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
 A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\color{#00f}{\large\sin^{4}\pars{x}} &= \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]&=\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.
A: Hint: Utilize the identities $\color{#C00000}{\sin(\pi/2-x)=\cos(x)}$ and $\color{#0000FF}{\sin(\pi/2+x)=\cos(x)}$:
$$
\begin{align}
&\int_0^{\pi/2}\left(\sin^2(x)+\cos^2(x)\right)^2\,\mathrm{d}x\\
&=\color{#C00000}{\int_0^{\pi/2}\sin^4(x)\,\mathrm{d}x}+\int_0^{\pi/2}2\sin^2(x)\cos^2(x)\,\mathrm{d}x+\color{#C00000}{\int_0^{\pi/2}\cos^4(x)\,\mathrm{d}x}\\
&=\color{#C00000}{2\int_0^{\pi/2}\sin^4(x)\,\mathrm{d}x}+\frac12\int_0^{\pi/2}\sin^2(2x)\,\mathrm{d}x\\
&=2\int_0^{\pi/2}\sin^4(x)\,\mathrm{d}x+\color{#0000FF}{\frac14\int_0^\pi\sin^2(x)\,\mathrm{d}x}\\
&=2\color{#00A000}{\int_0^{\pi/2}\sin^4(x)\,\mathrm{d}x}+\color{#0000FF}{\frac14\int_0^{\pi/2}\left(\sin^2(x)+\cos^2(x)\right)\,\mathrm{d}x}
\end{align}
$$
A: Try using the fact that $\sin^2(x) = (1 - \cos(2x))/2$.  You can deal similarly wit the $\cos^2$ integral that will appear.
A: Another way: rewrite the integrand as $\sin^2 x(1-\cos^2 x)=\sin^2x -(\sin x \cos x)^2$ to get two integrals. The first one is solved in the way ncmathsadist explained. The second becomes
$$
\frac{1}{4} \int_{0}^{\frac{\pi}{2}} (\sin 2 x)^2dx
$$ 
by using $2 \sin x \cos x = \sin 2 x$ and is solved in the exact same way (reducing the power).
A: Use the facts that $\sin x = (e^{ix}-e^{-ix})/2i$ and $(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4$ to get $$\sin^4 x = (e^{4ix}-4e^{3ix-ix}+6e^{2ix-2ix}-4e^{ix -3ix}+e^{-4ix})/(2i)^4,$$ which simplifies to $(e^{4ix}-4e^{2ix}+6-4e^{-2ix}+e^{-4ix})/16$. Combining conjugate terms again, we finally have rewritten the integrand as $$\sin^4 x =\frac18 \cos 4x -\frac12 \cos 2x+\frac38.$$ You should be able to integrate this with no problem.
The same technique works with any power of $\sin x$ or $\cos x$. (I hope I didn't make a typo in this!) EDIT: Found a typo, fixed it.
A: Hint: use integration by parts with $u= \sin^3 x $
