Question

# 27. [2 cOs- x dr0 cosx 4 sin2x

Open in App
Solution

## The given integral, I= ∫ 0 π 2 cos 2 x cos 2 x+4 sin 2 x dx . Simplify the function. I= ∫ 0 π 2 cos 2 x cos 2 x+4 sin 2 x dx = ∫ 0 π 2 cos 2 x cos 2 x+4( 1− cos 2 x ) dx = ∫ 0 π 2 cos 2 x cos 2 x+4−4 cos 2 x dx = ∫ 0 π 2 cos 2 x 4−3 cos 2 x dx Further simplify, I=− 1 3 ∫ 0 π 2 4−3 cos 2 x−4 4−3 cos 2 x dx =− 1 3 ( ∫ 0 π 2 4−3 cos 2 x 4−3 cos 2 x dx −4 ∫ 0 π 2 1 4−3 cos 2 x dx ) =− 1 3 ( ∫ 0 π 2 dx −4 ∫ 0 π 2 sec 2 x 4 sec 2 x−3 dx ) =− 1 3 ( [ x ] 0 π 2 −4 ∫ 0 π 2 sec 2 x 4( 1+ tan 2 x )−3 dx ) Now, the integration is, I=− π 6 + 2 3 ∫ 0 π 2 2 sec 2 x 1+4 tan 2 x dx (1) Let I 1 = ∫ 0 π 2 2 sec 2 x 1+4 tan 2 x dx Substitute 2tanx=t. Differentiate both sides with respect to x. 2 sec 2 xdx=dt When x=0,t=0 and when x= π 2 , t=∞. Now, integral becomes, I 1 = ∫ 0 ∞ dt 1+ t 2 = [ tan −1 t ] 0 ∞ = tan −1 ∞− tan −1 0 = π 2 From equation (1), we get, I=− π 6 + 2 3 × π 2 =− π 6 + π 3 = π 6 Thus, the solution of integral is π 6 .

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
MATHEMATICS
Watch in App
Join BYJU'S Learning Program