Intigrate : \(sinx/(cos^2x+1)(cos^2x+4) \)
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Kj Nikhil 7 years, 2 months ago
- 1 answers
Related Questions
Posted by Pinky Kumari 2 weeks ago
- 0 answers
Posted by Arpit Sahu 3 weeks, 6 days ago
- 0 answers
Posted by Dhruv Kumar 1 month, 2 weeks ago
- 1 answers
Posted by Suprith A 4 weeks ago
- 0 answers
Posted by Anupam Bind Anupam Bind 3 weeks, 3 days ago
- 0 answers
Posted by Durgesh Kuntal Nishantkuntal 1 week 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
Naveen Sharma 7 years, 2 months ago
Ans. \(\int{ {sin \space x }\over (cos^2 \space x+1 )(cos^2\space x+4)} dx \)
Put cos x = t
then -sin x dx = dt or sin x dx = -dt
\(\int{ {-dt }\over (t^2+1 )(t^2 +4)} \)
By Partial Fraction, We can Write
\({-1\over (t^2+1)(t^2+4)} = {-1\over 3 (t^2+1)} + {1\over 3 (t^2+4)} \)
So Now
\(\int {-1dt\over (t^2+1)(t^2+4)} ={\int {-1dt\over 3 (t^2+1)}} + {\int {1dt\over 3 (t^2+4)} }\)
=> \({-1\over 3}{\int {dt\over (t^2+1)}} + {{1\over 3}\int {dt\over (t^2+4)} }\)
=> \({-1\over 3}{tan^{-1}({t\over 1})} +{1\over 3}.{1\over 2}{tan^{-1}({t\over 2})} + C\) [\(Applying \int {1dx\over (x^2 +a^2)} = { 1\over a}{tan^{-1} ({x\over a})}\)]
=> \({-1\over 3}{tan^{-1}({cos x\over 1})} +{1\over 6}{tan^{-1}({cosx\over 2})} + C\)
0Thank You