Question
If Φ(x)=∫dxsin12x cos72x, then Φ(π4)−Φ(0)=
Answer: Option A
:
A
tanx=t⇒sec2xdx=dt∴f(x)=∫dxsin12xcos12x.cos4x=∫(1+tan2x)sec2x√tanxdx=∫(1+t2)√tdt=∫(t−1/2+t3/2)dt=2t1/2+25t5/2=2√tanx+25(tanx)5/2∴ϕ(π4)−ϕ(0)=2+25=125
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:
A
tanx=t⇒sec2xdx=dt∴f(x)=∫dxsin12xcos12x.cos4x=∫(1+tan2x)sec2x√tanxdx=∫(1+t2)√tdt=∫(t−1/2+t3/2)dt=2t1/2+25t5/2=2√tanx+25(tanx)5/2∴ϕ(π4)−ϕ(0)=2+25=125
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