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integrating from sin x to 1 (t^2f(t)dt=i-sinx for all x E R.THEN F(1/SQ.RT(3)) IS

ZZZ ZZZ , 13 Years ago
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Jitender Singh

Last Activity: 10 Years ago

Ans:
\int_{sinx}^{1}t^{2}f(t)dt = i - sinx
Differentiating both sides. Integral will be differentiated by Leibniz integral rule,
(1)^{2}f(1)\frac{\partial (1)}{\partial x}-(sinx)^{2}f(sinx)\frac{\partial (sinx)}{\partial x}+\int_{1}^{sinx}\frac{\partial (t^{2}f(t))}{\partial x}dx = -cosx
-sin^{2}x.(f(sinx)).cosx + 0 = -cosx
(f(sinx)) = \frac{1}{sin^{2}x}
f(\frac{1}{\sqrt{3}}) = \frac{1}{(\frac{1}{\sqrt{3}})^{2}}
f(\frac{1}{\sqrt{3}}) = 3
Thanks & Regards
Jitender Singh
IIT Delhi
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