Primitive of sin^-1x
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Akshat Mehta 6 years, 10 months ago
- 1 answers
Related Questions
Posted by Suprith A 1 month ago
- 0 answers
Posted by Digpal Singh Rathore 7 hours ago
- 0 answers
Posted by Pinky Kumari 3 weeks, 2 days ago
- 0 answers
Posted by Anupam Bind Anupam Bind 1 month ago
- 0 answers
Posted by Abinav Singh 5 days, 19 hours ago
- 0 answers
Posted by Durgesh Kuntal Nishantkuntal 2 weeks, 2 days ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Payal Singh 6 years, 10 months ago
{tex}\int sin^{-1}x \ dx{/tex}
= {tex}\int sin^{-1}x.1 \ dx{/tex}
Now Integrate by parts, Using ILATE first Function is sin-1 x and 2nd Function is 1.
= {tex}sin^{-1}x\int 1\ dx- \int {1\over \sqrt {1-x^2}}xdx{/tex}
{tex}= sin^{-1}x.x- I_1{/tex} .....(1)
Where {tex}I_1 = \int {1\over \sqrt {1-x^2}}xdx {/tex}
Put {tex}{1-x^2} = t{/tex}
-2xdx = dt
{tex}=> I_1 = -{1\over 2}\int {1\over \sqrt t}dt{/tex}
{tex}=> I_1 = - {\sqrt t}{/tex}
{tex}=> I_1 = -{\sqrt {1-x^2}}{/tex}
Put in (1), we get
= {tex}sin^{-1}x.x+\sqrt{1-x^2} + c{/tex}
0Thank You