CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
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+44 cos 2 x dx = 0 π 2 cos 2 x 43 cos 2 x dx

Further simplify,

I= 1 3 0 π 2 43 cos 2 x4 43 cos 2 x dx = 1 3 ( 0 π 2 43 cos 2 x 43 cos 2 x dx 4 0 π 2 1 43 cos 2 x dx ) = 1 3 ( 0 π 2 dx 4 0 π 2 sec 2 x 4 sec 2 x3 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 .


flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image