## Integration Worksheet6 solution5

In this page integration worksheet6 solution5 we are going to see solution of some practice question from the worksheet of integration.

Question 12

Integrate the following with respect to x,   sin⁻ ¹ x e^sin⁻ ¹ x/√(1 - x²)

Solution:

Here we are going to use the method partial differentiation to integrate the given question.For that first we are going to apply the substitution method before using partial differentiation.

∫ sin⁻ ¹ x e^sin⁻ ¹ x/√(1 - x²) dx

let t = sin⁻ ¹ x

dt = 1/√(1 - x²) dx

∫ sin⁻ ¹ x e^sin⁻ ¹ x/√(1 - x²) dx = ∫ t e^t dt

u = t             dv = e ^t

du = dt           v = e^t

∫ u dv = uv - ∫ vdu

= t e^t - ∫ e^t dt

= t e^t - e^t + C

= e^t (t - 1) + C

= e^sin⁻ ¹ x (sin⁻ ¹ x - 1) + C

Question 13

Integrate the following with respect to x,    x⁵ e^x²

Solution:

Here we are going to use the method partial differentiation to integrate the given question.For that first we are going to apply the substitution method before using partial differentiation.

∫  x⁵ e^x² dx = ∫(x²)² x e^x² dx

let t = x²

dt = 2 x dx

x dx = dt//2

∫(x²)² x e^x² dx = ∫t² e^t (dt/2)

= (1/2)∫t² e^t dt

u = t²             dv = e ^t

du = 2 t dt           v = e^t

∫ u dv = uv - ∫ vdu

= (1/2) [t² e^t - ∫ e^t 2 t dt]

= (1/2) [t² e^t - 2 ∫ t e^t dt]

u = t             dv = e^t

du = dt           v = e^t

= (1/2) {t² e^t - 2 [t e^t - ∫ e^t dt]}

= (1/2) t² e^t - [t e^t - e^t] + C

= [(1/2) t² e^t] - t e^t + e^t + C

= (1/2)e^t [t²- 2 t + 2]+ C

= (1/2)e^x² [(x²)²- 2 x² + 2]+ C

= (1/2)e^x² [x⁴- 2 x² + 2]+ C integration worksheet6 solution5 integration worksheet6 solution5