Question Statement

Evaluate the integral:

$\int -{\sec }^2\left(\frac{3}{2}y\right)dy$

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Background and Explanation

This problem involves integrating a composite trigonometric function where the argument is a linear function of $y$. You can approach this by either applying the standard integral formula for ${\sec }^2(ax)$ directly or by using substitution to handle the composite argument $\frac{3}{2}y$.

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Solution

Method 1: Direct Application of Standard Formula

We begin by factoring out the negative constant and applying the standard integration formula for ${\sec }^2(ax)$.

$\begin{array}{rl}I& =\int -{\sec }^2\left(\frac{3}{2}y\right)dy\\ & =-\int {\sec }^2\left(\frac{3}{2}y\right)dy\left(\text{using }\int {\sec }^2(ax)dx=\frac{\tan (ax)}{a}\right)\\ & =\frac{-\tan \left(\frac{3}{2}y\right)}{\frac{3}{2}}+c\\ & =\frac{-2}{3}\tan \left(\frac{3}{2}y\right)+c\end{array}$

Method 2: Substitution (U-Substitution)

Alternatively, we can use substitution to simplify the composite argument. Let $t=\frac{3y}{2}$.

First, set up the substitution: $\begin{array}{rl}I& =\int -{\sec }^2\left(\frac{3}{2}y\right)dy\\ & =-\int {\sec }^2\left(\frac{3y}{2}\right)dy\end{array}$

Put $\frac{3y}{2}=t$, then differentiate to find the relationship between $dy$ and $dt$: $\begin{array}{rl}\frac{3}{2}dy& =dt\\ dy& =\frac{2}{3}dt\end{array}$

Substitute into equation (1): $\begin{array}{rl}I& =-\int {\sec }^2(t)\cdot \frac{2}{3}dt\\ & =\frac{-2}{3}\int {\sec }^2(t)dt\\ & =\frac{-2}{3}\tan (t)+c\left(\because \int {\sec }^{2}xdx=\tan x\right)\\ & =\frac{-2}{3}\tan \left(\frac{3y}{2}\right)+c\end{array}$

Verification by Differentiation

To confirm our answer is correct, we differentiate the result with respect to $y$:

$\begin{array}{rl}\frac{d}{dy}\left(\frac{-2}{3}\tan \frac{3y}{2}+c\right)& =\frac{-2}{3}\frac{d}{dy}\left(\tan \frac{3y}{2}\right)+\frac{d}{dy}(c)\\ & =\frac{-2}{3}{\sec }^2\frac{3y}{2}\cdot \frac{d}{dy}\left(\frac{3y}{2}\right)+0\\ & =\frac{-2}{3}{\sec }^2\left(\frac{3y}{2}\right)\cdot \frac{3}{2}\\ & =\frac{-2}{3}\cdot \frac{3}{2}\cdot {\sec }^2\left(\frac{3y}{2}\right)\\ & =-{\sec }^2\left(\frac{3y}{2}\right)\end{array}$

Since differentiation yields the original integrand $-{\sec }^2\left(\frac{3}{2}y\right)$, the solution is verified.

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Key Formulas or Methods Used

• Standard Integral: $\int {\sec }^2(ax)dx=\frac{\tan (ax)}{a}+C$
• Substitution Rule: If $u=g(y)$, then $\int f(g(y))g^{\prime }(y)dy=\int f(u)du$
• Chain Rule (for verification): $\frac{d}{dy}[f(g(y))]=f^{\prime }(g(y))\cdot g^{\prime }(y)$
• Derivative of Tangent: $\frac{d}{dx}(\tan x)={\sec }^{2}x$

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Summary of Steps

1. Identify the form: Recognize the integral as $\int {\sec }^2(ay)dy$ where $a=\frac{3}{2}$, with a negative coefficient
2. Method A (Direct): Factor out the negative sign and apply the formula $\frac{\tan (ay)}{a}$ directly
3. Method B (Substitution): Let $t=\frac{3y}{2}$, compute $dy=\frac{2}{3}dt$, substitute and integrate $\int {\sec }^{2}tdt$
4. Simplify: Multiply the coefficients ($\frac{-2}{3}$ from the formula or substitution, times the constant from $dy$) to get the final coefficient $-\frac{2}{3}$
5. Verify: Differentiate the answer using the chain rule to confirm it matches the original integrand