Class 12 Integrals · Very Long Answer5 Marksdefinite_integration
Class 12 Maths · Integrals · Very Long Answer #119
Evaluate \(\int_0^{\pi/2}\sin2x\,\arctan(\sin x)dx\).
✅ Step-by-Step Solution
Final Answer: π/2−1
- Set u=sin x, du=cos x dx, and sin2x dx=2u du. The limits become 0,1.
- By parts, ∫₀¹2u arctan u du=[u²arctan u]₀¹−∫₀¹u²/(1+u²)du.
- The boundary term is π/4. Rewrite u²/(1+u²)=1−1/(1+u²).
- The remaining integral is [u−arctan u]₀¹=1−π/4; subtract it from the boundary term.
- Therefore the integral equals π/2−1.
📋 CBSE Stepwise Marking Scheme
- [1 mark] Set u=sin x, du=cos x dx, and sin2x dx=2u du. The limits become 0,1.
- [1 mark] By parts, ∫₀¹2u arctan u du=[u²arctan u]₀¹−∫₀¹u²/(1+u²)du.
- [1 mark] The boundary term is π/4. Rewrite u²/(1+u²)=1−1/(1+u²).
- [1 mark] The remaining integral is [u−arctan u]₀¹=1−π/4; subtract it from the boundary term.
- [1 mark] Therefore the integral equals π/2−1.
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