Q.
If the area bounded by f(x)=(tan)3x+tanx from x=0 to x=4π is k square units, then the maximum value of g(x)=ksinx is (∀x∈[0,4π])
3245
180
NTA AbhyasNTA Abhyas 2020Application of Integrals
Report Error
Solution:
The required area is A=∫04π((tan)3x+tanx)dx =∫04πtanx((tan)2x+1)dx =∫04πtanx⋅sec2xdx
Let tanx=t ⇒sec2xdx=dt ⇒I=∫01tdt=[2t2]01 =21 ∴g(x)=21sinx≤2⋅21