MCQ
$\int\limits_{\frac{\pi }{4}}^{\frac{{3\pi }}{4}} {\frac{x}{{1 + \sin x}}} dx$ મેળવો.
- A$\frac{\pi }{2}\left( {\sqrt 2 + 1} \right)$
- B$\pi \left( {\sqrt 2 - 1} \right)$
- C$2\pi \left( {\sqrt 2 - 1} \right)$
- D$\pi \sqrt 2 $
also let $K=\frac{x}{1+\sin x}$
Multiplying numerator and denominator by $(1-\sin x),$ we get
$K=\frac{x(1-\sin x)}{1-(\sin x)^{2}}$
$=\frac{x(1-\sin x)}{(\cos x)^{2}}$
$=x(1-\sin x) \sec ^{2} x$
$=x \sec ^{2} x-x \sin x \sec ^{2} x$
$=x \sec ^{2} x-x \tan$ $x \sec x$
Now, $I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} x \sec ^{2} x d x-\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} x \sec x \tan x d x$
$=\left[x \tan x-\int \frac{d x}{d x} \tan x d x\right]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}-\left[x \sec x-\int \frac{d x}{d x} \sec x d x\right]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}$
$=[x \tan x-\ln |\sec x|]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}$
$-[x \sec x-\ln |\sec x+\tan x|]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}+c$
$\Rightarrow I=\left\{\left[\frac{3 \pi}{4} \tan \frac{3 \pi}{4}-\ln \left|\frac{3 \pi}{4}\right|\right.\right.$
$\left.-\left[\frac{3 \pi}{4} \sec \frac{3 \pi}{4}-\ln | \sec \frac{3 \pi}{4}+\tan \frac{3 \pi}{4}\right]\right\}$
$-\left\{\left[\frac{\pi}{4} \tan \frac{\pi}{4}-\ln \left|\frac{\pi}{4}\right|\right.\right.$
$\left.-\left[\frac{\pi}{4} \sec \frac{\pi}{4}-\ln \left|\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right|\right]\right\}$
$=\frac{\pi}{2}(\sqrt{2}+1)$
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