Evaluate the integral $\int \sin^2 4x \cos^3 4x \, dx$
Added by Veronica B.
Close
Step 1
Recall that $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. So, we have: $\int f \sin^2(4x)\cos(4x) dx = \int f \left(\frac{1}{2}\sin(8x)\right) dx$ Now, we can integrate with respect to $x$: $\int f \left(\frac{1}{2}\sin(8x)\right) dx = \frac{1}{2} \int f Show more…
Show all steps
Your feedback will help us improve your experience
Atul Kumar and 79 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Evaluate the integral. J cos 4x 2 dx
Adi S.
Evaluate the integral. integral cos - 4 x dx
Madhur L.
Evaluate the integral: 6 sin^3 4x cos^3 4x dx 6 sin^3 4x cos^3 4x dx=
Drew S.
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Watch the video solution with this free unlock.
EMAIL
PASSWORD