a hard integral

DigitalComputer

New member
$\int \sin^{12}(7x) \ \cos^{3}(7x) \ dx$

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

sbhatnagar

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

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?

Last edited:

Alexmahone

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

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?