Integral Substitution 099144

1. **State the problem:** We want to calculate the indefinite integral $$\int \frac{7\cos(x)\sin(4x^2)}{8} \, dx.$$ 2. **Analyze the integrand:** The integral contains $\cos(x)$ and $\sin(4x^2)$. The half-angle formula $\sin(2x) = 2\sin(x)\cos(x)$ relates $\sin(2x)$ to $\sin(x)$ and $\cos(x)$, but here we have $\sin(4x^2)$, which is not directly in the form $\sin(2x)$. So applying the half-angle formula directly to $\sin(4x^2)$ is not straightforward. 3. **Check substitution $u=\sin(x)$:** The problem suggests using $u=\sin(x)$ after applying the half-angle formula. Let's try to rewrite the integrand to use this substitution. 4. **Rewrite $\sin(4x^2)$:** Note that $4x^2$ is not a linear function of $x$, so the half-angle formula for $\sin(2x)$ does not apply directly to $\sin(4x^2)$. The half-angle formula is for linear arguments. 5. **Re-examine the problem:** Possibly the problem intends to apply the half-angle formula to $\sin(2x)$ inside the integrand, but the integrand is $\frac{7\cos(x)\sin(4x^2)}{8}$. Since $4x^2$ is quadratic, the half-angle formula for $\sin(2x)$ is not applicable here. 6. **Conclusion:** The integral as given cannot be simplified by the half-angle formula $\sin(2x) = 2\sin(x)\cos(x)$ because the argument of sine is $4x^2$, not $2x$. Also, substitution $u=\sin(x)$ is not helpful because $\sin(4x^2)$ is independent of $\sin(x)$. 7. **Alternative approach:** Since the problem explicitly asks to apply the half-angle formula and substitution $u=\sin(x)$, it is likely there is a typo or misinterpretation in the problem statement. 8. **If the problem intended $\sin(4x)$ instead of $\sin(4x^2)$:** - Use the double-angle formula: $\sin(4x) = 2\sin(2x)\cos(2x)$. - Then apply $\sin(2x) = 2\sin(x)\cos(x)$. - Then substitute $u=\sin(x)$. 9. **Assuming the integrand is $\frac{7\cos(x)\sin(4x)}{8}$:** - $\sin(4x) = 2\sin(2x)\cos(2x)$. - $\sin(2x) = 2\sin(x)\cos(x)$. - So $\sin(4x) = 2 \times 2\sin(x)\cos(x) \times \cos(2x) = 4\sin(x)\cos(x)\cos(2x)$. - The integrand becomes $\frac{7\cos(x) \times 4\sin(x)\cos(x)\cos(2x)}{8} = \frac{7 \times 4}{8} \sin(x) \cos^2(x) \cos(2x) = \frac{7}{2} \sin(x) \cos^2(x) \cos(2x)$. - Substitute $u=\sin(x)$, then $du = \cos(x) dx$. - Express $\cos^2(x)$ and $\cos(2x)$ in terms of $u$ if possible. 10. **Since the original problem as stated is inconsistent, the integral cannot be solved as requested.**