## Integration Worksheet5 solution3

In this page integration worksheet5 solution3 we are going to see solution of some practice question from the worksheet of integration.

Question 8

Integrate the following with respect to x,cos^14 x sin x

Solution:

We are going to solve this problem by using substitution method. For that let us consider "t" as "cos x". The differentiation value of sin x is cos x. If we continue in this method we have to put (dt)^14.

t = cos x

differentiate with respect to x

dt = - sin x dx

sin x dx = - dt

= ∫ cos^14 x sin x  dx

= ∫ t^14  -dt

= t^(14+1)/(14+1) + C

= t^(15)/(15) + C

= (1/15) sin^15 x + C

Question 9

Integrate the following with respect to x,sin^5 x

Solution:

We are going to solve this problem by using substitution method. For that we are going to write sin^5x as sin^4x x sin x

= ∫ sin^5 x  dx

= ∫ sin^4 x sin x  dx

= ∫ (sin ²x)² sin x  dx

= ∫ (1 - cos ²x)² sin x  dx

t = cos x

dt = - sin x dx

sin x dx = - dt

= ∫ (1 - t²)² (-dt)

= ∫ (1 - t⁴ - 2t²) (-dt)

= ∫ (t⁴ + 2t² - 1) dt

= ∫ t⁴ dt + ∫ 2 t² dt - ∫1 dt

= ∫ t⁴ dt + 2∫ t² dt - ∫ dt

= t^(4+1)/(4+1) + 2t^(2+1)/(2+1)- t + C

= t⁵/5 + 2t³/3 - t + C

= (1/5)cos⁵x + (2/3)cos³x - cos x + C

Question 10

Integrate the following with respect to x,cos^7x

Solution:

We are going to solve this problem by using substitution method. For that we are going to write cos⁷x as cos⁶x x cos x

= ∫ cos⁷x dx

= ∫ cos⁶x x cos x   dx

= ∫ (cos² x)³ cos x  dx

= ∫ (1 - sin²x)³ cos x  dx

t = sin x

dt = cos x dx

= ∫ (1 - t²)³ dt

now we are going to use an algebraic identity to expand this

(a - b)³ = a³ - 3 a² b + 3 a b² - b³

= ∫ [(1³ - 3 1² t² + 3 (1) (t²)² - (t²)³] dt

= ∫ [(1 - 3 t² + 3 t⁴ - t⁶] dt

= ∫1 dt - 3∫ t² dt + 3∫ t⁴ dt - ∫ t⁶ dt

= t - 3t^(2+1)/(2+1) + 3t^(4+1)/(4+1)- t^(6+1)/(6+1) + C

= t - 3t³/3 + 3t⁵/5- t⁷/7 + C

= t - t³+ 3t⁵/5- t⁷/7 + C

= sin x - sin³x+ (3/5)sin⁵x - (1/7) sin⁷x + C

integration worksheet5 solution3 integration worksheet5 solution3