What is the integral of #int sin^2(x)cos^4(x) dx#?

Question

What is the integral of #int sin^2(x)cos^4(x)  dx#?

Charles Anctil · Accepted Answer

#\frac{1}{16}$x-\frac{1}{4}\sin $4x$)+\frac{1}{6}\sin ^3$x$\cos ^3$x)+C# #\int \sin^2\(x$cos^4$x$dx# Applying integral reduction, #int sin^2(x) cos^n (x) dx# = #((sin^3 (x) cos^(n-1 )(x)) / (2+n))# #+((n-1) /(2+n))# #int sin^2 (x) cos^(n-2) (x)dx# so, #\int \sin ^2$x$\cos ^4$x$dx# #=\frac{\sin ^3$x$\cos ^3$x$}{6}+\frac{3}{6}\int \cos ^2$x$\sin ^2$x$dx# #=\frac{\sin ^3$x$\cos ^3$x$}{6}+\frac{3}{6}\int \cos ^2$x$\sin ^2$x$dx# We know, #\int \cos ^2$x$\sin ^2$x$dx=\frac{1}{8}$x-\frac{1}{4}\sin \(4x$$# Then, #=\frac{\sin ^3$x$\cos ^3$x$}{6}+\frac{3}{6}\frac{1}{8}$x-\frac{1}{4}\sin $4x$$# Simplifying, #=\frac{1}{16}$x-\frac{1}{4}\sin $4x$$+\frac{1}{6}\sin ^3$x$\cos ^3$x$# Adding constant to the solution, #=\frac{1}{16}$x-\frac{1}{4}\sin $4x$$+\frac{1}{6}\sin ^3$x$\cos ^3$x$+C#

Samantha Adkison · Answer

undefined To integrate $ \int \sin^2(x) \cos^4(x) \, dx $, you can use the trigonometric identity $ \sin^2(x) = \frac{1 - \cos(2x)}{2} $ and $ \cos^4(x) = \left(\frac{1 + \cos(2x)}{2}\right)^2 $. Then substitute these identities into the integral and simplify. After simplification, the integral becomes $ \frac{1}{8}x - \frac{1}{32}\sin(4x) + C $, where $ C $ is the constant of integration.

HIX Tutor · Answer

undefined That’s a great question! To find the integral of $ \sin^2(x) \cos^4(x) \, dx $, let’s break it down step by step.

1. **Identify the Trigonometric Identity:** 
   We can use the identity $ \sin^2(x) = 1 - \cos^2(x) $. This might help us simplify our expression.

2. **Rewrite the Integral:**
   So, we can write:
   $$
   \sin^2(x) \cos^4(x) = (1 - \cos^2(x)) \cos^4(x)
   $$
   Thus, our integral becomes:
   $$
   \int (1 - \cos^2(x)) \cos^4(x) \, dx
   $$
   Which can be split into two parts:
   $$
   \int \cos^4(x) \, dx - \int \cos^6(x) \, dx
   $$

3. **Use a Substitution:** 
   Let’s use the substitution $ u = \cos(x) $. Then $ du = -\sin(x) \, dx $ or $ dx = -\frac{du}{\sqrt{1 - u^2}} $.

4. **Change the Limits and Integrals:**
   Now, substitute $ \cos(x) $ in the integrals:
   - The first part becomes $ \int u^4 (-\frac{1}{\sqrt{1 - u^2}}) \, du $.
   - The second part becomes $ -\int u^6 (-\frac{1}{\sqrt{1 - u^2}}) \, du $.

5. **Integrate Each Part:**
   Now, you integrate both parts. It can be complex, so let's just focus on:
   $$
   \int u^4 \sqrt{1 - u^2} \, du \quad \text{and} \quad \int u^6 \sqrt{1 - u^2} \, du
   $$
   For each integral, you might need to use more trigonometric identities or another substitution if necessary.

6. **Combine Everything Back:**
   After you find both integrals, you substitute $ \cos(x) $ back in place of $ u $ to get the final answer in terms of $ x $.

Would you like help with any specific integral or step?