Question Statement
Evaluate the integral:
β«Ο/6Ο/2βSinx(2+Sinx)Cosxβ,dx
Background and Explanation
To solve this integral, we need to understand the basic trigonometric identities and properties. Specifically:
- Cosecant (Cosecx=sinx1β) and cotangent (Cotx=sinxcosxβ) will help simplify the expression.
- The integral involves a rational trigonometric expression, which requires manipulating and simplifying using these identities.
- This type of integral can often be simplified by substitution or recognizing standard integral forms.
Solution
Step-by-step explanation:
- Rewrite the integrand:
We start by rewriting the integrand using trigonometric identities.
Sinx(2+Sinx)Cosxβ=(sinx2β+1)β1(β2CosecxCotx)
- Substitute:
Next, we simplify the expression by noting that Cosecx=sinx1β and Cotx=sinxcosxβ. This leads to:
β«Ο/6Ο/2β(2Cosecx+1)β1(β2CosecxCotx),dx
- Factor and simplify:
We can simplify the expression further, and focus on the integration:
=β61ββ«Ο/6Ο/2β(2Cosecx+1)β1(β2CosecxCotx),dx
- Apply integration rules:
Using standard integral rules for logarithmic and trigonometric functions, we get:
=β21βln((2Cosecx+1))βΟ/6Ο/2β
- Evaluate the bounds:
Finally, evaluate the logarithmic function at the limits x=Ο/2 and x=Ο/6:
=β21β[ln(sin(Ο/2)2β+1)βln(sin(Ο/6)2β+1)]
Simplifying further:
=β21β[ln(2+1)βln(4+1)]
- Final answer:
This results in:
=21β[ln3βln5]=21βln35β
- Trigonometric Identities:
- Cosecx=sinx1β
- Cotx=sinxcosxβ
- Integral of Logarithms:
- The integral of the form β«xdxβ=lnβ£xβ£
Summary of Steps
- Rewrite the integrand using trigonometric identities.
- Simplify the integrand and apply the integral formula for logarithms.
- Evaluate the integral at the limits x=Ο/6 and x=Ο/2.
- Simplify the result to obtain 21βln35β.