IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

kishore.subramanian.b

${sin}^{6}x+{cos}^{6}x=1-3{sin}^{2}x{cos}^{2}x=1-\frac{3}{4}{sin}^{2}2x$

$G.E=\int\frac{1}{1-\frac{3}{4}{sin}^{2}2x}$

$=\int\frac{4}{4-3{sin}^{2}2x}$ Put 2x=t , then 2dx=dt dx= dt/2

$=\int\frac{2}{4-3{sin}^{2}t}dt$ Divide Nr and Dr by ${cos}^{2}t$

$=\int\frac{2{sec}^{2}t}{4{sec}^{2}t-3{tan}^{2}t}$

$=\int\frac{2{sec}^{2}t}{4+{tan}^{2}t}$ put

$=\int\frac{2}{4+{v}^{2}}dv$

= $2\frac{1}{2}{tan}^{-1}\frac{v}{2}+c$

= ${tan}^{-1}\frac{tant}{2}+c$

= ${tan}^{-1}\frac{tan2x}{2}+c$

P.S: add dx or dt whereever necessary.