Question

integration of sin^3(2x+1)​

Answer 1

I=∫sin

3

(2x+1)dx

Let, 2x+1=t

2dx=dt

I=

2

1

∫sin

3

(t)dt

We know that,

sin(3t)=3sint−4sin

3

t

4sin

3

t=3sint−sin(3t)

sin

3

t=

4

1

[3sint−sin(3t)]

I=

2

1

[∫

4

1

(3sintdt)−∫

4

1

sin(3t)dt]

I=

2

1

×

4

1

[3cost−

3

cos3t

]+c. substitute t=2x+1

I=

24

1

[9cos(2x+1)−cos(6x+3)]+c.