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JEE Advanced 2021 Paper 1 · Mathematics · Indefinite and Definite Integrals · hard

Definite integrals has (have) the stated value?

One or more options correct

Which of the following definite integrals has (have) the stated value?
  1. A. ∫0π/2sin⁡xsin⁡x+cos⁡x dx=π4\displaystyle\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx=\frac{\pi}{4}
  2. B. ∫−11x21+ex dx=13\displaystyle\int_{-1}^{1}\frac{x^2}{1+e^{x}}\,dx=\frac13
  3. C. ∫0πxsin⁡x1+cos⁡2x dx=π22\displaystyle\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}\,dx=\frac{\pi^2}{2}
  4. D. ∫01ln⁡(1+x)1+x2 dx=πln⁡24\displaystyle\int_0^{1}\frac{\ln(1+x)}{1+x^2}\,dx=\frac{\pi\ln2}{4}
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Answer: A, B

(A) Use x→π2−xx\to\frac\pi2-x: the integral II equals the same with cos⁡\cos in the numerator, so 2I=∫0π/21 dx2I=\int_0^{\pi/2}1\,dx and I=π4I=\frac\pi4. True. (B) Use x→−xx\to-x and add: 2I=∫−11x2(11+ex+11+e−x)dx=∫−11x2dx=232I=\int_{-1}^{1}x^2\left(\frac1{1+e^x}+\frac1{1+e^{-x}}\right)dx=\int_{-1}^{1}x^2dx=\frac23, so I=13I=\frac13. True. (C) With x→π−xx\to\pi-x, I=π2∫0πsin⁡x1+cos⁡2xdx=π2[−arctan⁡(cos⁡x)]0π=π2⋅π2=π24I=\frac\pi2\int_0^\pi\frac{\sin x}{1+\cos^2x}dx=\frac\pi2\left[-\arctan(\cos x)\right]_0^\pi=\frac\pi2\cdot\frac\pi2=\frac{\pi^2}{4}, not π22\frac{\pi^2}{2}. False. (D) With x=tan⁡θx=\tan\theta and θ→π4−θ\theta\to\frac\pi4-\theta the value is πln⁡28\frac{\pi\ln2}{8}, half of the stated value. False.
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