1\int_{0}^{\pi /2}{(1-sin^{2}\theta )^{5/2}cos\theta }d\theta putting x=sin\theta
\int_{0}^{\pi /2}{cos ^{6}\theta } d\theta = \int_{0}^{\pi /2}({cos ^{3}\theta)^{2} } d\theta
cos 3\theta = {4cos^{3}\theta - 3cos \theta } \Rightarrow cos^{3}\theta = cos 3\theta + 3cos\theta
SORRY DIVIDED BY 4 NAHIN KIYA
ab replace it and expand.
AB SAMAJH TA HOON HO JAYEGA
1{\color{blue} \int ( cos ^2 \theta ) ^ 3 = \int \left( \frac{1+ cos2\theta }{2}}\right)^{3} manmay's method is shorter
manmay divided by 4 bhool gaye kya
1CUBE KAROGE TO THODA COMPLICATED HO SAKTA HAI. YEHI EASY RAHEGA SAYAD.
AGAR CUBE KAROGE TO COS3 2X TERM TO AAYEGA NA PHIR MERA WALA FORMULA USE KARNA PADEGA.
1or use reduction formulae for cos θ
without usin waali or gamma function
1\int_{0}^{pi/2}{(1-cos^2\theta)^{5/2}cos\theta d\theta}=\int_{0}^{pi/2}{cos^6\theta d\theta}
can u plzz explain me the stepp..............
1srry but i didnt find pi anywhere in the latexx srry.....firstly used it..............but the main thing i didnt undersatnd it is how u wrote {(1-cos^2\theta)^{5/2}cos\theta d\theta}={cos^6\theta d\theta}
29@Sanchit ..put x = sinθ ..then dx/dθ = cosθ ..
manmay has done a mistake there..that shud be (1 - sin2θ)5/2
