यदि $I = \int_0^{\pi /4} \sin^2 x \, dx$ और $J = \int_0^{\pi /4} \cos^2 x \, dx$ है,तो $I = $

  • A
    $\frac{\pi}{4} - J$
  • B
    $2J$
  • C
    $J$
  • D
    $\frac{J}{2}$

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Similar Questions

यदि $f$ एक सतत फलन है,तो निम्नलिखित में से कौन सा सत्य है?

$\int_0^\pi \frac{x \tan x}{\sec x + \tan x} \,dx = $

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

समाकलन $\int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{-101}} d x$ का मान किसके बराबर है?

मान लीजिए $f(x) = 7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$ सभी $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ के लिए है। तो सही व्यंजक है/हैं:
$(A) \int_0^{\pi/4} x f(x) dx = \frac{1}{12}$
$(B) \int_0^{\pi/4} f(x) dx = 0$
$(C) \int_0^{\pi/4} x f(x) dx = \frac{1}{6}$
$(D) \int_0^{\pi/4} f(x) dx = 1$

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