx or d x = du , so

Q12: Find

Sol:
We write

x2 - 2x + 5 = (x2 — 2x + ) + 5 = (x2 — 2x + 1) + 4
= (x - 1)2 + 4 = u2 + a2 ,
where u = x - 1 and a = 2. Then x = u + 1, dx = du, and we have

The second integral here is the one considered in Example 2 in Section 10.4, so we have

and therefore

Q13: Find

Sol:
We have

so
2x3 + x2 + 2x - 1 = A (x - 1)(x2 + 1) + 5 (x + 1)(x2 + 1) + Cx(x2 - 1) + D(x2 — 1).
Now put

Equating coefficients of x3 gives
2 = A + 5 + C, so C = 0.
Our partial fractions decomposition is therefore

so

## Integration by Parts

Example 14: Find ∫ ln x dx.
Sol:
Here our only choice is
w = ln x, dv = dx,
so

and we have

Example 15: Find ∫ x2ex dx.
Sol: If we put
u = x2, dv = ex dx,
then du = 2x dx,  v = ex
and (1) gives
....(2)
Here the second integral is easier than the first, so we are encouraged to continue in the same way. When the second integral is integrated by parts with
u = x, dv = ex dx,

so that
du = dx, v = ex,
then we get

xex - ex.
When this is in serted in (2), our final result is
∫x2ex dx = x2ex - 2xex + 2ex + c.

Example 16: Find ∫ex cos x dx.

Sol: For convenience we denote this integral by J. If we put
u = ex, dv = cos x dx,
then
du = ex dx, v = sin x,
and (1) yields
J = ex sin x - ex sin x dx.   .....(3)
Now we come to the interesting part of this problem. Even though the new integral is no easier than the old, it turns out to be fruitful to apply the same method again to the new integral. Thus, we put
u = ex, dv = sin x dx,
so that
du = ex dx, v = -cos x,
and obtain
ex sin x dx = -ex cos x + ex cos x dx.     ....(4)
The integral on the right is J again , so (4) can be written
ex sin x dx = - ex cos x + J.     ......(5)
In spite of appearances, we are not going in a circle, because substituting (5) in (3) gives

J = ex sin x + ex cos x - J.

It is now easy to solve for J by writing
2J = ex sin x + ex cos x or J = 1/2 (ex sin x + ex cos x), and all that remains is to insert the constant of integration:

Exmaple 17: Find a reduction formula for Jn = ∫ sinn x dx.
Sol:
We integrate by parts with
u = sinn-1x,   dv = sin x dx,
so that du = {n — 1) sinn-2 x cos x dx, v = —cos x,
and therefore

We now transpose the term involving Jn and obtain
nJn = - sinn-1 x cos x + (n - 1)Jn-2,
so that

or equivalently,
....(6)

Example 18: Calculate

Sol:
For convenience we write

By formula (6) we have

so

We apply this formula with n = 8, then repeat with n = 6, n = 4, n = 2:

Therefore

The document Solved Examples on Methods of Integration | Physics for JEE Main & Advanced is a part of the JEE Course Physics for JEE Main & Advanced.
All you need of JEE at this link: JEE

## Physics for JEE Main & Advanced

289 videos|635 docs|179 tests

## Physics for JEE Main & Advanced

289 videos|635 docs|179 tests

### Up next

 Explore Courses for JEE exam
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;