Question

Integral from 0 to 1/2 x sin^-1 (x) dx

Answer 1

@sirm3d

Answer 2

Integration by parts, and some subs, switching boundaries.trigs..all in 1 :D

Answer 3

and brb, washroom

Answer 4

\[\int\limits_{0}^{1}x \sin^{-1} x dx\]
the choice is \[u=sin^{-1}x, \quad dv = x dx\]\[du=\frac{1}{\sqrt {1-x^2}},\;v=\frac{1}{2}x^2\]

Answer 5

type error. upper limit of integration=1/2

Answer 6

My work, but dont think u can understand it :P

Answer 7

I was stucked in x^2 / 2 sqrt (1-x^2) for at least 15mins.. hard to decide to use subs, parts, trig, or whatever..

Answer 8

seeing \[\sqrt{a^2-x^2}\quad \sqrt{a^2+x^2}\quad \sqrt{x^2-a^2}\] calls for trigonometric substitution EVEN IF a simple substitution will work, because a trigonometric substitution always work.

Answer 9

There is one w/ sqrt (1 - w^2) < this is much harder to use trig subs

Answer 10

The problem is, I only have 6 mins per ques on the exam >.>

Answer 11

for the limits, \(x=\sin \theta\)
\[x=0: \theta = 0\]\[x=1/2: \theta = \pi/6\]

Answer 12

@lovekblue u have dne the right substitution
after substitution u will get
integral of 1/2 sin^2θ dθ