The integral \( \int \sin^{12}(7x) \cos^{3}(7x) \ dx \) can be solved using the substitution \( u = \sin(7x) \). This transforms the integral into \( \frac{1}{7} \int u^{12}(1-u^2) \ du \). The discussion highlights a common mistake where the powers of sine are incorrectly noted as 13 instead of 12. The correct formulation is essential for accurate integration.
PREREQUISITESStudents of calculus, mathematics educators, and anyone looking to enhance their skills in solving trigonometric integrals.
sbhatnagar said:\( \displaystyle \int \sin^{12}(7x) \cos^{3}(7x) \ dx = \int \sin^{13}(7x) \{ 1-\sin^2(7x)\}\cos(7x) \ dx\)
Now substitute \( u=\sin(7x) \).
\( \displaystyle \int \sin^{13}(7x) \{ 1-\sin^2(7x)\}\cos(7x) \ dx = \frac{1}{7}\int u^{12}(1-u^2) \ du\)
Can you take it from here?