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
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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.
Hint: Utilize the identities $\color{#C00000}{\sin(\pi/2-x)=\cos(x)}$ and $\color{#0000FF}{\sin(\pi/2+x)=\cos(x)}$: $$ \begin{align} &\int_0^{\pi/2}\left(\sin^2(x)+\cos^2(x)\right)^2\,\mathrm{d}x\\ &=\color{#C00000}{\int_0^{\pi/2}\sin^4(x)\,\mathrm{d}x}+\int_0^{\pi/2}2\sin^2(x)\cos^2(x)\,\mathrm{d}x+\color{#C00000}{\int_0^{\pi/2}\cos^4(x)\,\mathrm{d}x}\\ &=\color{#C00000}{2\int_0^{\pi/2}\sin^4(x)\,\mathrm{d}x}+\frac12\int_0^{\pi/2}\sin^2(2x)\,\mathrm{d}x\\ &=2\int_0^{\pi/2}\sin^4(x)\,\mathrm{d}x+\color{#0000FF}{\frac14\int_0^\pi\sin^2(x)\,\mathrm{d}x}\\ &=2\color{#00A000}{\int_0^{\pi/2}\sin^4(x)\,\mathrm{d}x}+\color{#0000FF}{\frac14\int_0^{\pi/2}\left(\sin^2(x)+\cos^2(x)\right)\,\mathrm{d}x} \end{align} $$
Try using the fact that $\sin^2(x) = (1 - \cos(2x))/2$. You can deal similarly wit the $\cos^2$ integral that will appear.
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).
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.
Hint: use integration by parts with $u= \sin^3 x $
