निश्चित समाकलन का मान ज्ञात कीजिए $- \int_{0}^{\frac\pi 2} \sin 2x \tan^{-1 }(\sin x) dx$
Miscellaneous Exercise-31
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माना $I = \int_{0}^{\frac\pi 2} \sin 2x \tan^{-1 }(\sin x) dx = \int_{0}^{\frac\pi 2} 2 \sin x \cos x \tan^{-1}(\sin x) dx (\therefore \sin 2x = 2 \sin x \cos x)$
$\sin x = t$ रखने पर,
$\Rightarrow$ जब $x = 0 \Rightarrow t = 0$ और जब $x = \frac{\pi}{2} \Rightarrow t = \sin \frac{\pi}{2} = 1$
$\therefore I=\int_{0}^{1} 2 t \cos x \tan^{−1 }t \frac{d t}{\cos x}=2 \int_{0}^{1} t \cdot \tan^{−t }t dt$
$=2\left[\left\{\left(\tan ^{-1} t\right) \frac{t^{2}}{2}\right\}_{0}^{1}-\int_{0}^{1} \frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} d t\right] \ ($खण्डशः समाकलन के प्रयोग से$)$
$=2\left(\frac{\tan ^{-1} 1}{2}-0\right)-\int_{0}^{1} \frac{t^{2}}{1+t^{2}} d t =2\left(\frac{\pi}{8}\right)-\int_{0}^{1} \frac{\left(1+t^{2}\right)-1}{1+t^{2}} d t$
$=\frac{\pi}{4}-\int_{0}^{1}\left(1-\frac{1}{1+t^{2}}\right) d t =\frac{\pi}{4}-\left[t-\tan ^{-1} t\right]_{0}^{1}$
$=\frac{\pi}{4}-\left[1-\tan ^{-1} 1-(0-0)\right] =\frac{\pi}{4}-1+\frac{\pi}{4}=\frac{\pi}{2}-1$
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