# Evaluate the following : ∫01t51-t2⋅dt - Mathematics and Statistics

Sum

Evaluate the following : int_0^1 t^5 sqrt(1 - t^2)*dt

#### Solution

Let I = int_0^1 t^5 sqrt(1 - t^2)*dt

Put t = sin θ
∴ dt = cos θ dθ

When t = 1, θ = sin–11 = pi/(2)

When t = 0, θ = sin–10 = 0

∴ I = int_0^(pi/2) sin^5 theta sqrt(1 - sin^2 theta)cos theta*d theta

I = int_0^(pi/2) sin^5 theta*cos theta* cos theta*d theta

= int_0^(pi/2) sin^5 theta(1 - sin^2 theta)*d theta

= int_0^(pi/2) (sin^5 theta - sin^7 theta)*d theta

= int_0^(pi/2) sin^5 theta*d theta - int_0^(pi/2) sin^7 thetad theta.
Using Reduction formula, we get

I = 4/5*2/3 - 6/7*4/5*2/3

= (8)/(15)[1 - 6/7]

= (8)/(15) xx (1)/(7)

= (8)/(105).

Concept: Fundamental Theorem of Integral Calculus
Is there an error in this question or solution?
Chapter 4: Definite Integration - Miscellaneous Exercise 4 [Page 176]

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 4 Definite Integration
Miscellaneous Exercise 4 | Q 2.05 | Page 176
Share