Question

Evaluate ∫dx/(sinx+sin2x)

Answer 1

Explanation:

∫

x

3

−x

2

−x+1

3x+5

​

dx

∫

x

2

(x−1)−(x−1)

3x+5

​

dx=∫

(x−1)(x

2

−1)

3x+5

​

dx

∫

(x−1)(x+1)(x−1)

3x+5

​

dx=∫

(x−1)

2

(x+1)

3x+5

​

dx

Let

(x−1)

2

(x+1)

3x+5

​

=

x−1

A

​

+

(x−1)

2

B

​

+

(x+1)

C

​

3x+5=A(x

2

−1)+B(x+1)+C(x−1)

2

=Ax

2

−A+Bx+B+Cx

2

−2Cx+C

A+C=0,B−2C=3,−A+B+C=5

⇒A=

2

−1

​

, B=4, C=

2

1

​

2

−1

​

∫

x−1

1

​

dx+4∫

(x−1)

2

1

​

+

2

1

​

∫

x+1

1

​

=

2

−1

​

log∣x−1∣+4×

(x−1)

−1

​

+

2

1

​

log∣x+1∣+C

=

2

1

​

log

∣

∣

∣

∣

∣

​

x−1

x+1

​

∣

∣

∣

∣

∣

​

−

(x−1)

4

​

+C

Answer 2

∫dx/(sinx+sin2x)

Explanation:

To evaluate the definite integral ∫dx/(sinx+sin2x), we can use the substitution u = sin x. This substitution allows us to rewrite the integrand as follows:

∫dx/(sinx+sin2x) = ∫du/(u+u^2)

We can then rewrite the integrand as a partial fraction:

∫du/(u+u^2) = ∫(1/u - 1/(u+1))du

Integrating each of these terms separately, we get:

∫(1/u - 1/(u+1))du = ln|u| - ln|u+1| + C

Substituting back in for u, we get:

ln|sin x| - ln|sin x + 1| + C

This is the final answer for the definite integral ∫dx/(sinx+sin2x).