ધારો કે $[\cdot]$ એ મહત્તમ પૂર્ણાંક વિધેય દર્શાવે છે. તો $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{12(3+[x])}{3+[\sin x]+[\cos x]} \right) dx$ ની કિંમત શોધો:

  • A
    $15\pi+4$
  • B
    $11\pi+2$
  • C
    $13\pi+1$
  • D
    $12\pi+5$

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Similar Questions

$n \in N$ માટે,જો $I_n = \int \frac{\sin nx}{\sin x} dx = \frac{2}{n-1} \sin(n-1)x + I_{n-2}$ અને $\int_0^\pi \frac{\sin nx}{\sin x} dx = \frac{k\pi}{2}$ હોય,તો $k =$

ધારો કે $I = \int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} dx$. તો

$\int_{-1}^{3} \left( \tan^{-1} \left( \frac{x}{x^2+1} \right) + \tan^{-1} \left( \frac{x^2+1}{x} \right) \right) dx$ નું મૂલ્ય શોધો.

$\int_{-2}^{\pi} \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}} \,dx$ ની કિંમત શોધો. (જ્યાં $[\cdot]$ એ મહત્તમ પૂર્ણાંક વિધેય છે.)

વિધાન $(A)$: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} = \frac{\pi}{12}$
કારણ $(R)$: $\int_{a}^{b} \frac{f(x) dx}{f(x)+f(a+b-x)} = \frac{b-a}{2}$

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