Question

the question is given below​

Answer 1

Consider the given integral.

I=∫

(1−sinx)(1+sin

2

x)

2cosx

dx

Let t=sinx

dt=cosxdx

Therefore,

I=∫

(1−t)(1+t

2

)

2

dt

I=∫(

1−t

1

+

t

2

+1

t+1

)dt

I=−∫

t−1

1

dt+

2

1

∫

t

2

+1

2t

dt+∫

t

2

+1

1

dt

I=−ln(t−1)+

2

1

ln(t

2

+1)+tan

−1

t+C

On putting the value of t, we get

I=−ln(sinx−1)+

2

1

ln(sin

2

x+1)+tan

−1

(sinx)+C

Hence, this is the answer