Solved Examples on Methods of Integration | Physics for JEE Main & Advanced PDF Download

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 ∫ x²e^x 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 = e^x dx,

so that
du = dx, v = e^x,  
then we get

xe^x - e^x.
When this is in serted in (2), our final result is
∫x²e^x dx = x²e^x - 2xe^x + 2e^x + c.

Example 16: Find ∫e^x cos x dx.

Sol: For convenience we denote this integral by J. If we put
u = e^x, dv = cos x dx,
then
du = e^x dx, v = sin x,
and (1) yields
J = e^x sin x - ∫e^x 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 = e^x, dv = sin x dx,
so that
du = e^x dx, v = -cos x,
and obtain
∫ex sin x dx = -e^x cos x + ∫ e^x cos x dx.     ....(4)
The integral on the right is J again , so (4) can be written
∫e^x sin x dx = - e^x cos x + J.     ......(5)
In spite of appearances, we are not going in a circle, because substituting (5) in (3) gives

J = e^x sin x + e^x cos x - J.

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


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

We now transpose the term involving J_n and obtain
nJ_n = - sin^n-1 x cos x + (n - 1)J_n-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