Question Statement
Evaluate the following integral:
β«02β(1+Cosx)(2+Cosx)Sinxβ,dx
Background and Explanation
To solve this integral, we use a substitution method. Understanding the following concepts is important:
- Trigonometric identities: sinx and cosx are the core functions involved.
- Substitution: Weβll make a substitution to simplify the integral.
- Integration of rational functions: After substitution, the integrand becomes a rational function, which can be split and integrated easily.
Solution
Step-by-step explanation of the solution:
- Substitute:
Letβs make the substitution to simplify the integral. Set:
u=1+Cosx
Now, calculate the limits of u:
- When x=0, u=1+cos(0)=2.
- When x=2Οβ, u=1+cos(2Οβ)=1.
- Express du:
Differentiate u=1+cosx with respect to x:
du=βsinx,dxorβdu=sinx,dx
- Substitute into the integral:
Now substitute u and du into the integral:
β«0Ο/2β(1+cosx)(2+cosx)sinxβ,dx=β«21βu(1+u+1)βduβ
Simplifying the integrand:
=β«21β(uβ1β+u+11β),du
- Separate the integrals:
Split the integral into two parts:
=ββ«21βu1β,du+β«21βu+11β,du
- Integrate:
The integrals are simple logarithmic integrals:
=βlnuβ21β+ln(u+1)β21β
- Evaluate the bounds:
Now evaluate the logarithmic expressions:
=β(ln2βln1)+(ln(2+1)βln(1+1))
Simplifying further:
=βln2+0+ln3βln2
- Final answer:
Combine the terms:
=ln3βln4=ln43β
- Substitution: u=1+cosx, du=βsinx,dx
- Logarithmic Integrals:
- β«u1β,du=lnβ£uβ£
Summary of Steps
- Make the substitution u=1+cosx.
- Determine the new limits for u.
- Substitute u and du into the integral.
- Split the integral into two simpler parts.
- Integrate each part using logarithmic rules.
- Evaluate the integrals at the limits.
- Simplify the final expression to ln43β.