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Definite Integration question

2021 · 26 Aug · Shift 2 · Q36
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  5. /2021 · 26 Aug · Shift 2 · Q36

Definite Integration question

2021 · 26 Aug · Shift 2 · Q36

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The value of ∫−π2π2(1+sin⁡2x1+πsin⁡x) dx\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{1 + {{\sin }^2}x} \over {1 + {\pi ^{\sin x}}}}} \right)} \,dx−2π​∫2π​​(1+πsinx1+sin2x​)dx is
  1. A
    π2{\pi \over 2}2π​
  2. B
    5π4{{5\pi } \over 4}45π​
  3. C
    3π4{{3\pi } \over 4}43π​
  4. D
    3π2{{3\pi } \over 2}23π​
View written solutionFree

Correct answer: C

  1. Let I=∫−π/2π/21+sin⁡2x1+πsin⁡x dx.I=\int_{-\pi/2}^{\pi/2} \frac{1+\sin^2 x}{1+\pi^{\sin x}}\,dx.I=∫−π/2π/2​1+πsinx1+sin2x​dx.

We use the standard property: If I=∫abf(x) dx,I=\int_a^b f(x)\,dx,I=∫ab​f(x)dx, then after substituting x↦−xx\mapsto -xx↦−x, I=∫−b−af(−x) dx.I=\int_{-b}^{-a} f(-x)\,dx.I=∫−b−a​f(−x)dx. Since the interval is symmetric, we can combine f(x)f(x)f(x) and f(−x)f(-x)f(−x).

  1. Define f(x)=1+sin⁡2x1+πsin⁡x.f(x)=\frac{1+\sin^2 x}{1+\pi^{\sin x}}.f(x)=1+πsinx1+sin2x​. Then f(-x)=\frac{1+\sin^2(-x)}{1+\pi^{\sin(-x)}}= rac{1+\sin^2 x}{1+\pi^{-\sin x}}.

Now simplify f(−x)f(-x)f(−x): 1+π−sin⁡x=1+1πsin⁡x=1+πsin⁡xπsin⁡x.1+\pi^{-\sin x}=1+\frac{1}{\pi^{\sin x}}=\frac{1+\pi^{\sin x}}{\pi^{\sin x}}.1+π−sinx=1+πsinx1​=πsinx1+πsinx​. So f(−x)=1+sin⁡2x(1+πsin⁡x)/πsin⁡x=(1+sin⁡2x)πsin⁡x1+πsin⁡x.f(-x)=\frac{1+\sin^2 x}{(1+\pi^{\sin x})/\pi^{\sin x}}=\frac{(1+\sin^2 x)\pi^{\sin x}}{1+\pi^{\sin x}}.f(−x)=(1+πsinx)/πsinx1+sin2x​=1+πsinx(1+sin2x)πsinx​.

  1. Add f(x)f(x)f(x) and f(−x)f(-x)f(−x): f(x)+f(−x)=1+sin⁡2x1+πsin⁡x+(1+sin⁡2x)πsin⁡x1+πsin⁡x.f(x)+f(-x)=\frac{1+\sin^2 x}{1+\pi^{\sin x}}+\frac{(1+\sin^2 x)\pi^{\sin x}}{1+\pi^{\sin x}}.f(x)+f(−x)=1+πsinx1+sin2x​+1+πsinx(1+sin2x)πsinx​. Factor out (1+sin⁡2x)(1+\sin^2 x)(1+sin2x): f(x)+f(−x)=(1+sin⁡2x)1+πsin⁡x1+πsin⁡x=1+sin⁡2x.f(x)+f(-x)=(1+\sin^2 x)\frac{1+\pi^{\sin x}}{1+\pi^{\sin x}}=1+\sin^2 x.f(x)+f(−x)=(1+sin2x)1+πsinx1+πsinx​=1+sin2x.

  2. Hence, using symmetry,

\int_{-\pi/2}^{\pi/2}(1+\sin^2 x)\,dx.$$ Therefore, $$I=\frac12\int_{-\pi/2}^{\pi/2}(1+\sin^2 x)\,dx.$$ 5. Now evaluate the two parts: $$\int_{-\pi/2}^{\pi/2}1\,dx=\pi.$$ Also, $$\int_{-\pi/2}^{\pi/2}\sin^2 x\,dx=2\int_0^{\pi/2}\sin^2 x\,dx=2\cdot \frac{\pi}{4}=\frac{\pi}{2}.$$ So, $$\int_{-\pi/2}^{\pi/2}(1+\sin^2 x)\,dx=\pi+\frac{\pi}{2}=\frac{3\pi}{2}.$$ Thus, $$I=\frac12\cdot \frac{3\pi}{2}=\frac{3\pi}{4}.$$ 6. Therefore the correct option is $$\boxed{\frac{3\pi}{4}}$$ which is option $\boxed{\text{C}}$.
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