## Integration Worksheet6 solution3

In this page integration worksheet6 solution3 we are going to see solution of some practice question from the worksheet of integration.

Question 7

Integrate the following with respect to x,    x cos ² x

Solution:

Here we are going to use the method partial differentiation to integrate the given question.

∫ x cos ² x dx

now we are going to use the trigonometric formula for cos ² x

∫ x cos ² x dx   = ∫ x (1 + cos 2x)/2 dx

= (1/2) ∫ x (1 + cos 2x) dx

= (1/2) ∫ (x + x cos 2x) dx

= (1/2) [∫ x dx + ∫ x cos 2x dx]

= (1/2) [(x²/2)+ ∫ x cos 2x dx]

Let I = ∫ x cos 2x dx

u = x             dv = cos 2x

du = dx           v = sin 2x/2

= x (sin 2 x)/2 - ∫(sin 2x/2) dx

= x (sin 2 x)/2 - (1/2) ∫ sin 2x dx

= x (sin 2 x)/2 + (1/2) (cos 2x/2) + C

= (x/2) (sin 2 x) + (1/4) (cos 2x) + C

= (1/2) [(x²/2)+ (x/2) (sin 2 x) + (1/4) (cos 2x) + C

Question 8

Integrate the following with respect to x,    x cos 5 x cos 2 x

Solution:

Here we are going to use the method partial differentiation to integrate the given question.

∫ x cos 5 x cos 2 x dx

now we are going to apply the trigonometric formula 2 cos A cos B

∫ x cos 5 x cos 2 x dx = (2/2)∫ x cos 5 x cos 2 x dx

= (1/2)∫ x 2 cos 5 x cos 2 x dx

= (1/2)∫ x [cos (5 x + 2x) + cos (5 x - 2x)] dx

= (1/2)∫ x [cos 7x + cos 3x] dx

= (1/2){∫ [x cos 7x] dx + ∫ [x cos 3x] dx}

now we are going to integrate x cos 7x and x cos 3x separately

∫ [x cos 7x] dx

u = x          dv = cos 7 x

du = dx        v = sin 7x/7

= x (sin 7x/7) - ∫(sin 7x/7) dx

= (x/7) (sin 7x) - (1/7)∫sin 7x dx

= (x/7) (sin 7x) + (1/7) (cos 7x/7) + C

= (x/7)(sin 7x) + (1/49) (cos 7x) + C

∫ [x cos 3x] dx

u = x          dv = cos 3 x

du = dx        v = sin 3x/3

= x (sin 3x/3) - ∫(sin 3x/3) dx

= (x/3) (sin 3x) - (1/3)∫sin 3x dx

= (x/3) (sin 3x) + (1/3) (cos 3x/3) + C

= (x/3)(sin 3x) + (1/9) (cos 3x) + C

= (1/2){(x/7)(sin 7x) + (1/49) (cos 7x)+(x/3)(sin 3x)+(1/9)(cos 3x)}+ C

integration worksheet6 solution3 integration worksheet6 solution3