π 4

∫

0

sinx+cosx

16+9sin2x

dx
Now, sin2x = 2sinx cosx
∴ 1- sin2x = 1 - 2sinx cosx
∴ 1 - sin2x = (sinx−cosx)2
Put sin x - cos x = t
⇒ (sin x + cos x) dx = dt
x =

π

4

, t = 0
x = 0 , t = - 1
So,

0

∫

−1

1

16−9(1−t2)

dt
=

0

∫

−1

1

16−9−9t2

dt
=

0

∫

−1

1

25−9t2

dt
=

1

9

0

∫

−1

1

25 9 −t2

dt
=
=

1

30

(log|1|−log‌

1

4

)
=

1

30

log 4
OR
Given

3

∫

1

(x2+3x+ex) dx
⇒ a = 1 , b = 3
⇒ h =

3−1

n

and f (x) = x2 + 3x + ex
I =

3

∫

1

(x2+3x+ex) dx
I =

lim

h→0

h [f (1) + f (1 + h) + ... + f (1 + (n - 1)h)]
I =

lim

h→0

h [4 + e + [(1+h)2 + 3 (1 + h) + e1+h]] + [(1+2h)2 + 3 (1 + 2h) + e1+2h] + ... + [(1+(n−1)h)2 + 3 (1 + (n - 1) h + e1+(n−1)h)]
I =

lim

h→0

h
I =

lim

n→∞

4n ×

2

n

+ e ×

2

n

+

8

n3

(n−1)(n−2)(n−3)

6

+

5×4

n2

(

n2−n

2

) + e

2

n+1

(e

2

n

(n−1)−1)
I = 8 +

8

3

+ 10 + e (e2 - 1)
I =

62

3

+ e3 - e