How to integrate $\int^{\pi/2}_0\sin^4xdx$ Sorry if the question is lame but here it

The following was given in my textbook $$\int^{\pi/2}_0\sin^4xdx$$

so i integrated it this way

$$\implies\int^{\pi/2}_0\sin^4xdx = \int^{\pi/2}_0\frac{\sin^5x}{5}(-cosx)$$$

and then substituted the values but i am getting a wrong answer which 0 and probably my approach is wrong because in my textbook a totally different approach is there. So what and were i am wrong please help me.

Thanks

Akash
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\begin{align}
\color{#00f}{\large\sin^{4}\pars{x}} &amp;= \bracks{1 - \cos\pars{2x} \over 2}^{2}
={1 \over 4}\bracks{1 - 2\cos\pars{2x} + \cos^{2}\pars{2x}}
={1 \over 4} - \half\,\cos\pars{2x} + {1 \over 4}\,{1 + \cos\pars{4x} \over 2}
\\[3mm]&amp;=\color{#00f}{\large{3 \over 8} -\half\,\cos\pars{2x} + {1 \over 8}\,\cos\pars{4x}}
\end{align}
Now, integrate each term in the right hand side.
Hint: use integration by parts with $u= \sin^3 x $