Question Statement

(a) Evaluate the integral: $\int \left({\tan }^{2}2\theta +{\cot }^{2}2\theta \right)d\theta$

(b) Evaluate the integral: $\int {\sin }^2\left(\frac{11}{2}y\right)dy$

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Background and Explanation

These problems involve integrating trigonometric functions using fundamental identities to convert non-standard forms into integrable expressions. Both questions require applying trigonometric identities—Pythagorean identities for part (a) and the half-angle identity for part (b)—followed by standard integration techniques including substitution.

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Solution

Part (a)

Let $I=\int \left({\tan }^{2}2\theta +{\cot }^{2}2\theta \right)d\theta$

Split the integral into two separate integrals: $I=\int {\tan }^{2}2\theta d\theta +\int {\cot }^{2}2\theta d\theta$

Apply the Pythagorean identities ${\tan }^{2}x={\sec }^{2}x-1$ and ${\cot }^{2}x={\csc }^{2}x-1$ (since ${\sec }^2\theta -{\tan }^2\theta =1$ and ${\csc }^2\theta -{\cot }^2\theta =1$):

$I=\int \left({\sec }^{2}2\theta -1\right)d\theta +\int \left({\csc }^{2}2\theta -1\right)d\theta$

Expand the integrals: $I=\int {\sec }^{2}2\theta d\theta -\int 1d\theta +\int {\csc }^{2}2\theta d\theta -\int 1d\theta$

Integrate each term using standard formulas $\int {\sec }^2(ax)dx=\frac{\tan (ax)}{a}$ and $\int {\csc }^2(ax)dx=-\frac{\cot (ax)}{a}$:

$I=\frac{\tan 2\theta }{2}-\theta +\left(\frac{-\cot 2\theta }{2}\right)-\theta +c$

Combine like terms: $I=\frac{1}{2}\tan 2\theta -\frac{1}{2}\cot 2\theta -2\theta +c$

Part (b)

Let $I=\int {\sin }^2\left(\frac{11}{2}y\right)dy$

Use substitution: Let $t=\frac{11}{2}y$

Differentiate to find $dy$: $\frac{11}{2}dy=dt\Rightarrow dy=\frac{2}{11}dt$

Substitute into the integral: $I=\int {\sin }^{2}t\cdot \frac{2}{11}dt=\frac{2}{11}\int {\sin }^{2}tdt$

Apply the half-angle identity ${\sin }^{2}x=\frac{1-\cos 2x}{2}$: $I=\frac{2}{11}\int \frac{1-\cos 2t}{2}dt=\frac{1}{11}\int (1-\cos 2t)dt$

Split and integrate term by term: $I=\frac{1}{11}\int 1dt-\frac{1}{11}\int \cos 2tdt=\frac{1}{11}t-\frac{1}{11}\cdot \frac{\sin 2t}{2}+c$

Simplify: $I=\frac{1}{11}t-\frac{1}{22}\sin 2t+c$

Substitute back $t=\frac{11}{2}y$: $I=\frac{1}{11}\cdot \frac{11}{2}y-\frac{1}{22}\sin \left(2\cdot \frac{11}{2}y\right)+c$

Simplify the final expression: $I=\frac{y}{2}-\frac{1}{22}\sin (11y)+c$

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Key Formulas or Methods Used

• Pythagorean Identities: ${\tan }^2\theta ={\sec }^2\theta -1$ and ${\cot }^2\theta ={\csc }^2\theta -1$
• Half-Angle Identity: ${\sin }^{2}x=\frac{1-\cos 2x}{2}$
• Standard Integrals: $\int {\sec }^2(ax)dx=\frac{\tan (ax)}{a}$ and $\int {\csc }^2(ax)dx=-\frac{\cot (ax)}{a}$
• Substitution Method: $u$-substitution to simplify the argument of trigonometric functions
• Integration of Cosine: $\int \cos (ax)dx=\frac{\sin (ax)}{a}$

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Summary of Steps

For Question (a):

1. Split the integral into two separate integrals for ${\tan }^2$ and ${\cot }^2$
2. Apply Pythagorean identities to convert to ${\sec }^2$ and ${\csc }^2$ forms
3. Expand and integrate each term using standard trigonometric integral formulas
4. Combine the $-\theta$ terms to get $-2\theta$ and add the constant of integration

For Question (b):

1. Substitute $t=\frac{11}{2}y$ to simplify the argument, finding $dy=\frac{2}{11}dt$
2. Apply the half-angle identity ${\sin }^{2}t=\frac{1-\cos 2t}{2}$ to convert to integrable form
3. Split into two simple integrals and evaluate: $\int 1dt$ and $\int \cos 2tdt$
4. Substitute back to the original variable $y$ and simplify the coefficients