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a hard integral

DigitalComputer

New member
Feb 10, 2012
5
\[ \int \sin^{12}(7x) \ \cos^{3}(7x) \ dx \]

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

sbhatnagar

Active member
Jan 27, 2012
95
\( \displaystyle \int \sin^{12}(7x) \cos^{3}(7x) \ dx = \int \sin^{12}(7x) \{ 1-\sin^2(7x)\}\cos(7x) \ dx\)

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
Jan 26, 2012
268
\( \displaystyle \int \sin^{12}(7x) \cos^{3}(7x) \ dx = \int \sin^{13}(7x) \{ 1-\sin^2(7x)\}\cos(7x) \ dx\)

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?