# Integration Formulas

```IMPORTANT FORMULAS IN INTEGRATION
Introduction and Definition of Integration
Consider a function π(π₯). When we differentiate it with respect to π₯, we get:
π
ππ₯
[π (π₯ )] = π ′ (π₯) ; where π ′ (π₯) is the derivative of π(π₯ ).
Now, when we integrate π ′ (π₯), i.e.
∫ π ′ (π₯) ππ₯ = π (π₯ ) + π; where π (π₯ ) is called the integral of π ′ (π₯) and c is the
constant of integration. The operator ππ₯ indicates that we are integrating w.r.t. π₯.
We need the constant of integration because the integral of any term is not definite
i.e. the constant lost in differentiation will not be recovered on integration.
For example, if we differentiate a function π(π₯ ) = 3π₯ 2 + 5π₯ + 7, we get π ′ (π₯ ) =
6π₯ + 5. Now, while integrating π ′ (π₯), we get ∫ π ′ (π₯) ππ₯ = ∫(6π₯ + 5)ππ₯ which will
be equal to 3π₯ 2 + 5π₯. In other words, the constant ‘7’ which is lost during the
differentiation process is not retrieved in the integration process. To compensate this,
we add an arbitrary constant c known as constant of integration.
The process of determining an integral of a function is called integration and the
function to be integrated is called the integrand.
Since the integral of any function is not completely defined, it is known as Indefinite
Integration.
Therefore, we see that integration is the reverse process of differentiation.
Basic Integrals of Algebraic Functions
1.
2.
π
∫ π₯ ππ₯ =
π₯ π+1
π+1
π
+ π; π ≠ −1
∫(ππ₯ + π) ππ₯ =
(ππ₯+π)π+1
π.(π+1)
+ π; π ≠ −1
Integrals of a Combination of Functions
Theorem 1: The integral of the product of a constant and a function is equal to the
product of the constant and the integral of the function.
∫ π. π (π₯ )ππ₯ = π ∫ π (π₯ )ππ₯
Theorem 2: The integral of a sum or difference of two functions is equal to the sum
or difference of their integrals.
∫[π(π₯ ) &plusmn; π(π₯ )]ππ₯ = ∫ π (π₯ )ππ₯ &plusmn; ∫ π(π₯ )ππ₯
Integrals of Trigonometric Functions
1.
∫ sin π₯ ππ₯ = − cos π₯ + π
2.
∫ cos π₯ ππ₯ = sin π₯ + π
3.
∫ sec 2 π₯ ππ₯ = tan π₯ + π
4.
∫ cosec 2 π₯ ππ₯ = − cot π₯ + π
5.
∫ sec π₯ tan π₯ ππ₯ = sec π₯ + π
6.
∫ πππ ππ π₯ cot π₯ ππ₯ = −πππ ππ π₯ + π
7.
∫ tan π₯ ππ₯ = log|sec π₯ | + π
8.
∫ cot π₯ ππ₯ = log|sin π₯ | + π
9.
∫ sec π₯ ππ₯ = log|sec π₯ + tan π₯ | + π = log |tan ( 4 + 2)| + π
π
π₯
π₯
10. ∫ cosec π₯ ππ₯ = log|cosec π₯ − cot π₯ | + π = log |tan | + π
2
Formulae and Identities of Trigonometrical Functions
The following identities and relations can be used to simplify the integrand when it contains higher powers of
sin x, cos x and other t-functions and hence bring it into any of the standard forms:
1.
Fundamental Identities
(a) Square Relations: 1) ππππ π½ + ππππ π½ = π ⇒ ππππ π½ = π − ππππ π½ ⇒ ππππ π½ = π − ππππ π½
2) π + ππππ π½ = ππππ π½ ⇒ ππππ π½ = ππππ π½ − π ⇒ ππππ π½ − ππππ π½ = π
3) π + ππππ π½ = ππππππ π½ ⇒ ππππ π½ = ππππππ π½ − π ⇒ ππππππ π½ − ππππ π½ = π
(b)Basic Relations: 1)
πππ π
ππππ π
ππππ
3)
2.
πππ π
= πππ π πππ π
= πππ π
4)
2)
ππππ
πππππ π
Converting products into sums or differences
Using Corollaries from Multiple Angle Forms
(i) ππππ π¨ =
π−πππ ππ¨
(ii) ππππ π¨ =
π+πππ ππ¨
π
π
π−πππ ππ¨
(iii) π+πππ ππ¨ = ππππ π¨ = ππππ π¨ − π
(iv) √π &plusmn; πππ ππ¨ = πππ π¨ &plusmn; πππ π¨
(v)
π− πππ π¨
(vi)
πππ π¨
π+ πππ π¨
πππ π¨
=
π ππππ
π¨
π
π¨
π
π πππ πππ
=
π¨
π
π ππππ
π¨
π
π¨
π¨
π πππ πππ
π
π
π
π
π¨
π
π¨
πππ
π
πππ
=
=
= πππ
π¨
π
π¨
πππ
π
πππ
π¨
(vii) ) π+πππ ππ¨ = π ππππ π
π
π
π¨
(viii) π−πππ ππ¨ = π ππππππ π
π
(ix) ππππ π¨ = π [ππππ π¨ − πππ ππ¨]
π
(x) ππππ π¨ = π [πππ ππ¨ + π πππ π¨]
π¨
π
= πππ
π¨
π
= πππππ π πππ π
= πππ π
3.
πππ π
ππππ π
π
Integrals of ,
π
π ππ+π
, ππ , ππ πππ πππ+π
1
1.
∫ π₯ ππ₯ = log|π₯ | + π
2.
∫ ππ₯+π ππ₯ = a log|ππ₯ + π| + π
3.
∫ π π₯ ππ₯ = π π₯ + π
4.
∫ π π₯ ππ₯ = log π + π
5.
∫ π ππ₯+π ππ₯ =
1
1
ππ₯
π ππ₯+π
π
+π
Integrals of Inverse Trigonometric Functions
1. ∫
2. ∫
3. ∫
4. ∫
5. ∫
6. ∫
1
√1−π₯ 2
1
1+π₯ 2
ππ₯ = sin−1 π₯ + π = − cos −1 π₯ + π
ππ₯ = tan−1 π₯ + π = − cot −1 π₯ + π
1
|π₯|√π₯ 2 −1
1
ππ₯ = sec −1 π₯ + π = − cosec −1 π₯ + π
π₯
√π2 −π₯
1
π2 +π₯
ππ₯ = sin−1 + π
2
π
1
π₯
a
π
−1
+π
2 ππ₯ = tan
1
|π₯|√π₯ 2 −π
1
π₯
a
π
ππ₯ = sec −1 + π
2
Note: Learn all properties of inverse trigonometric functions very well.
STANDARD METHODS OF INTEGARTION
When the given function cannot be integrated directly by using standard formulae, we
try other methods of integration. The process of integration is largely of tentative
nature and no systematic procedure can be given as in differentiation. However the
following are two important methods of integration:
(i) Integration by substitution
(ii) Integration by parts
Integration by Substitution
Consider the following complicated integrand. We notice that the given integrand
cannot be integrated directly by using any of the known standard forms. So, we use
the method of substitution to change the variable of integration so that the integrand
becomes simple enough to be integrated by applying a known standard formula.
∫[π(π₯ )]π π ′ (π₯ )ππ₯ ; πΏππ‘ π (π₯ ) = π‘
π ′ (π₯ )ππ₯ = ππ‘
The integrand now takes on the form,
∫ π‘ π ππ‘ =
π‘ π+1
π+1
+π =
[π(π₯)]π+1
π+1
+π
Note: There are no hard and fast rules for making suitable substitution. Practice will
help acquire the necessary skill. However, the following guidelines will be found
useful:
(i)
If the integrand contains
√1 − π₯ 2 ; π π’ππ π‘ππ‘π’π‘π π₯ = sin π ππ π₯ = cos π
√π2 − π₯ 2 ; π π’ππ π‘ππ‘π’π‘π π₯ = asin π ππ π₯ = acos π
(ii)
If the integrand contains
(1 + π₯ 2 ) ππ √1 + π₯ 2 ; π π’ππ π‘ππ‘π’π‘π π₯ = tan π ππ π₯ = cot π
(π2 + π₯ 2 ) ππ √π2 + π₯ 2 ; π π’ππ π‘ππ‘π’π‘π π₯ = atan π ππ π₯ = acot π
(iii) If the integrand contains
√π₯ 2 − 1 ; π π’ππ π‘ππ‘π’π‘π π₯ = sec π ππ π₯ = cosec π
√π₯ 2 − π2 ; π π’ππ π‘ππ‘π’π‘π π₯ = asec π ππ π₯ = acosec π
(iv)
If the integrand contains
√1 &plusmn; π₯ ; π π’ππ π‘ππ‘π’π‘π π₯ = sin 2π ππ π₯ = cos 2π
ππ π π’ππ π‘ππ‘π’π‘π (1 &plusmn; π₯ ) = π‘ 2 ππ √1 &plusmn; π₯ = π‘
√π &plusmn; π₯ ; π π’ππ π‘ππ‘π’π‘π π₯ = asin 2π ππ π₯ = acos 2π
ππ π π’ππ π‘ππ‘π’π‘π (π &plusmn; π₯ ) = π‘ 2 ππ √π &plusmn; π₯ = π‘
(v)
If the integrand contains
√π−π₯
; π π’ππ π‘ππ‘π’π‘π
√π+π₯
(vi)
π₯ = acos 2π
If the integrand contains some linear expression of x, put that value= π‘.
(vii) ∫
cos π₯−sin π₯
cos π₯+sin π₯
ππ₯ ; ππ’π‘ (cos π₯ + sin π₯) = π‘ ⇒ (cos π₯ − sin π₯)ππ₯ = ππ‘
π π₯ −π −π₯
(viii) ∫ π₯ −π₯ ππ₯ ; ππ’π‘ (π π₯ + π −π₯ ) = π‘ ⇒ (π π₯ − π −π₯ )ππ₯ = ππ‘
π +π
(ix) If the integrand contains √π₯ or similar terms, we can put √π₯ = π‘ so that
ππ₯ = 2π‘ππ‘
Some Important Formulas
1. ∫[π(π₯
2. ∫
)]π ′ (
π ′ (π₯)
π(π₯)
π π₯ )ππ₯ (π ≠ −1) =
[π(π₯)]π+1
π+1
+π
ππ₯ = log π (π₯ ) + π
Note: Do not directly use the above two formulae. Make the necessary substitution,
integrate in terms of the new variable and finally re-substitute back the answer in
terms of the starting variable.
Integration by Parts
If π and π are two functions of π, then
ππ
∫ ππ ππ = π ∫ π ππ − ∫ [ ππ . ∫ π ππ] ππ
i.e.
π°πππππππ ππ πππ πππππππ ππ πππ πππππππππ = πππ ππππππππ &times;
π°πππππππ ππ πππ − π°πππππππ ππ [π«πππππππππππ ππ πππ &times; π°πππππππ ππ πππ ]
If the two functions are of different types, take the first function to be the function
which comes first in the word ‘ILATE’ where
I – Inverse Trigonometric Functions
L – Logarithmic Functions
A – Algebraic Functions
T – Trigonometric Functions
E – Exponential Functions
Note:
ο The rule of integration by parts is useful in integrating single functions like
logarithmic and inverse t-functions. For this purpose, unity (i.e. 1) is taken as the
second function.
ο Here also, we see that ILATE Rule is not completely foolproof or hard and fast.
However it will found to be useful in most cases. Sometimes, not following ILATE
Rule is found to be simpler.
ο In many cases, substitution precedes the use of the by-parts formula.
Integrals of the form ∫ πππ πππ (ππ + π) ππ or ∫ πππ πππ (ππ + π) ππ
In such cases, we first put πΌ = ∫ π ππ₯ π ππ (ππ₯ + π) ππ₯ or πΌ = ∫ π ππ₯ πππ  (ππ₯ + π) ππ₯.
Then we integrate the given function by parts taking πππ as first function and
πππ (ππ + π) or πππ (ππ + π) as the second function. We then notice that on
integrating by parts the second time, the integrand contains a term of πΌ. By taking all
the πΌ terms to one side, we get the value of πΌ, which is the necessary integral.
Integrals of the form ∫ ππ [π(π) + π′ (π)] ππ
Method:
1. First split the given integral into two parts as follows:
∫ π π₯ [π(π₯ ) + π ′ (π₯ )] ππ₯ = ∫ π π₯ π(π₯ ) ππ₯ + ∫ π π₯ π ′ (π₯ ) ππ₯
2. Integrate only the first integral i.e. ∫ π π₯ π(π₯ ) ππ₯ by parts taking π(π₯ ) as first
function and π π₯ as second function.
3. Now, the integrand becomes π(π₯ )π π₯ − ∫ π π₯ π ′ (π₯ ) ππ₯ + ∫ π π₯ π ′ (π₯ ) ππ₯ . The
last two integrals will cancel each other.
Integrating a Complex Fraction by Decomposing into Partial Fractions
There are various methods of decomposing a given proper rational fraction into a
group of simple fractions each having as its denominator one of the factors of the
denominator of the original rational fraction. These simple fractions are called partial
fractions. Partial fractions method can be used in the following cases:
Case 1: When the denominator of the original complex fraction contains linear
unrepeated factors.
π₯+π
π΄
π΅
=
+
(π₯ + π)(π₯ + π) (π₯ + π) (π₯ + π)
We then solve the given expression by substituting various values of π₯ to obtain the
values of A and B.
This method can be extended and the given function can be decomposed to more
number of partial fractions depending upon the number of linear factors present in
the denominator of the given function.
Case 2: When the denominator of the original complex fraction contains linear
repeated factors.
π₯+π
π΄
π΅
πΆ
=
+
+
(π₯ + π)(π₯ + π)2 (π₯ + π) (π₯ + π) (π₯ + π)2
We then solve the given expression by substituting various values of π₯ to obtain the
values of A, B and C.
This method can be extended and the given function can be decomposed to more
number of partial fractions depending upon the number of linear factors present in
the denominator of the given function.
Case 3: When the denominator of the original complex fraction contains a quadratic
equation not resolvable into linear factors.
π₯+π
π΄
π΅π₯ + πΆ
=
+
(ππ₯ + π)(ππ₯ 2 + ππ₯ + π ) (ππ₯ + π) (ππ₯ 2 + ππ₯ + π)
We then solve the given expression by substituting various values of π₯ to obtain the
values of A, B and C. We can also equate the corresponding coefficients on both
sides to find the above values.
This method can be extended and the given function can be decomposed to more
number of partial fractions depending upon the number of linear or quadratic factors
present in the denominator of the given function.
Case 4: When there are polynomials in the numerator and denominator and the
degree of the numerator is greater than the degree of the denominator i.e. the given
fraction is an improper fraction, then we first use polynomial division to convert it
into a proper fraction and then apply any of the above cases if needed.
INTEGRALS OF THE FORM ∫
ππ₯
1
π₯
ππ₯
1
π₯−π
ππ₯
1
π+π₯
ππ
ππ +π
π,∫
ππ
ππ −π
π,∫
ππ
ππ −ππ
AND APPLICATIONS
1. ∫ 2 2 = tan−1 + π
π₯ +π
π
π
2. ∫ 2 2 = πππ | | + π
π₯ −π
2π
π₯+π
3. ∫ 2 2 = πππ | | + π
π −π₯
2π
π−π₯
Note: Before applying the above standard forms, always make the coefficient of π₯ 2
unity.
Method for solving integrals of the form ∫
ππ &plusmn;π
ππ +π
ππ:
Working Rule:
1. Divide the numerator and denominator by π₯ 2 .
1
2. It now takes the form ∫
1&plusmn; 2
π₯
1
π₯ 2+ 2
π₯
1
1
π₯
π₯2
ππ₯. Now substitute π₯ β = π‘ ⇒ (1 &plusmn;
2
3. To get an expression for (π₯ +
1
π₯
1 2
1
π₯
π₯2
2
2
2 ), (π₯ β ) = π‘ ⇒ (π₯ +
4. Rewrite the given integrand in terms of t and integrate using ∫
) ππ₯ = ππ‘.
) = π‘ 2 &plusmn; 2.
ππ₯
π₯ 2 +π2
,∫
ππ₯
π₯ 2 −π2
form.
Note: The method used above can be generalized to integrate integrals of the form
π₯ 2 &plusmn;π
∫ π₯ 4 +ππ₯ 2 +π ππ₯.
Method of Completing Squares
This method can be used to integrate integrals consisting of quadratic equations in
the denominator such as ∫
ππ₯
ππ’πππππ‘ππ
i.e. ∫
ππ₯
ππ₯ 2 +ππ₯+π
etc.
This method is especially useful in integrating quadratic equations which are not
resolvable into linear factors.
Working Rule:
1. Make the coefficient of π₯ 2 unity by taking the numerical coefficient of π₯ 2 outside.
ππ₯
1
∫ ππ₯ 2 +ππ₯+π = π ∫
ππ₯
π
π
π₯ 2 + π₯+
π
π
.
2. Complete squares in terms containing π₯ by adding and subtracting the square of
half of the coefficient of π₯ and put the denominator in the form {(π₯ + πΌ )2 &plusmn; π½ 2 }
1
∫
π
1
ππ₯
π
π
π₯ 2 +ππ₯+π
1
= ∫
π
ππ₯
π
π2 π2 π
π₯ 2 + π₯+ − +
π
4π 4π π
1
ππ₯
= ∫
π
π 2
π2
(π₯+2π) +(π−4π)
which is of the form
ππ₯
.
∫
π (π₯+πΌ)2 &plusmn;π½2
3. Apply the appropriate standard form ∫
ππ₯
π₯ 2 +π
Method to evaluate integrals of the form ∫
2,∫
ππ₯
π₯ 2 −π
ππ+π
πππ +ππ+π
2,∫
ππ₯
π2 −π₯ 2
ππ i.e. ∫
.
ππππππ
πππππππππ
ππ
Working Rule:
1. Split the given integral into two parts as follows:
ππ₯+π
π₯
1
∫ ππ₯ 2 +ππ₯+π ππ₯ = π ∫ ππ₯ 2 +ππ₯+π ππ₯ + π ∫ ππ₯ 2 +ππ₯+π ππ₯ .
2. The first integral can be integrated by the method of substitution by putting ππ₯ 2 +
ππ₯ + π = π‘ while the second integral can be integrated directly by applying the
method of completing squares and applying the appropriate standard form.
INTEGRALS OF THE FORM ∫
ππ₯
ππ
√ππ +π
,∫
π
ππ
√ππ −π
,∫
π
ππ
&amp; APPLICATIONS
√ππ −ππ
π₯
1. ∫ 2 2 = π ππ−1 + π
π
√π −π₯
ππ₯
2. ∫ 2 2 = log|π₯ + √π₯ 2 + π2 | + π
√π₯ +π
ππ₯
3. ∫ 2 2 = log|π₯ + √π₯ 2 − π2 | + π
√π₯ −π
Note: Before applying the above standard forms, always make the coefficient of π₯ 2
unity.
Method to evaluate integrals of the type ∫
ππ
√πππ +ππ+π
i.e ∫
ππ
√πππππππππ
Working Rule:
1. Make the coefficient of π₯ 2 unity by taking the numerical coefficient of π₯ 2 outside.
ππ₯
∫ √ππ₯ 2 +ππ₯+π =
1
∫
π
√
π
π
π
π
ππ₯
π
π
√π₯ 2 + π₯+
π
π
.
2. Put π₯ 2 + π₯ + in the form {(π₯ + πΌ )2 &plusmn; π½ 2 } by method of completing squares.
3. Apply the appropriate standard form ∫
Method to evaluate integrals of the type ∫
ππ₯
√π₯ 2 +π
,
2 ∫
ππ+π
√πππ +ππ+π
ππ₯
√π₯ 2 −π
,
2 ∫
ππ i.e. ∫
ππ₯
√π2 −π₯ 2
.
ππππππ
√πππππππππ
ππ
Working Rule:
1. Put ππ₯ + π = π΄.
π
ππ₯
(ππ₯ 2 + ππ₯ + π) + π΅ i.e. ππππππ = π΄.
π
ππ₯
(ππ’πππππ‘ππ ) + π΅.
2. Find A and B by equating the coefficients of the corresponding terms on both
sides.
3. Rewrite the given integral by substituting the numerator,
ππ₯ + π = π΄.
π
ππ₯
(ππ₯ 2 + ππ₯ + π) + π΅.
4. Now the given integral takes the form ∫
π΄.
π
(ππ₯ 2 +ππ₯+π)+π΅
ππ₯
√ππ₯ 2 +ππ₯+π
ππ₯.
5. Now, split the above integral into two integrals which can be easily evaluated:
π΄∫
π
(ππ₯ 2 +ππ₯+π)
ππ₯
√ππ₯ 2 +ππ₯+π
ππ₯ + π΅ ∫
ππ₯
√ππ₯ 2 +ππ₯+π
.
6. The first integral can be integrated by the method of substitution by putting ππ₯ 2 +
ππ₯ + π = π‘ while the second integral can be integrated directly by applying the
method of completing squares and applying the appropriate standard form.
INTEGRALS OF THE FORM ∫ √ππ − ππ ππ, ∫ √ππ + ππ ππ, ∫ √ππ − ππ ππ &amp;
APPLICATIONS
1
1
1
1
1
1
π₯
1. ∫ √π2 − π₯ 2 ππ₯ = π₯√π2 − π₯ 2 + π2 π ππ−1 + π
2
2
π
2. ∫ √π₯ 2 + π2 ππ₯ = π₯√π₯ 2 + π2 + a2 log|π₯ + √π₯ 2 + π2 | + π
2
2
3. ∫ √π₯ 2 − π2 ππ₯ = π₯√π₯ 2 + π2 − a2 log|π₯ + √π₯ 2 − π2 | + π
2
2
Aid to Memory:
1
1
∫(πΌππ‘ππππππ) ππ₯ = 2 π₯(πΌππ‘ππππππ) &plusmn; 2 a2 (πΌππ‘πππππ ππ ππππππππππ ππ πΌππ‘ππππππ) + π.
Method to find integrals of the type ∫ √πππ + ππ + π ππ i.e. ∫ √πππππππππ ππ
Working Rule:
1. Make the coefficient of π₯ 2 unity by taking the numerical coefficient of π₯ 2 outside.
π
π
∫ √ππ₯ 2 + ππ₯ + π ππ₯ = √π ∫ √π₯ 2 + π π₯ + π ππ₯ .
π
π
π
π
2. Put π₯ 2 + π₯ + in the form {(π₯ + πΌ )2 &plusmn; π½ 2 } by method of completing squares.
3. Apply the appropriate standard form ∫ √π2 − π₯ 2 ππ₯, ∫ √π₯ 2 + π2 ππ₯, ∫ √π₯ 2 − π2 ππ₯.
INTEGRALS OF THE FORM:
ππ
ππ
ππ
ππ
∫ π+π ππ¨π¬ π , ∫ π+π π¬π’π§ π , ∫ π ππ¨π¬ π+π π¬π’π§ π , ∫ π ππ¨π¬ π+π π¬π’π§ π+π πππ.
Working Rule:
π₯
1. Put sin π₯ =
2 tan2
π₯
π₯
1+tan2 2
2π₯
2. Transfer 1 + tan
2
π₯
2
Replace 1 + tan
2
, cos π₯ =
1−tan2 2
.
π₯
1+tan2 2
to the numerator after taking L.C.M. of the denominator.
π₯
= sec 2 in the numerator.
2
π₯
1
π₯
2
2
2
3. Put tan = π‘ so that sec 2 ππ₯ = ππ‘.
4. The new integrand will now be in the form ∫
ππ‘
ππ‘ 2 +ππ‘+π
which can be evaluated by
the method of completing squares followed by applying the appropriate standard
forms.
INTEGRALS OF THE FORM: ∫
π πππ π+π πππ π
π ππ¨π¬ π+π π¬π’π§ π
ππ
Working Rule:
1. Put π πππ  π₯ + π π ππ π₯ = π΄.
ππ’πππππ‘ππ = π΄.
π
ππ₯
π
ππ₯
(π cos π₯ + π sin π₯ ) + π΅. (π cos π₯ + π sin π₯ ) i.e.
(πππππππππ‘ππ ) + π΅. (πππππππππ‘ππ ).
2. Find A and B by equating the coefficients of cos π₯ and sin π₯ separately on both
sides.
3. Rewrite the given integral by substituting the numerator,
(π πππ  π₯ + π π ππ π₯ ) = π΄.
π
ππ₯
(π cos π₯ + π sin π₯ ) + π΅(π cos π₯ + π sin π₯ ).
π
4. Now the given integral takes the form ∫
π΄.ππ₯(π cos π₯+π sin π₯)+π΅(π cos π₯+π sin π₯)
π cos π₯+π sin π₯
ππ₯.
5. Now, split the above integral into two integrals as follows:
π΄∫
π
(π cos π₯+π sin π₯)
ππ₯
π cos π₯+π sin π₯
ππ₯ + π΅ ∫ ππ₯ .
6. The first integral can be integrated by the method of substitution by putting
(π cos π₯ + π sin π₯) = π‘ while the second integral can be integrated directly
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