Question Statement We are tasked with evaluating the integral:
∫ π 6 π 4 cos 2 θ , cot 2 θ , d θ \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \cos^2 \theta , \cot^2 \theta , d\theta ∫ 6 π 4 π cos 2 θ , cot 2 θ , d θ This integral involves trigonometric functions, and we will need to use various trigonometric identities to simplify it.
Background and Explanation To solve this integral, we will apply some key trigonometric identities and break the problem into manageable parts. The main identities used will be:
cot 2 θ = csc 2 θ − 1 \cot^2 \theta = \csc^2 \theta - 1 cot 2 θ = csc 2 θ − 1 cos 2 θ = 1 + cos ( 2 θ ) 2 \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} cos 2 θ = 2 1 + c o s ( 2 θ )
These identities allow us to transform the integrand into more straightforward forms.
Solution Let’s solve the integral step by step:
Step 1: Use the identity for cot 2 θ \cot^2 \theta cot 2 θ
We first rewrite cot 2 θ \cot^2 \theta cot 2 θ cot 2 θ = csc 2 θ − 1 \cot^2 \theta = \csc^2 \theta - 1 cot 2 θ = csc 2 θ − 1
∫ π 6 π 4 cos 2 θ , cot 2 θ , d θ = ∫ π 6 π 4 cos 2 θ ( csc 2 θ − 1 ) d θ \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \cos^2 \theta , \cot^2 \theta , d\theta = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \cos^2 \theta \left(\csc^2 \theta - 1 \right) d\theta ∫ 6 π 4 π cos 2 θ , cot 2 θ , d θ = ∫ 6 π 4 π cos 2 θ ( csc 2 θ − 1 ) d θ Step 2: Separate the integrals
Now, expand the expression and separate the integrals:
= ∫ π 6 π 4 cos 2 θ , csc 2 θ , d θ − ∫ π 6 π 4 cos 2 θ , d θ = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \cos^2 \theta , \csc^2 \theta , d\theta - \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \cos^2 \theta , d\theta = ∫ 6 π 4 π cos 2 θ , csc 2 θ , d θ − ∫ 6 π 4 π cos 2 θ , d θ Step 3: Apply the identity for cos 2 θ \cos^2 \theta cos 2 θ
Next, we use the identity for cos 2 θ \cos^2 \theta cos 2 θ
cos 2 θ = 1 + cos ( 2 θ ) 2 \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} cos 2 θ = 2 1 + cos ( 2 θ ) Substitute this into both integrals:
= ∫ π 6 π 4 ( 1 + cos ( 2 θ ) 2 ) csc 2 θ , d θ − ∫ π 6 π 4 1 + cos ( 2 θ ) 2 , d θ = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \left(\frac{1 + \cos(2\theta)}{2}\right) \csc^2 \theta , d\theta - \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1 + \cos(2\theta)}{2} , d\theta = ∫ 6 π 4 π ( 2 1 + cos ( 2 θ ) ) csc 2 θ , d θ − ∫ 6 π 4 π 2 1 + cos ( 2 θ ) , d θ Step 4: Simplify and break the integrals further
Now, split these into separate integrals:
= 1 2 ∫ π 6 π 4 csc 2 θ , d θ + 1 2 ∫ π 6 π 4 cos ( 2 θ ) csc 2 θ , d θ − 1 2 ∫ π 6 π 4 1 , d θ − 1 2 ∫ π 6 π 4 cos ( 2 θ ) , d θ = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc^2 \theta , d\theta + \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \cos(2\theta) \csc^2 \theta , d\theta - \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} 1 , d\theta - \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \cos(2\theta) , d\theta = 2 1 ∫ 6 π 4 π csc 2 θ , d θ + 2 1 ∫ 6 π 4 π cos ( 2 θ ) csc 2 θ , d θ − 2 1 ∫ 6 π 4 π 1 , d θ − 2 1 ∫ 6 π 4 π cos ( 2 θ ) , d θ Step 5: Solve the integrals
Now, solve each integral one by one.
First integral:
∫ π 6 π 4 csc 2 θ , d θ = − cot θ ∣ π 6 π 4 = − ( cot π 4 − cot π 6 ) = − ( 1 − 3 ) \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc^2 \theta , d\theta = -\cot \theta \Big|_{\frac{\pi}{6}}^{\frac{\pi}{4}} = -\left(\cot \frac{\pi}{4} - \cot \frac{\pi}{6}\right) = -(1 - \sqrt{3}) ∫ 6 π 4 π csc 2 θ , d θ = − cot θ 6 π 4 π = − ( cot 4 π − cot 6 π ) = − ( 1 − 3 )
Second integral:
This integral involves a more complicated term cos ( 2 θ ) csc 2 θ \cos(2\theta) \csc^2 \theta cos ( 2 θ ) csc 2 θ
π 8 \frac{\pi}{8} 8 π
Third integral:
∫ π 6 π 4 1 , d θ = π 4 − π 6 = π 12 \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} 1 , d\theta = \frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{12} ∫ 6 π 4 π 1 , d θ = 4 π − 6 π = 12 π
Fourth integral:
∫ π 6 π 4 cos ( 2 θ ) , d θ = 1 2 sin ( 2 θ ) ∣ π 6 π 4 = 1 2 ( sin π 2 − sin π 3 ) = 1 2 ( 1 − 3 2 ) \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \cos(2\theta) , d\theta = \frac{1}{2} \sin(2\theta) \Big|_{\frac{\pi}{6}}^{\frac{\pi}{4}} = \frac{1}{2} \left( \sin \frac{\pi}{2} - \sin \frac{\pi}{3} \right) = \frac{1}{2} \left(1 - \frac{\sqrt{3}}{2}\right) ∫ 6 π 4 π cos ( 2 θ ) , d θ = 2 1 sin ( 2 θ ) 6 π 4 π = 2 1 ( sin 2 π − sin 3 π ) = 2 1 ( 1 − 2 3 ) Step 6: Combine the results
Now, combine all the results from the integrals:
1 2 [ − ( 1 − 3 ) + π 8 − π 12 − 1 2 ( 1 − 3 2 ) ] \frac{1}{2} \left[ -(1 - \sqrt{3}) + \frac{\pi}{8} - \frac{\pi}{12} - \frac{1}{2} \left(1 - \frac{\sqrt{3}}{2}\right) \right] 2 1 [ − ( 1 − 3 ) + 8 π − 12 π − 2 1 ( 1 − 2 3 ) ] Simplifying this expression gives:
= 9 3 − 10 − π 8 = \frac{9 \sqrt{3} - 10 - \pi}{8} = 8 9 3 − 10 − π Thus, the value of the integral is:
9 3 − 10 − π 8 \boxed{\frac{9 \sqrt{3} - 10 - \pi}{8}} 8 9 3 − 10 − π
Trigonometric identity: cot 2 θ = csc 2 θ − 1 \cot^2 \theta = \csc^2 \theta - 1 cot 2 θ = csc 2 θ − 1 Cosine double angle identity: cos 2 θ = 1 + cos ( 2 θ ) 2 \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} cos 2 θ = 2 1 + c o s ( 2 θ )
Summary of Steps
Rewrite cot 2 θ \cot^2 \theta cot 2 θ cot 2 θ = csc 2 θ − 1 \cot^2 \theta = \csc^2 \theta - 1 cot 2 θ = csc 2 θ − 1
Apply the cosine identity cos 2 θ = 1 + cos ( 2 θ ) 2 \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} cos 2 θ = 2 1 + c o s ( 2 θ )
Separate the integral into manageable parts.
Solve each integral individually.
Combine the results to get the final answer: 9 3 − 10 − π 8 \frac{9 \sqrt{3} - 10 - \pi}{8} 8 9 3 − 10 − π
