∫3+tanx2−tanxdx=∫3cosx+sinx2cosx−sinxdx2cosx−sinx=A(3cosx+sinx)+B(cosx−3sinx)3A+B=2A−3B=−1 ⇒A=21,B=21∴∫3cosx+sinx2cosx−sinxdx=2x+21ln∣3cosx+sinx∣+C=21(x+ln∣3cosx+sinx∣)+C=21(αx+ln∣βsinx+γcosx∣)+Cα=1,β=1,γ=3∴α+βγ=1+13=4
Let ∫3+tanx2−tanx dx=21(αx+loge∣βsinx+γcosx∣)+C, where C is the constant of integration. Then α+βγ is equal to :
Held on 9 Apr 2024 · Verified 6 Jul 2026.
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