cos2x42, COS2dx is equal to(sinx +cosx)(A)+C(B)log lsin x +cosxl+Csin x+ cosx(C) log lsin x -cosx+C(sin x +cosx)

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Solution

The given integral is, I= ∫ cos2x ( cosx+sinx ) 2 dx .

Solve the integral,

I= ∫ cos 2 x− sin 2 x ( cosx+sinx ) 2 dx = ∫ ( cosx+sinx )( cosx−sinx ) ( cosx+sinx )( cosx+sinx ) dx = ∫ ( cosx−sinx ) ( cosx+sinx ) dx

Let, ( cosx+sinx )=t.

Differentiate both sides with respect to t,

( cosx−sinx ) dx dt =1 ( cosx−sinx )dx=dt

Now, integral becomes,

I= ∫ dt t =log| t |+C (1)

Substitute value of t in equation (1),

I=log| ( cosx+sinx ) |+C

Thus, option (B) is correct.

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