MathematicsIntegration by SubstitutionJEE Advanced 1989Moderate
Visualized Solution (Hindi)
Rewrite in Terms of sinx and cosx
- Evaluate I=∫(tanx+cotx)dx
- Rewrite using tanx=cosxsinx and cotx=sinxcosx:
- I=∫(cosxsinx+sinxcosx)dx
Combine Using Common Denominator
- Take the LCM of the denominators:
- I=∫sinxcosx(sinx⋅sinx)+(cosx⋅cosx)dx
- I=∫sinxcosxsinx+cosxdx
The 2 Manipulation
- Multiply and divide by 2 to create sin2x in the denominator:
- I=2∫2sinxcosxsinx+cosxdx
- Using 2sinxcosx=sin2x:
- I=2∫sin2xsinx+cosxdx
Choosing the Right Substitution
- Look for a function whose derivative is (sinx+cosx):
- Let t=sinx−cosx
- Differentiating both sides with respect to x:
- dxdt=cosx−(−sinx)=cosx+sinx
- So, dt=(sinx+cosx)dx
Relating sin2x to t
- Square the substitution to find sin2x:
- t2=(sinx−cosx)2
- t2=sin2x+cos2x−2sinxcosx
- Substitute sin2x+cos2x=1 and 2sinxcosx=sin2x:
- t2=1−sin2x⟹sin2x=1−t2
Substitute and Integrate
- Substitute dt and sin2x into the integral:
- I=2∫1−t2dt
- Apply the standard integral formula ∫1−x21dx=sin−1x+C:
- I=2sin−1t+C
Final Back-substitution
- Substitute t=sinx−cosx back into the result:
- I=2sin−1(sinx−cosx)+C
- Note: This is equivalent to the form 2tan−1(2tanx−cotx)+C as requested in some contexts.
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