Question

Arrange the following in ascending order based on their values:
A. \(\displaystyle \int_0^{\frac{\pi}{2}}\frac{1}{1+\sin x}\,dx\),
B. \(\displaystyle \int_1^2 x^2\,dx\),
C. \(\displaystyle \int_0^{\frac{\pi}{2}}\sin x\,dx\),
D. \(\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^3x\,dx\).

• A. C, D, B, A
• B. D, C, A, B
• C. D, C, B, A
• D. C, D, A, B

Hint:

For symmetric limits \([-a,a]\), the integral of an odd function is always \(0\).

Answer 1

The Correct Option is B

Concept:
To arrange definite integrals by value, first evaluate each integral separately.

Step 1: Evaluate A. \[ A=\int_0^{\frac{\pi}{2}}\frac{1}{1+\sin x}\,dx \] Multiply numerator and denominator by \(1-\sin x\): \[ A=\int_0^{\frac{\pi}{2}}\frac{1-\sin x}{1-\sin^2x}\,dx \] \[ A=\int_0^{\frac{\pi}{2}}\frac{1-\sin x}{\cos^2x}\,dx \] \[ A=\int_0^{\frac{\pi}{2}}(\sec^2x-\sec x\tan x)\,dx \] \[ A=[\tan x-\sec x]_0^{\frac{\pi}{2}} \] The limiting value is: \[ A=1 \]

Step 2: Evaluate B. \[ B=\int_1^2 x^2\,dx \] \[ B=\left[\frac{x^3}{3}\right]_1^2 \] \[ B=\frac{8}{3}-\frac{1}{3} \] \[ B=\frac{7}{3} \]

Step 3: Evaluate C. \[ C=\int_0^{\frac{\pi}{2}}\sin x\,dx \] \[ C=[-\cos x]_0^{\frac{\pi}{2}} \] \[ C=0-(-1) \] \[ C=1 \]

Step 4: Evaluate D. \[ D=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^3x\,dx \] Since \(\sin^3x\) is an odd function and the limits are symmetric, \[ D=0 \]

Step 5: Arrange in ascending order. \[ D=0,\quad C=1,\quad A=1,\quad B=\frac{7}{3} \] Therefore, \[ D,C,A,B \] \[ \therefore \text{Correct Answer is (B)} \]