I am currently working on an integral using u substitution and have hit a wall at the above point. The integral in question is $\int$$\sin^7(x)$$\cos^9(x)$ using $u=sin(x)$.

After finding $du=cos(x) dx$, I am asked to express $\cos^8(x)$ in terms of $u$, at which point I am stuck. My current attempt was to reduce the power down from 8 to 2, and express it as $(1-u^2)$, but that gives me a wrong answer (this assignment checks for us before submission).

Any help is appreciated.


  • $\begingroup$ What did you find the integral to be (your "wrong" answer)? $\endgroup$ – Henry Oct 18 '15 at 9:55
  • $\begingroup$ I hadn't got to that point yet as the next few steps required it expressed in terms of u still. $\endgroup$ – Dwayne H Oct 18 '15 at 9:56
  • $\begingroup$ OK, your "wrong" answer in terms of $u$? $\endgroup$ – Henry Oct 18 '15 at 9:57
  • $\begingroup$ I'd only gotten to the point of $u^7(1-u^2)$, but seeing as the assignment said it was wrong I didn't try going ahead and solving for that. $\endgroup$ – Dwayne H Oct 18 '15 at 9:59
  • $\begingroup$ $\displaystyle \int u^7(1-u^2)^4\,du$ with a power of $4$ seems more likely $\endgroup$ – Henry Oct 18 '15 at 10:00

If $$\cos^2 x = 1-\sin^2 x$$ then $$\cos^8 x = (1-\sin^2 x)^4.$$


Using $u=\sin x$, you have $du=\cos x\,dx$, and also $\cos^2x=1-u^2$, so $\cos^8x=(1-u^2)^4$. Thus your integral becomes $$ \int u^7(1-u^2)^4\,du $$


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