Question

"Evaluate int sin^5(7x)dx", I tried using the power reduction formulas but my answer still turned out wrong (which I'll post following this).

Answer 1

\[\int\limits\limits \sin ^{5}(7x) dx\] like this right?

Answer 2

I ended up with \[(-1/35)\cos(7x)\sin^(4)(7x)+(5/63)\sin^(2)(7x)\cos(7x)-(5/63)\cos(7x)+c\]

Answer 3

yes that's the problem @user2486

Answer 4

my answer typed wrong before it was like this: \[\frac{ -1 }{ 35 } \cos(7x)\sin^4(7x)+\frac{ 5 }{ 63 } \sin^2(7x)\cos(7x)-\frac{ -5 }{ 63 } \cos(7x)+c\]

Answer 5

\[\begin{align*}\sin^57x&=\sin^47x\sin 7x\\
&=\left(1-\cos^27x\right)^2\sin7x
\end{align*}\]
Letting \(u=\cos7x\), you have \(-\dfrac{1}{7}~du=\sin 7x~dx\).
\[-\frac{1}{7}\int\left(1-u^2\right)^2~du\]

Answer 6

@SithsAndGiggles I tried that way too earlier and got stuck, thanks for clearing it up.