To evaluate the given integral

Asked Aug 16 '17 at 09:45

Active Apr 09 '18 at 21:43

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If $0<a<1$ and $ A = \int_{0}^a \cos^{-1}x~\mathrm dx $, then $$\int_{-a}^a(\cos^{-1}x -\sin^{-1}\sqrt {1-x^2})~\mathrm dx =\pi a -\lambda A $$ Then find the value of $\lambda$

My doubt :

$\sin^{-1}\sqrt {1-x^2}=\cos^{-1}x $ if we write this then the value of requires integral will zero.

$\pi a-\lambda A =0 $.

Can anyone explain how to find the value of $\lambda$?

edited Apr 09 '18 at 21:43

Yuriy S

31,474

asked Aug 16 '17 at 09:45

Girish Kumar Chandora

1

your condition on $a$ is wrong. It's not enough that a>0. – FDP Aug 16 '17 at 13:10
@FDP can you expain it further – Girish Kumar Chandora Aug 16 '17 at 13:11
where is defined $\cos^{-1}x$? – FDP Aug 16 '17 at 13:12
1

@FDP its understood the value of $a \in [-1,1]$ – Girish Kumar Chandora Aug 16 '17 at 13:14
By the way, what is the parity of $\cos^{-1}x$? – FDP Aug 16 '17 at 13:17
@FDP sorry i didnt get you – Girish Kumar Chandora Aug 16 '17 at 13:18
If f is even what can you say about $\int_{-a}^{a} f(x)dx$ if you know the value of $\int_{0}^{a} f(x)dx$? ($f$ is defined on $[-a,a]$) – FDP Aug 16 '17 at 13:40
@FDP it will be two times of that value – Girish Kumar Chandora Aug 16 '17 at 13:42

To evaluate the given integral

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