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