Question

evaluate: integration of log(2+sinx/2-sinx) dx from π/2 to -π/2​

Answer 1

Answer:

Let I=∫

sinx

log(tan

2

x

)

dx

Let logtan

2

x

=t. Then,

tan

2

x

1

sec

2

2

x

×

2

1

dx=dt

dx=sinx dt

Putting logtan

2

x

=t and dx=sinx dt, we get,

I=∫

sinx

t

sinx dt=∫t dt=

2

t

2

+C=

2

(logtan

2

x

)

2

+C

Answer 2

Answer:

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