Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Definite Integration

Question:

$I_1=\int\limits_0^{π/2}\frac{\sin x-\cos x}{1+\sin x.\cos x}dx,I_2=\int\limits_0^{2π}\cos^6x\,dx,I_3=\int\limits_{-π/2}^{π/2}\sin^3x\,dx,I_4=\int\limits_{0}^{1}ln(\frac{1}{x}-1)dx$. Then:

Options:

$I_2=I_3=I_4=0, I_1 ≠0$

$I_1=I_2=I_3=0, I_4 ≠0$

$I_1=I_3=I_4=0, I_2 ≠0$

None of these

Correct Answer:

$I_1=I_3=I_4=0, I_2 ≠0$

Explanation:

$I_1=\int\limits_0^{π/2}\frac{\sin x-\cos x}{1+\sin x.\cos x}dx$ (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)

$I_4=\int\limits_{0}^{1}ln(\frac{x}{1-x})dx⇒2I_4=\int\limits_{0}^{1}ln\,1\,dx=0$