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Question : 26
Total: 29
Evaluate:
t a n − 1
OR
Evaluate:
OR
Evaluate:
Solution:
Consider the given integral
I =
t a n − 1
Let t = sinx
⇒ dt = cos x dx
When x =
When x = 0 , t = 0
Now, ∫ 2 sin x cos xt a n − 1
= ∫ 2tt a n − 1
=[ t a n − 1 t ] [
( t a n − 1 t ) ∫ 2 t d t ]
[ t a n − 1 t ] [ 2 .
] (
× 2 .
)
=t 2 t a n − 1
=t 2 t a n − 1 [ 1 −
]
=t 2 t a n − 1
∴ I =
t a n − 1
=[ t 2 t a n − 1 t − t + t a n − 1 t ] 0 1
=[ 1 2 t a n − 1 1 − 1 + t a n − 1 1 ] [ 0 2 t a n − 1 0 − 0 + t a n − 1 0 ]
=[ 1 ×
− 1 +
]
=
− 1 +
=
OR
I =
Using the property
I =
⇒ I =
Adding (1) and (2)
2I =
⇒ I =
|
|
=
|
|
Putt a n 2
∴ 2 tan xs e c 2
⇒ tan xs e c 2
When x = 0 , z = 0 and when x =
∴ I =
⇒ I =
=
| t a n − 1 ( z ) | 0 ∞
=
t a n − 1 ∞ − t a n − 1 0
=
(
− 0 )
=
I =
Let t = sinx
⇒ dt = cos x dx
When x =
When x = 0 , t = 0
Now, ∫ 2 sin x cos x
= ∫ 2t
=
=
=
=
∴ I =
=
=
=
=
=
OR
I =
Using the property
I =
⇒ I =
Adding (1) and (2)
2I =
⇒ I =
=
Put
∴ 2 tan x
⇒ tan x
When x = 0 , z = 0 and when x =
∴ I =
⇒ I =
=
=
=
=
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