A) \[\frac{9}{8}\]
B) \[\frac{-9}{8}\]
C) \[-2\]
D) 2
Correct Answer: C
Solution :
[c] Let \[\sin x=t\] \[\Rightarrow \cos xdx=dt\] \[\therefore I=\int{\frac{dt}{{{t}^{3}}{{(1+{{t}^{6}})}^{2/3}}}=\int{\frac{dt}{{{t}^{7}}{{\left( 1+\frac{1}{{{t}^{6}}} \right)}^{2/3}}}}}\] Put \[\frac{1}{{{t}^{6}}}+1=k\] \[\Rightarrow \frac{-6}{{{t}^{7}}}dt=dk\] \[\therefore I=\frac{-1}{6}\int{\frac{dk}{{{k}^{2/3}}}=\frac{-1}{6}\frac{{{k}^{-\frac{2}{3}+1}}}{-\frac{2}{3}1}+c}\] \[=\frac{-1}{2}{{k}^{1/3}}+c\] \[=\frac{-1}{2}{{(1+si{{n}^{6}}x)}^{1/3}}\cdot \text{cose}{{\text{c}}^{\text{2}}}\text{x}+c\] \[\therefore \lambda =3\] and \[f\left( \frac{\pi }{3} \right)=\left( \frac{4}{3} \right)\left( \frac{-1}{2} \right)\Rightarrow \lambda f(\pi /3)=-2\]
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