Question

20. The value of 2 (r3 + x cosx + tan5x +1) dx is(A) 0(B) 2(C) π(D)I

Answer 1

The given integral is I= ∫ − π 2 π 2 ( x 3 +xcosx+ tan 5 x+1 )dx .

Let, I 1 = ∫ − π 2 π 2 x 3 dx

Since, ( −x ) 3 =− x 3 , so x 3 is an odd function. Thus, I 1 =0.

Let, I 2 = ∫ − π 2 π 2 xcosxdx

Since, ( −x )cos( −x )=−xcosx, so xcosx is an odd function. Thus, I 2 =0.

Let, I 3 = ∫ − π 2 π 2 tan 5 xdx

Since, tan 5 ( −x )=− tan 5 x, tan 5 x is an odd function. Thus, I 3 =0.

Let, I 4 = ∫ − π 2 π 2 1⋅dx

Solve the integral.

I 4 = ∫ − π 2  π 2 1⋅dx = [ x ] − π 2 π 2 = π 2 + π 2 =π

Substitute all the values in given integral.

I= I 1 + I 2 + I 3 + I 4 =0+0+0+π =π

Thus, the correct option is (C).