Find the integral $\int\left(w^2+\cos\left(4w^3-3\right)^2\right)\sec\left(4w^3-3\right)^2dw$

Got another answer? Verify it here!

Learn how to solve integration techniques problems step by step online. Find the integral int((w^2+cos(4w^3-3)^2)sec(4w^3-3)^2)dw. Rewrite the integrand \left(w^2+\cos\left(4w^3-3\right)^2\right)\sec\left(4w^3-3\right)^2 in expanded form. Expand the integral \int\left(w^2\sec\left(4w^3-3\right)^2+\cos\left(4w^3-3\right)^2\sec\left(4w^3-3\right)^2\right)dw into 2 integrals using the sum rule for integrals, to then solve each integral separately. Applying the trigonometric identity: \cos\left(\theta\right)\cdot\sec\left(\theta\right)=1. The integral \int w^2\sec\left(4w^3-3\right)^2dw results in: \frac{1}{12}\tan\left(4w^3-3\right).