$I_1=\int\limits_0^{π/2}\frac{\sin x-\cos x}{1+\sin x.\cos x}dx,I_2=\int\limits_0^{2π}\cos^6x\,dx,I_3=\int\limits_{-π/2}^{π/2}\sin^3x\,dx,I_4=\int\limits_{0}^{1}ln(\frac{1}{x}-1)dx$. Then:
Answer & explanation
Correct answer: option 3
$I_1=\int\limits_0^{π/2}\frac{\sin x-\cos x}{1+\sin x.\cos x}dx$ (apply (a – x) property)
$⇒I_1=\int\limits_0^{π/2}\frac{\sin x-\cos x}{1+\sin x.\cos x}dx⇒2I_1=0⇒I_1=0$
$I_2=\int\limits_0^{2π}\cos^6x≠0$ (From the graph)
$I_3=\int\limits_{-π/2}^{π/2}\sin^3x\,dx=0$ [as sin3 x is an odd function]
$I_4=\int\limits_{0}^{1}ln(\frac{1}{x}-1)dx$ (apply (a – x) property)
$I_4=\int\limits_{0}^{1}ln(\frac{x}{1-x})dx⇒2I_4=\int\limits_{0}^{1}ln\,1\,dx=0$
