# Evaluate the definite integral$\int\limits_0^\frac{\Large \pi}{2}\cos^2xdx$

$\begin{array}{1 1}\large \frac{\pi}{4} \\\large \frac{\pi}{2} \\ \large \frac{\pi}{6} \\ \large \frac{\pi}{8} \end{array}$

Toolbox:
• (i)&#x222B;abf(x)dx=F(b)&#x2212;F(a)" role="presentation" style="position: relative;">abf(x)dx=F(b)F(a)∫abf(x)dx=F(b)−F(a)\int \limits_a^b f(x)dx=F(b)-F(a)
• (ii)cos2&#x2061;x=1+cos&#x2061;2x2" role="presentation" style="position: relative;">cos2x=1+cos2x2cos2⁡x=1+cos⁡2x2 \cos ^2 x=\large\frac{1+\cos 2x}{2}
• (iii)&#x222B;cos&#x2061;axdx=1asin&#x2061;ax+c" role="presentation" style="position: relative;">cosaxdx=1asinax+c∫cos⁡axdx=1asin⁡ax+c\int \cos ax dx=\frac{1}{a} \sin ax+c
Given I=&#x222B;0&#x03C0;2cos2&#x2061;xdx" role="presentation" style="position: relative;">I=0π2cos2xdxI=∫0π2cos2⁡xdx I=\int\limits_0^\frac{\Large \pi}{2}\cos^2xdx

cos2" role="presentation" style="position: relative;">cos2cos2cos ^2 can be written as 1+cos&#x2061;2x2" role="presentation" style="position: relative;">1+cos2x21+cos⁡2x2\large\frac{1+\cos 2x}{2}

Hence I=&#x222B;0&#x03C0;21+cos2&#x2061;x2dx" role="presentation" style="position: relative;">I=0π21+cos2x2dxI=∫0π21+cos2⁡x2dx I=\int\limits_0^\frac{\Large \pi}{2}\large\frac{1+\cos^2x}{2}dx

On splitting te terms we get,

&#x222B;0&#x03C0;/212dx+&#x222B;0&#x03C0;/2cos&#x2061;2x2dx" role="presentation" style="position: relative;">0π/212dx+0π/2cos2x2dx∫0π/212dx+∫0π/2cos⁡2x2dx\int \limits_0^{\pi/2} \large\frac{1}{2} dx +\int \limits_0^{\pi/2}\frac{\cos 2x}{2}dx

=12&#x222B;0&#x03C0;/2dx+12&#x222B;0&#x03C0;/2cos&#x2061;2xdx" role="presentation" style="position: relative;">=120π/2dx+120π/2cos2xdx=12∫0π/2dx+12∫0π/2cos⁡2xdx=\frac{1}{2} \int \limits_0^{\pi/2} \large dx +\frac{1}{2} \int \limits_0^{\pi/2}\cos 2x dx

On itegrating we get,

12[x]0&#x03C0;/2+12[sin&#x2061;2x2]0&#x03C0;/2" role="presentation" style="position: relative;">12[x]π/20+12[sin2x2]π/2012[x]0π/2+12[sin⁡2x2]0π/2\large\frac{1}{2}[x]_0^{\pi/2}+\frac{1}{2}\bigg[\frac{\sin 2x}{2}\bigg]_0^{\pi/2}

On applying limits,

12[&#x03C0;2&#x2212;0]+14[sin&#x2061;2.&#x03C0;2&#x2212;sin&#x2061;0]" role="presentation" style="position: relative;">12[π20]+14[sin2.π2sin0]12[π2−0]+14[sin⁡2.π2−sin⁡0]\large\frac{1}{2}[\frac{\pi}{2}-0]+\frac{1}{4}[\sin2.\frac{\pi}{2}-\sin 0]

&#x03C0;4+0(Becausesin&#x2061;0=0andsin&#x2061;&#x03C0;=0)" role="presentation" style="position: relative;">π4+0(Becausesin0=0andsinπ=0)π4+0(Becausesin⁡0=0andsin⁡π=0)\large\frac {\pi}{4}+0\qquad(Because\;\sin0=0\;and\;\sin \pi=0)

&#x222B;0&#x03C0;/2cos2&#x2061;xdx=&#x03C0;4" role="presentation" style="position: relative;">0π/2cos2xdx=π4∫0π/2cos2⁡xdx=π4\int \limits_0^{\pi/2} \cos ^2xdx=\large\frac{\pi}{4}
edited Apr 11, 2016 by meena.p