# Integral of $\sin^5x\cdot\cos^{14}x$

I just don't understand what I'm doing wrong. I'm doing it exactly like it was taught to me but I'm getting a completely different answer from the correct answer.

$$\int\limits_0^{\frac{\pi}{2}}\sin^5(x)\cos^{14}(x)dx$$

My work:

$$\int\limits_0^{\frac{\pi}{2}}[\sin^2(x)]^2\cos^{14}(x)\sin(x)dx$$

$$\int\limits_0^{\frac{\pi}{2}}[1-\cos^2(x)]^2\cos^{14}(x)\sin(x)dx$$

substitute $$\cos(x)$$ for $$u$$.
$$du=\sin(x)dx$$

$$\int\limits_0^{\frac{\pi}{2}}(1-u^2)^2u^{14}du$$

$$\int\limits_0^{\frac{\pi}{2}}(1-2u+u^4)u^{14}du$$

$$\int\limits_0^{\frac{\pi}{2}}u^{14}-2u^{16}+u^{18}du$$

$$\frac{1}{15}u^{15}-\frac{2}{17}u^{17}+\frac{1}{19}u^{19}\Bigg|_{-1}^0=$$

$$\frac{1}{15}\cos^{15}\left(\frac{\pi}{2}\right)-\frac{2}{17}\cos^{17}\left(\frac{\pi}{2}\right)+\frac{1}{19}\cos^{19}\left(\frac{\pi}{2}\right) - \left(\frac{1}{15}\cos^{15}(0)-\frac{2}{17}\cos^{17}(0)+\frac{1}{19}\cos^{19}(0)\right)$$

$$0-0+0-(1-1+1)$$

$$-1+1-1=-1$$

The correct answer is $$\frac{8}{4845}$$

Where did I go wrong?

It is in fact $$u=-\cos x$$ and by substitution we get $$\frac{1}{15}u^{15}-\frac{2}{17}u^{17}+\frac{1}{19}u^{19}\Bigg|_{-1}^0={1\over 15}-{2\over 17}+{1\over 19}={8\over 4845}$$You where only wrong in substitution...
$$u = \cos x$$ $$du = -\sin x \,dx$$
(and not $$du = \sin x \,dx$$, as you wrote).
There are other errors in your calculation. But they cancel each other out. If you make a substitution and you have limits of integration then you have to transform these limits too. So if the substitution is $$u=\cos(x)$$ then the limit of integration change to
$$\int\limits_{\color{red}{0}}^\color{red}{{\frac{\pi}{2}}}f(\cos(x))(-\sin(x))dx=\int\limits_{\color{red}{\cos0}}^\color{red}{{\cos\frac{\pi}{2}}}f(u)du={F(u){\LARGE|}}_{\color{red}{\cos0}}^\color{red}{{\cos\frac{\pi}{2}}}=F(\cos(x)){\LARGE|}_{\color{red}{0}}^\color{red}{{\frac{\pi}{2}}}$$