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Formula Sheet
Indefinite Integration
SECOND
EDITION
22.
I N D E F I N I T E
I N T E G R A T I O N
1. INTRODUCTION
Integration is a reverse process of differentiation. The integral or primitive of a function f(x) with respect to x is a
differential function φ(x) such that the derivative of φ(x) with respect to x is the given function f(x). It is expressed
symbolically as ∫f(x)dx^ = φ(x)
Thus.
d f(x)dx (x) (x) f(x) dx
= φ ⇔ φ =
The process of finding the integral of a function is called Integration and the given function is called Integrand.
Now, it is obvious that the operation of integration is the inverse operation of differentiation. Hence the integral of
a function is also named as the anti-derivative of that function.
Further we observe that
( )
( )
( )
2
2 2
2
d x 2x dx d x 2 2x 2xdx x constant dx
d x k 2x dx
So we always add a constant to the integral of function, which is called the constant of Integration. It is generally
denoted by c. Due to the presence of this arbitrary constant such an integral is called an Indefinite Integral.
2. ELEMENTERY INTEGRATION
The following integrals are directly obtained from the derivatives of standard functions.
(a) 0.dx =c
(b) 1.dx = x +c
(c) k.dx = kx + c(k ∈R)
(d)
n 1 n x x dx c(n 1) n 1
= + ≠ −
(e) (^) e
dx log x c x
∫ =^ +
(f)
x x
∫e dx^ =^ e^ +c
Mathematics | 22.
Illustration 3: Evaluate:
4
∫ sin^ x^ dx (JEE MAIN)
Sol: Here as we know, 2
1 cos2x sin x 2
= , Now by putting this in the above integration and solving we will get the
term
(1 2cos2x cos 2x)dx 4
∫ −^ + , After that by using the formula
2 1 cos 4x cos 2x 2
= we can solve the problem given above.
4 sin x dx
2 1 cos2x (^12) dx (1 2cos2x cos 2x)dx 2 4
=^ −^ +
1 1 cos 4x 1 1 2cos2x dx (3 4 cos2x cos 4x)dx 4 2 8
−^ +^ =^ −^ +
∫ ∫ =^
1 sin 4x 3x 2sin2x C 8 4
−^ +^ +
Illustration 4: If
3 4
f '(x) 4x x
= − such that f(2)=0,then, find f(x) (JEE ADVANCED)
Sol: Here
3 4
f '(x) 4x x
= − (^) therefore 3 4
f(x) 4x dx x
hence by splitting this integration and solving we will get
the result.
We have,
3 4
d 3 f(x) 4x dx (^) x
3 3 3 4 4 4
f(x) 4x dx 4x dx dx 4 x dx 3 x dx x x
3 1 4 1 4 3
x x 1 4 3 C x C (^3 1 4 1) x
…(i)
Given 4 3
f(2) 2 C 0 2
0 16 C C
Putting the value of C in (i), we get 4 3
f(x) x x^8
4. METHODS OF INTEGRATION
When the integrand can’t be reduced into some standard form then integration is performed using following
methods
4.1 Integration by Substitution
4.1.1 Integrand is a Function of Another Function
If the integral is of the form
'
∫ f^ ^ φ(x)^ φ(x)dx, then we put^ φ(x)^ =^ tso that^
'
φ (x)dx=dt. Now integral is reduced ∫ f(t)dt.
In this method the function is broken into two factors so that one factor can be expressed in terms of
the function whose differential coefficient is the second factor.
In case of objective questions in which direct indefinite integration is asked, function being very
complicated to integrate, then try differentiating the options.
MASTERJEE CONCEPTS
22.4 | Indefinite Integration
If
dx I , sin(x a)cos(x b)
then I is Equal to
(a)
1 sin(x a) log C sin(a b) cos(x b)
(b)
1 sin(x a) log C cos(a b) cos(x b)
(c)
1 sin(x a) log C sin(a b) cos(x b)
(d)
1 sin(x a) log C cos(a b) cos(x b)
Vaibhav KrishnanJEE 2009, AIR 22
Illustration 5: Evaluate:
2 2
∫ x tanx^ sec x dx (JEE MAIN)
Sol: This problem is based on integration using substitution method. In this we can put 2 x =t and therefore 2x
dx=dt and then solving we will get the result.
Let 2 x =t
⇒ 2x dx=dt ⇒ x dx =
dt ∴
2 2 x tanx sec x dx
∫ =^
tant sec t dt sec t c sec x c 2 2 2
4.1.2 Integrand is the Product of Function and its Derivative
If the integral is of the form I= ' f (x)
f(x) dx we put f(x) = t and convert it into a standard integral.
Illustration 6: Evaluate:
2 tanx sec x dx
∫ (JEE MAIN)
Sol: Here 2 sec x is a derivatives of tanx hence we can put tan x = t and 2 sec x.dx = dt thereafter we can solve the
given problem.
Let tan x =t ⇒ 2 sec x.dx =dt
2 I = tanx sec x dx = t dt
2 2 t tan x c c 2 2
4.1.3 Integrand is a Function of the Form f(ax+b)
Here we put ax+b =t and convert it into a standard integral. Now if,
∫ f(x)dx^ = φ(x),then^
f(ax b)dx (ax b) a
∫^ +^ =^ φ^ +
Illustration 7: Evaluate: cos3x cos5x dx
(JEE MAIN)
Sol: By multiplying and dividing by 2 in the given integration and using the formula
2cos A.cosB = cos A ( + B (^) ) + cos A( − B)we can solve it.
I= cos3x cos5x dx ∫ (^ )
cos8x cos2x dx sin8x sin2x c 2 2 8 2
22.6 | Indefinite Integration
m n sin x cos x dx
∫ , where m, n ϵ N
⇒ If m is odd put cos x=t
If n is odd put sin x = t
If both m and n are odd, put sin x=t if m ≥ n and cos x=t otherwise.
If both m and n are even, use power reducing formulae
2 1 cos 2x^21 cos 2x sin x or cos x 2 2
If m+n is a negative even integer, put tan x=t
Shrikant Nagori (JEE 2009, AIR 30)
MASTERJEE CONCEPTS
Illustration 10: Evaluate:
dx sinx +cos x
(JEE ADVANCED)
Sol: As we know, if integration is in the form of
dx
asinx +bcosx
∫ then we can put
a=r cos θ and b=r sin θ hence the integration will be
1 x log tan c r 2
+ θ +
Here a=1 & b=
So
dx sinx +cos x
1 x 1 1 1 x logtan tan 1 c logtan c 1 1 2 2 2 2 8
(^) − π +^ +^ =^ +^ +
4.1.5 Standard Substitutions
The following standard substitutions will be useful
Integrand form Substitutions
2 2 a − x or^ 2 2
a −x
x = a sin θ or x = a cos θ
2 2 x + a or^ 2 2 x
+a
x = a tan θ or x = a cot θ or x=a sinh θ
2 2 x − a or^ 2 2
x −a
x = a sec θ or x = acosec θ
x
a + x
or
a x
x
or x(a + x) or
x(a +x)
x=a
2 tan θ
x
a − x
or
a x
x
or x(a − x)or
x(a −x)
x = a
2 sin θ or x= a cos 2 θ
Mathematics | 22.
x
x − a
or
x a
x
or x(x − a)or
x(x −a)
x=a
2 sec θ or x=a
2 cosec θ
a x
a x
or
a x
a x
x = a cos2 θ
x
x
− α
β −
or (x − α)( β − x) ( β > α)
x = 2 2 α cos θ + β sin θ
Some Standard Integrals
(a) tanxdx = logsec x + c = −logcos x +c
(b) ∫cot xdx = logsinx + c = −logcosec x +c
(c) ∫ sec xdx = log(sec x + tanx) +c
x log(sec x tanx) c logtan c 4 2
π = − − + = (^) + (^) +
(d) (^) cosecxdx = − log(cosecx + cot x) +c
x log(cosecx cotx) c log tan c 2
(e) sec x tanxdx = sec x +c
(f) ∫cosec x cot x dx = − cosec x +c
(g)
2
∫^ sec^ xdx^ =^ tanx^ +c
(h) 2 cosec xdx = − cot x +c
(i) (^) ( )
x logxdx xlog c x logx 1 c e
If the integral is of the form
(^1 1 ) p q (^) r R x , x , x ...... dx
∫ , where R is a rational function then,
Let a = lcm of (p,q,r,…….) and put x =
a t
Nitish Jhawar (JEE 2009, AIR 7)
MASTERJEE CONCEPTS
Illustration 11: Prove that:
2 2 2 2
dx n(x x a ) C x a
log 2 2 2 2
dx n(x x a ) C x a
∫ ^
(JEE ADVANCED)
Sol: By putting x = a sec θ ⇒ dx = a sec θ tan dθ θ , we can solve the problem given above.
Let x = a sec θ ⇒ dx = a sec^ θ^ tan dθ θ^ ⇒ 2 2
dx
x −a
∫ =^
asec tan d sec d atan
θ θ θ = θ θ θ
∫ ∫ = logn(sec^ θ +^ tan^ θ +)^ C
= log
2 2 x a x n( ) C a a
+ + = log
2 2 n(x + x + a ) +C'
Mathematics | 22.
4 4 2 2 2 2
4 4
(sin x cos x)(sin x cos x)(sin x cos x) dx (sin x cos x)
∫ =^
2 2 2 2 1.(sin x − cos x)dx = − cos2xdx ^ cos2x = cos x −sin x
sin 2x C 2
4.2 Integration by Parts
If u and v are two functions of x, then
( ) ( )
du (u.v)dx u v dx v dx dx dx
This is also known as uv rule of integration. This method of integrating is called integration by parts.
- From the first letter of the words inverse circular, logarithmic, Algebraic, Trigonometric, Exponential functions, we get a word ILATE. Therefore the preference of selecting the u function will be according
to the order ILATE.
- In some problems we have to give preference to logarithmic function over inverse trigonometric
functions. Hence sometimes the word LIATE is used for reference.
- For the integration of Logarithmic or Inverse trigonometric functions alone, take unity (1) as the v
function.
Shivam Agarwal (JEE 2009, AIR 27)
MASTERJEE CONCEPTS
Illustration 15: Evaluate: (1 +x)logxdx
(JEE MAIN)
Sol: Here we can integrate the given problem by using Integration by parts i.e.
( ) ( )
du (u.v)dx u v dx v dx dx dx
Here u = logx^ and v = (1^ +^ x).
Let I= ∫(1 +x)logxdx
Integrating by parts, taking log x as 1 st function, (by LIATE rule) we get
I= (^) ( )
d logx 1 x dx (logx). (1 x)dx dx dx
2 2 x 1 x logx x. x dx 2 x 2
+^ −^ +
2 x x x logx 1 dx 2 2
+^ −^ +
^
2 2 x x x logx x C 2 4
Illustration 16: Evaluate:
3
∫ sec^ xdx (JEE ADVANCED)
Sol: Here we can solve by integrating by parts, taking sec x as the first function.
I =
3
∫ sec^ xdx=^
2
∫ sec x.sec^ xdx Let^
2 u = sec x & v =sec x
I = sec x tan x − (^) ∫ ( sec x tan x. tan x dx) = 2 sec x tanx − sec x tan xdx
∫ =^
2 sec x tanx − sec x(sec x −1)dx
22.10 | Indefinite Integration
3 I = sec x tanx − sec xdx + sec xdx
∫ ∫ ⇒^
I = sec x + tan x − I + sec x dx
⇒ 2 I = sec x. tan x + log (^) ( sec x + tan x (^) )+ C
I sec x tanx log(sec x tanx) C 2
Illustration 17: Evaluate :
1 2 (sin x) dx
−
(JEE ADVANCED)
Sol: We can write the given integration as
1 2 (sin x) .1dx
− ∫ and^ then^ taking^ (^ )
2 1 u sin x & v 1
− = =
solving by integration by parts.
I=
1 2 d 1 2 (sin x) .x (sin x) .x dx dx
− ^ −
= (^) ( ) 1 2 1 2
(sin x) .x 2 sin x. .x dx 1 x
− −
Now, putting
1 sin x
− = t ⇒ x = sin t so that 2
dx
1 −x
= dt
⇒ I =
1 2 x(sin x) 2 t.sintdt − − ∫ =^ {^ }
1 2 x(sin x) 2 t cos t cos tdt
−
− − + ∫ (again Integrating by parts)
= (^) { } 1 2 x(sin x) 2 t cos t sint C − − − + + = 1 2 x(sin x) 2t cos t 2sint C −
- − + = 1 2 1 2 x(sin x) 2sin x. 1 x 2x C − −
- − − +
Illustration 18: Evaluate:
1 1
1 1
sin x cos x dx sin x cos x
− −
− −
(JEE ADVANCED)
Sol: By using the formula 1 1 sin x cos x 2
− − π
- = , we can solve the above problem.
Let I=
1 1
1 1
sin x cos x dx sin x cos x
− −
− −
( )
1 1 sin x ( / 2) sin x dx ( / 2)
− − − π −
π
1 1 sin x cos x 2
(^) − − π +^ =
2sin x dx sin xdx 1dx 2
(^) − π − −^ =^ − π (^) π
sin xdx x − − π
… (i)
Putting
1 sin x
− = θ ⇒ x=
2 sin θ so that dx = 2 sin θ. cos θ d θ = sin 2 θ d θ.
1 sin xdx .sin2 d
−
∫ =^ ∫ θ^ θ θLet^ u^ = θ^ & V^ =^ sin 2θ^ , then integing by parts we get
( )
1 2sin 2 sin .cos 2 4
− θ = − θ + θ θ
cos2 1 1
. cos2 d cos2 sin 2 2 2 4
θ θ −θ + θ θ = − θ + θ
(1 2sin ) sin 1 sin 2 2
− θ − θ + θ − θ
(sin x )(1 2 x) x. 1 x C 2 2
− − − + − + (^) … (ii)
From (i) and (ii), we get
I =
(1 2x)sin x x x x C 2 2
− −^ +^ −^ −^ +
π
= (^) { }
x x (1 2x)sin x x C − − − − − + π
4.2.1 Integration by Cancellation
Illustration 19: Evaluate :
3x tan x sec dx x x
∫ (JEE MAIN)
22.12 | Indefinite Integration
4.3 Integration of Rational Functions
4.3.1 When the Denominator can be Factorized (Using Partial Fraction)
Let the integrand be of the form
f(x)
g(x)
, where both f(x) and g(x) are polynomials. If degree of f(x) is greater than
degree of g(x), then first divide f(x) by g(x) till the degree of the remainder becomes less than the degree of g(x).
Let Q(x) be the quotient and R(x), the remainder then
f(x) R(x) Q(x) g(x) g(x)
Now in R(x)/g(x), factorize g(x) and then write partial fractions in the following manner:
(a) For every non-repeated linear factor in the denominator. Write
1 A B
(x a)(x b) x a x b
(b) For every repeated linear factor in the denominator. Write
3 2 3
1 A B C D
(x a) (x b) (x^ a)^ (x a) (x a) (x^ b)
− − −^ − − −
(c) For every non-repeated quadratic factor in the denominator. Write
2 2
1 Ax B C
(ax bx c)(x d) ax bx c x^ d
(d) For every repeated quadratic factor in the denominator. Write
2 2 2 2 2
1 Ax B Cx^ D E
(ax bx c) (x d) (ax bx c) ax bx c x^ d
+^ +
Consider f(x) as the function we need to factorize
- For non- repeated linear factor in the denominator.
Let f(x)=
1 A B
(x a)(x b) (x a) (x b)
To obtain the value of A remove (x−a) from f(x) and find f(a).
Similarly, to obtain value of B, remove (x-b) from f(x) and find f(b).
- For repeated linear factor in the denominator.
Let f(x) = 3 2 3
1 A B C D
(x a) (x b) (x^ a)^ (x a) (x a) (x^ b)
− − −^ − − −
To obtain value of D remove (x−b) from f(x) and find f(b).
To obtain value of c remove (x−a)^3 from f(x) and find f(a).
Now that we have reduced the number of unknowns from 4 to 2, we can find A and B easily by equating.
MASTERJEE CONCEPTS
Mathematics | 22.
Now let’s try this method for
4 3 2
2 2 3
x x 2x x 4
x(x 2)(x 1)
Partial fraction will be of the form
4 3 2
2 2 3 2 2 2 2 2 3
x x 2x x 4 A Bx C Dx E Fx G Hx I
x(x 2)(x 1) x (x 2) (x 1) (x 1) (x 1)
Now remove x and put x=0, we get A=
Now remove
2 3 (x + 1) and put
2 x =-1 i.e. x = i (you can also substitute x = −i ).
We get Hi+I = –3i –2. Hence H = –3 and I = –2.
Now remove
2 (x + 2)and put x= (^) 2i. We get B ( 2i ) +C=2 2i +3. Hence B = 2 and C = 3
Now the number of unknowns have reduced from 9 to 4 and the remaining unknowns can be solved
easily.
This method very useful instead of solving for all the unknowns at the same time.
Also remember that substituting an imaginary number for x is not discussed anywhere in NCERT. So, use
this method only for competitive exams.
Ravi Vooda (JEE 2009, AIR 71)
Illustration 23: Evaluate : 2
x dx x − x − 2
(JEE MAIN)
Sol: Here the given integration is in the form of
(x − a)(x − b)
, hence by using partial fractions we can split it as
A B
(x a) (x b)
and then by solving we will get the required result.
Here I =
x dx (x − 2)(x +1)
∫ =^
dx 3 x 2 x 1
−^ +
∫ =^
2log(x 2) log(x 1) c log (x 2) (x 1) c 3 3
− + + + = ^ − + +
Illustration 24: Evaluate : 4 2
xdx
3x − 18x + 11
∫ (JEE ADVANCED)
Sol: Here simply by putting t= 2 x ⇒ dt = 2x dx and then by using partial fractions we can solve the given problem.
I=
4 2
xdx
3x − 18x + 11
∫ dx^ =^2
dt 2
3t − 18t + 11
∫ (Put t=^
2 x ⇒ dt = 2x dx)
( )
2 2 2 2
1 dt 1 dt 1 dt
(^6) t 6t (11 / 3) 6 (t 3) (16 / 3) 6 (t 3) 4 / 3
1 1 (t 3) (4 / 3) 3 3t 3 3 4 log C log C (^6 2) (4 / 3) (t 3) (4 / 3) 48 3t 3 3 4
× − + − +
2
2
3 3x 3 3 4 log C (^48) 3x 3 3 4
Mathematics | 22.
I =
2 2
(^2 2 )
1 (1 / x ) 1 (1 / x ) dx dx x (1 / x ) x 1 / x 2
+ ^
Now taking x-1/x=t ⇒[1+1/
2 x ]dx = dt, we get
I =
1 2
dt 1 t tan c t 2 2 2
− ^
2 1 1 x 1 tan c 2 2x
−
4.4 Integration of Irrational Functions
If any one term in numerator or denominator is irrational then it is made rational by a suitable substitution. Also
if the integral is of the form
2
dx
ax + bx +c
or 2
∫ ax^ +^ bx^ +cdx
Then we integrate it by expressing 2 2 2 ax + bx + c = (x + α ) + β
Also for integrals of the form 2
px q dx ax bx c
or
2 (px + q) ax bx +c dx
First we express px+q in the form
px+q = A [
d 2 (ax bx c) dx
- ]+B and then proceed as usual with standard form.
Illustration 26: Evaluate :
x
x 2x
e dx 5 − 4e −e
(JEE MAIN)
Sol: Simply by putting
x e (^) = t , then ex dx = dt, we can solve the given problem.
Put
x e = t , then
x e dx = dt
x
x 2x
e dx 5 − 4e −e
2
dt
5 − 4t −t
2
dt
5 − (t +4t)
∫ =^
2
dt
5 − (t + 4t + 4) + 4
( )
2
dt
9 − t + 2
2 2
dt
(3) − (t +2)
∫ =^
1 t^2 sin C 3
− ^ +^
x 1 e^2 sin C 3
−
Illustration 27: Evaluate :
dx (x − a)(x −b)
∫ (JEE ADVANCED)
Sol: Here first expand (^) (x − a)(x − b) and then adding and subtracting by
2 a b
2
, we can reduce the above
integration. After that by putting
a b x u 2
, we can solve the given problem.
Let, I =
dx (x − a)(x −b)
2
dx x − (a + b) x +ab
( ) ( )
2 2 2
dx
x − (a + b) x + (a + b) / 2 − (a + b) / 2 +ab
( ( )) (^) ( )
(^2 2 )
dx
x − (a + b) / 2 − ^ (a + b + 2ab) / 4 −ab
(^2 )
dx
x (a b) / 2 a b / 4
(^2 )
dx
x − (a + b) / 2 − (a −b) / 2
∫ … (i)
22.16 | Indefinite Integration
On putting
a b x u 2
so that dx = du in (i), we get
I =
( )
2 2
du
u − (a −b) / 2
2 2 2 2
dx log x x a x a
2 2 a^ b log u u C 2
c
Putting u=
a b x 2
^
, we get
I =
2 2 a b a b a b log x x C 2 2 2
−^ +^ −^ −^ +
^ ^ ^ ^ ^
c = (^) ( )( )
a b log x x a x b C 2
− + − − + c
5. STANDARD INTEGRALS
(a)
1 2 2
1 1 x dx tan c x a a^ a
− ^
(b) 2 2
1 1 x a dx log c x a 2a^ x^ a
(c) (^) 2 2
1 1 a x dx log c a x 2a^ a^ x
(d) 1 1 2 2
1 x x dx sin c cos c a a a x
− ^ ^ − ^
− ^ ^ ^
(e) 1 2 2 2 2
1 x dx sinh c log x x a c a x a
− ^ ^
^
(f) 1 2 2 2 2
1 x dx cosh c log x x a c a x a
− ^ ^
^
(g)
2 2 2 x^2 2 a^1 x a x dx a x sin c 2 2 a
− − = − + +
(Substitute x = acosθ or x = asinθ and proceed)
(h)
2 2 2 x^2 2 a 2 2 x a dx x a n x x a c 2 2
∫ +^ =^ +^ +^ log+^ +^ +
2 2 2 x^2 2 a 2 2 x a dx x a n x x a c 2 2
∫ +^ =^ +^ +^ ^ +^ +^ + (Substitute^ x^ =^ atanθ^ or^ x^ =^ acotθ^ and proceed)
(i)
2 2 2 x^2 2 a 2 2 x a dx x a n x x a c 2 2
∫ −^ =^ −^ −^ log +^ −^ +
2 2 2 x^2 2 a 2 2 x a dx x a n x x a c 2 2
∫ −^ =^ −^ −^ ^ +^ −^ + (Substitute^ x^ =^ asecθ^ or^ x^ =^ acosecθ^ and proceed)
( j)
1 2 2
1 1 x dx sec c x x a a^ a
− = + −
∫ (Valid for x > a > 0)
(k)
ax ax 2 2
e e sinbx dx (asinbx bcosbx) c a b
ax 1 2 2
e b sin bx tan c a a b
−^ +
+ ^ ^
22.18 | Indefinite Integration
m (^2)
dx
ax + b ax + bx +c
∫ ax + b =1 / t
Illustration 28: Evaluate :
dx
(x + 1) x − 2
(JEE MAIN)
Sol: Simply by putting
2 x − 2 = t, ∴dx = 2t dtwe can solve the given problem by using the appropriate formula.
dx
(x + 1) x − 2
Put
2 x − 2 =t
∴ dx =2t dt
∴ I =
2
2t dt
(t +3)t
∫ =^2
dt 2 t +( 3)
∫ =^
2 1 t tan c 3 3
− ^
2 1 x 2 tan c 3 3
−
(∵ t = (x − 2))
Illustration 29: Evaluate : 2
dx
(x −4) x
∫ (JEE MAIN)
Sol: Here first put 2 x = t therefore dx = 2t dt and then using partial fractions we reduce the given integration in
standard form. After that by solving we will get the result.
Let I = 2
dx
(x −4) x
Put
2 x = t ∴ dx = 2t dt then I = 4
2t dt (t −4)t
2 2
dt
(t + 2)(t −2)
Put 2 t = z∴ 2 2
(t 2)(t 2) (z^ 2)(z^ 2)
+ − +^ −
A B
z 2 z 2
A =
− (^) and B =
2 2 2 2
(t 2)(t 2) 4(t 2) 4(t 2)
∴ I =
2 2 2 2
1 1 dt 1 dt 2 (t 2)(t 2) 2 t 2 2 t 2
∫ ∫ ∫ =^
1 1 t 1 t 2 tan log c 2 2 2 4 2 t 2
− ^ ^ −
1 1 x 1 x 2 tan log C 2 2 2 4 2 x 2
− ^ ^ −
c (∵ t = x)
6. SPECIAL TRIGONOMETRIC FUNCTIONS
Here we shall study the methods for evaluation of the following types of integrals.
Type 1
(i) 2
dx
a +bsin x
(ii) 2
dx
a +bcos x
(iii) 2 2
dx
acos x + bsinx cos x +csin x
(iv) 2
dx
∫(asinx^ +bcos x)
Method: Divide the numerator and denominator by cos^2 x in all such types of integrals and then put tan x=t
Mathematics | 22.
Illustration 30: Evaluate : (^2)
dx
1 +3sin x
∫ (JEE MAIN)
Sol: Here dividing the numerator and denominator by cos 2 x we can solve it.
I =
2
2 2
sec xdx
sec x +3tan x
2 1 2
sec x dx 1 tan (2tanx) c 1 4 tan x^2
− = +
Type 2
(i)
dx
a +bcos x
(ii)
dx
a +bsinx
(iii)
dx
acos x +bsinx
(iv)
dx
asinx + bcos x +c
Method: In such types of integrals we use the following substitutions
( )
( )
2 2
2tan x / 2 (^) 2t sinx 1 tan x / 2 1 t
( )
( )
(^2 )
2 2 2
1 tan x / 2 (^1) t 2dt cos x ; dx 1 tan x / 2 1 t 1 t
and integrate another method for the evaluation of the integral.
Illustration 31: Evaluate:
dx dx 5 +4 cos x
(JEE MAIN)
Sol: Here by putting
( )
( )
2
2
1 tan x / 2 cos x 1 tan x / 2
and then by taking tan (x/2) = t we can solve the given problem
I =
2 2
dx
5 + 4 (1^ − tan (x / 2)) / (1 +tan (x / 2))
2
2
sec (x / 2) dx 9 +tan (x / 2)
2 2
dt 2 3 +t
where tan (x/2) = t
1 1 t 2 tan C 3 3
^
1 1 tan x / 2 2 tan C 3 3
Type 3
(i)
psinx qcos x dx asinx bcos x
(ii)
psinx dx asinx +bcos x
(iii)
qcos x dx asinx +bcos x
For their integration, we first express numerator as follows-
Numerator = A (denominator) + B (derivative of denominator)
Then integral = Ax + B log (denominator) + C
Illustration 32: Evaluate :
6 3sinx 14 cos x dx 3 4 sinx 5cos x
∫ (JEE ADVANCED)
Sol: By using partial fractions, we can reduce the given integration to the standard form.
6 3sinx 14 cos x dx 3 4 sinx 5cos x
⇒ 6 + 3sinx + 14 cos x = A ( 3 + 4 sinx + 5cos x) + B ( 4 cos x − 5sinx) + c
Solving R.H.S. & comparing both sides, we get 4A – 5B = 3 5A + 4B = 14
Also, 3A+C=6 ∴
A(3 4 sinx 5cos x) B(4 cosx 5sinx) c
3 4 sinx 5cos x
