sinx1 cos x12.

Open in App

Solution

The given function is sin 2 x 1+cosx .

∫ sin 2 x 1+cosx dx (1)

Also, sinx=2sin x 2 cos x 2 and 1+cosx=2 cos 2 x 2 (2)

From (1) and (2),

sin 2 x 1+cosx = ( 2sin x 2 cos x 2 ) 2 2 cos 2 x 2 =2 sin 2 x 2 =1−cosx (3)

Put (3) in (1), we get,

∫ sin 2 x 1+cosx dx = ∫ ( 1−cosx )dx = ∫ 1dx− ∫ cosxdx =x−sinx+c

Thus, the integral of the function sin 2 x 1+cosx is x−sinx+c.

Suggest Corrections

Similar questions

√12+√12+√12+√12+√12•••••••••••••• infinite

√12+√12-√12+√12-√12 •••••••••••••• infinite

c o s \frac{π}{12} + c o s \frac{3 π}{12} + c o s \frac{9 π}{12} + c o s \frac{11 π}{12} =

\frac{(a + 1)^{2} + (a - 1)^{2}}{(a + 1)^{2} - (a - 1)^{2}}

f (x) = 4 x^{8}, then

f' (\frac{1}{2}) = f' (- \frac{1}{2})

f (\frac{1}{2}) = - f' (- \frac{1}{2})

f (- \frac{1}{2}) = f (- \frac{1}{2})

f (\frac{1}{2}) = f' (- \frac{1}{2})

Join BYJU'S Learning Program