Question

27. [2 cOs- x dr0 cosx 4 sin2x

Answer 1

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 .