Difinite Intgrals Laq Iib 2015-2016

8/20/2019 Difinite Intgrals Laq Iib 2015-2016

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A i m s

MATHEMATICS-IIB

DEFINITE INTEGRALS Q.NO 23

1. ∫0

π /4sinx+cosx

9+16sin2 x dx

sol : let

diff .w. r. t ' x '

L. L : x=0⇨t =sino−coso=0−1=−1

U .L : x=π

4

⇨t =sin

(

π

4

) – cos

(

π

4

)=

1

√ 2−

1

√ 2

sinx−cosx=t S.O.B

⇨(sinx−cosx)2=t 2

⇨ sin2 x+cos2 x−2 sinxcosx=t

2

⇨ 1−sin2 x=t 2

⇨

I =∫0

π /4sinx+cosx

9+16sin2 x dx

¿∫−1

01

9+16(1−t 2)

dt

¿∫−1

01

9+16−16 t

2 dt

¿∫−1

01

25−16 t 2 dt

¿∫−1

01

(5)2−(4 t )2 dt

¿[ 12 [5 ]

log|5+4 t

4−5 t |]4

|5+05−0|−¿ log|

5−15+1|

log ¿

¿ 1

40¿

1−¿ log 1

9

log¿

¿ 1

40 ¿

¿ 1

40[0−log3

−2 ]

¿ 1

40[2log3 ]

¿ 1

20

log3

2. ∫0

1log(1+ x )

(1+ x2) dx

sol : let

diff .w .r .t ' x '

L. L : x=0⇨θ=0

U .L : x=1⇨ θ=π

4

∫0

1

log

(1+ x)

(1+ x2) dx

Aims tutorial | 9000 68 76 00

sinx−cosx=t

(cosx+sinx ) dx=dt

1−t 2

=sin2 x

∴∫ 1a2− x2

dx= 12a

log|a+ xa− x|

log1=0

x=tanθ

dx=sec2θdθ



A i m s

MATHEMATICS-IIB

¿∫0

π /4log (1+tanθ)

(1+ tan2θ)

sec2θdθ

I =∫0

π / 4

log (1+tanθ ) dθ……… ..(1)

I =∫0

π /4

log(1+ tan [ π

4−θ])dθ

I =∫0

π /4

log

(1+ 1−tanθ

1+tanθ )dθ

I =∫0

π /4

log( 2

1+ tanθ )dθ

I =∫0

π /4

log2dθ−∫0

π /4

log(1+tanθ)dθ

I + I =∫0

π /4

log2dθ

2 I =log2∫0

π /4

(1)dθ

2 I =log2 [θ ]

2 I =log2 [ π

4−0]

I =π

8 log 2

3. ∫0

π x . sinx

1+sinx dx

Sol:

I =∫0

π

x . sinx1+sinx

dx

I =∫0

π (π − x ). sin (π − x)

1+sin (π − x ) dx

I =∫0

π

(π − x ).sinx1+sinx

dx

I =∫0

π πsinx

1+sinx dx−∫

0

π x . sinx

1+sinx dx

I + I =π ∫0

π sinx

1+sinx dx

2 I =π ∫0

π sinx(1−sinx)

(1+sinx )(1−sinx)dx

2 I =π ∫0

π sinx−sin

2 x

(1−sin2 x)

dx

2 I =π ∫0

π sinx−sin

2 x

(cos2 x) dx

2 I =π ∫0

π

[ sinx

cosx . 1

cosx−

sin2 x

cos2 x ]dx

2 I =π ∫0

π

[tanx . secx− tan2 x ]dx


f ( x ) dx=¿∫0

a

f (a− x )d

a .b

¿=log a−log b

=a

−



A i m s

MATHEMATICS-IIB

2 I =π ∫0

π

tanx.secx.dx -

π ∫0

π

tan2 x dx

2 I =π [ secx ] -

sex

(¿¿2 x−1)dx

π ∫0

π

¿

2 I =π [ secπ −sec0 ] - π [ tanx− x ]

2 I =π [−1−1 ] - π [0−0 ]+π [ π ]

2 I =−2π +π 2

I =¿ π

2

2 −π

4. ∫0

π x . sinx

1+cos2 x

dx

Sol :

I =∫0

π x . sinx

1+cos2 x

dx

I =∫0

π (π − x ). sin (π − x)

1+cos2 (π − x)

dx

I =∫0

π (π − x ).sinx

1+cos2 x

dx

I =∫0

π πsinx

1+cos2 x

dx−∫0

π x . sinx

1+cos2 x

dx

I + I =π ∫0

π sinx

1+cos2 x dx

2 I =π ∫0

π sinx

1+cos2 x

dx

let cosx=t ⇨−sixdx=dt

¿

L. L : x=0⇨ t =1

U .L : x=π ⇨ t =−1

⇨ 2 I =π ∫1

−1−1

1+t 2 dt

⇨ 2 I =π ∫−1

11

1+t 2 dt

2 I =π [ tan−1t ]

2 I =π [ tan−1 (1 )− tan−1(−1)]

2 I =π

[

π

4

+ π

4

]

2 I =π [2. π

4 ] ¿π [. π

4 ]

I =π

2

4


∫ tanx.secx.dx=secx

∫ sex2 x dx=tanx+c

=a

−

sinxdx=−dt



A i m s

MATHEMATICS-IIB

. ∫0

π /2 x

cosx+sinx dx

sol :

I =∫0

π /2 x

cosx+sinx dx

I =∫0

π /2 (π

2− x)

cos (

π

2− x)+ sin (

π

2− x)

dx

I =∫0

π /2 (π

2− x)

sinx+cosx dx

¿∫0

π /2π

2

cosx+sinx

dx−∫0

π /2 x

cosx+sinx

dx

I + I =π

2∫0

π /21

cosx+sinx dx

2 I =π

2∫0

π /21

cosx+sinx dx

Multiply∧divide By

√ a2+b

2∈ !

[ 1√ 2 ]

2 I = π

2√ 2∫0

π /21

cosx. 1

√ 2+sinx .

1

√ 2

dx

2 I = π

2√ 2∫0

π /21

cosx.cos (π

4

)+sinx. sin(π

4

)dx

2 I = π

2√ 2∫0

π /21

cos ( x−π

4) .

dx

2 I = π 2√ 2∫0

π

sec ( x−π 4 ). dx

2 I = π

2√ 2log|sec( x−π

4 )+ tan ( x−π

4 )|

2 I = π

2√ 2log|sec( π

2−

π

4 )+tan (π

2−

π

4)|

−π

2√ 2log|sec(0−

π 4 )+ tan (0− π

4)|

2 I = π

2√ 2[log|√ 2+1|− log|√ 2−1|]

2 I = π

2√ 2log|√ 2+1√ 2−1|

2 I = π 2√ 2

log|(√ 2+1)(√ 2+1)

(√ 2−1)(√ 2+1)|

2 I = π

2√ 2log

(√ 2+1)2

√ 22−1

2

2 I = 2π

2√ 2log

(√ 2+1)2−1

I = π 2√ 2

log(√ 2+1)

6. ∫0

π /2sin

2 x

cosx+sinx dx

sol :


=a

−



A i m s

MATHEMATICS-IIB

I =∫0

π /2sin

2 x

cosx+sinx dx …(1)

I =∫0

π /2 sin2(

π

2− x )

cos (π

2− x)+ sin (

π

2− x)

dx

I =∫0

π /2cos

2 x

sinx+cosx dx … (2)

"ddin#(1)∧(2)

¿∫0

π /2sin

2 x

cosx+sinx dx+∫

0

π /2cos

2 x

cosx+sinx dx

I + I =∫0

π /2sin

2 x+cos

2 x

cosx+sinx dx

2 I =∫0

π /21

cosx+sinx dx

Let t= tan ( x2 )$ dx =

2dt

1+t 2 $

cos x=1−t

2

1+t 2 sinx =

2t

1+t 2

2 I =∫0

11

(1−t

2

1+t 2

)+( 2 t

1+t 2 )

( 2dt

1+t 2)

2 I =∫0

11

1−t 2+2 t

1+t 2

( 2dt

1+ t 2)

2 I =2∫0

11

−(t 2−2t −1) dt

I =∫0

11

−[t 2−2 t +(1 )2−(1)2−1 ] dt

I =∫0

11

−[(t −1)2−(√ 2)2 ]

dt

I =∫0

11

[(√ 2)

2−(t −1)2

]

dt

2 I = 1

2 (√ 2 ) [ log|√ 2−t +1

√ 2+ t −1|]

I = 1

2√ 2 [log|√ 2√ 2|−log|√ 2+1√ 2−1|]

I = π

2√ 2log|(√ 2+1)(√ 2+1)

(√ 2−1)(√ 2+1)|

I = π

2√ 2log

(√ 2+1)2

√ 22−1

2

I = 2 π

2√ 2log

(√ 2+1)2−1

I = 1

√ 2log(√ 2+1)

7. ∫3

7

√7− x

x−3dx .

Sol :

let x=3cos2

θ+7sin2

θ dx=8 sinθcosθ.dθ


=a

−



A i m s

MATHEMATICS-IIB

L. L : x=3⇨θ=0

U .L : x=7⇨ θ=π

2

7− x=7−3cos2θ−7sin

2θ

¿7(1−sin2θ)−3cos

2θ

¿7(cos2θ)−3cos

2θ

¿4 cos2θ

x−3=3cos2θ+7sin2

θ−3

¿7sin2θ−3(1−cos

2θ)

¿7sin2θ−3(sin2

θ)

¿4 sin2θ

∫3

7

√7− x

x−3dx

¿∫0

π /2

√ 4cos

2θ

4sin2θ8sinθcosθdθ

¿8∫0

π /2

cos2θdθ

¿8.( 12 ) π

2=2π

8. ∫0

π

x .sin7 x cos

6 xdx

Sol :

I =∫0

π

x . sin7 xcos

6 x dx

I =∫0

π

(π − x)sin7(π − x )cos6(π − x )dx

I =∫0

π

(π − x)sin7 x cos

6 x dx

I =∫0

π

π sin7 x cos

6 x dx -

∫0

π

x sin7 x cos

6 x dx

I =∫0

π

π sin7 x cos

6 x dx − I

I + I =π ∫0

π

sin7 xcos

6 x dx

2 I =2π ∫0

π /2

sin7 xcos

6 x dx

I =π ∫0

π /2

sin7 x cos

6 x dx

cosn x sin

% x dx=¿

∫0

π /2

¿

( n−1 ) (n−3 ) (n−5 ) … (%−1 ) (%−3 ) …( %+n ) (%+n−2 ) ( %+n−4 ) …

I =π 6.4.2.5.3.1

13.11.9.7.5.3.1

I =π 6.4.213.11.9.7


∫

π /2

cos

n

x dx=

n−1

n .

n−3

n−2 .

n−5

n−4 …

π

2

f ( x ) dx=¿

a

f (a− x )dx

=a



A i m s

MATHEMATICS-IIB

I = 16 π

3003


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Difinite Intgrals Laq Iib 2015-2016