Question

Evaluate integral 0 to π/2 fn5xdx​

Answer 1

Answer:

Let I=∫

0

2

π

4+5cosx

1

dx

Substitute tan(

2

x

)=t⇒

2

1

sec

2

(

2

x

)dx=dt

So, x→0 ⇒t→0 and x→

2

π

⇒t→1

gives cosx=

1+t

2

1−t

2

,dx=

1+t

2

2dt

⇒ I=∫

0

1

4+5

1+t

2

1−t

2

1

1+t

2

2dt

=2∫

0

1

9−t

2

1

dt

⇒ I=

3

2

∫

0

1

1−

9

t

2

1

dt

Substitute

3

t

=u⇒dt=3du

So, t→0 ⇒u→0 and t→1 ⇒u→

3

1

⇒ I=

3

2

∫

0

3

1

1−u

2

1

du=

3

2

[

2

1

log

1−u

1+u

]

0

3

1

⇒ I=

3

1

(log2−0)=

3

1

log2

Answer 2

Answer:

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Step-by-step explanation:

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