MHB Yes, the 13's should be 12's. My mistake.

[Blandongstein]

\[ \int \sin^{12}(7x) \ \cos^{3}(7x) \ dx \]

Ho do I solve this Integral? What can I substitute??

 

[sbhatnagar]

\( \displaystyle \int \sin^{12}(7x) \cos^{3}(7x) \ dx = \int \sin^{12}(7x) \{ 1-\sin^2(7x)\}\cos(7x) \ dx\)

Now substitute \( u=\sin(7x) \).

\( \displaystyle \int \sin^{12}(7x) \{ 1-\sin^2(7x)\}\cos(7x) \ dx = \frac{1}{7}\int u^{12}(1-u^2) \ du\)

Can you take it from here?

 

[alexmahone]

\( \displaystyle \int \sin^{12}(7x) \cos^{3}(7x) \ dx = \int \sin^{13}(7x) \{ 1-\sin^2(7x)\}\cos(7x) \ dx\)

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?

Shouldn't those 13's be 12's?