Question

Use the reduction formulas in a table of integrals to evaluate the following integrals. $$\int \tan ^{4} 3 y d y$$

Short answer

Answer: The integral of \(\tan^{4}(3y) dy\) is \(\frac{1}{9}\tan^2(3y) - \frac{1}{3} + y + C\), where \(C\) is the integration constant.

Step-by-step solution

Identify the parameters

In this problem, we have \(n = 4\) (the power of tangent function), \(a = 3\) (the constant coefficient), and the variable \(x\) is replaced by \(y\).

Apply the reduction formula

Now, using the reduction formula, we get: $$\int \tan^{4}(3y) dy = \frac{1}{3} \cdot \frac{\tan^2(3y)}{3} - \int \tan^2(3y) dy$$

Simplify the integral

Simplify the first term on the right-hand side: $$\int \tan^{4}(3y) dy = \frac{1}{9}\tan^2(3y) - \int \tan^2(3y) dy$$

Apply the reduction formula again to the remaining integral

For the remaining integral of \(\tan^2(3y)\), we apply the reduction formula again, as we did before. This time, \(n = 2\), \(a = 3\), and the variable is still \(y\). So we get: $$\int \tan^2(3y) dy = \frac{1}{3} \cdot \frac{\tan^0(3y)}{1} - \int \tan^0(3y)dy$$ Now, \(\tan^0(3y) = 1\), so the integral becomes: $$\int \tan^2(3y) dy = \frac{1}{3} \cdot \frac{1}{1} - \int 1 dy$$

Simplify further and evaluate the remaining integral

Simplify the expression and evaluate the remaining integral: $$\int \tan^{2}(3y) dy = \frac{1}{3} - \left(\int 1 dy\right) = \frac{1}{3} - y$$

Substitute the result back into the original expression

Now, substitute the result from Step 5 back into the expression from Step 3: $$\int \tan^{4}(3y) dy = \frac{1}{9}\tan^2(3y) - \left(\frac{1}{3} - y\right)$$

Add the integration constant

Finally, add the integration constant, \(C\), to complete the solution: $$\int \tan^{4}(3y) dy = \frac{1}{9}\tan^2(3y) - \frac{1}{3} + y + C$$