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Question Number 87585 by Power last updated on 05/Apr/20

	Answered by redmiiuser last updated on 05/Apr/20
	$arcsin x+arccos x=(π/2) arcsin x.arccos x=(π/2)arcsin x−(arcsin x)^2 arcsin x=t dx=cos t.dt x=sin t ∫_0 ^(π/2) ((π/2)t−t^2 )cos t.dt =(π/2)∫_0 ^(π/2) tcos t.dt−∫_0 ^(π/2) t^2 cos t.dt =(π/2)[t∫_0 ^(π/2) cos t.dt−∫_(0 ) ^(π/2) {∫_0 ^(π/2) cos t.dt}.dt]−[∫_0 ^(π/2) cos t.dt−∫_0 ^(π/2) 2tsin t.dt] =(π/2)[tsin t+cos t]_0 ^(π/2) −[t^2 sin t+2tcos t−2sin t]_0 ^(π/2) =(π/2)[(π/2)−1]−[(π^2 /4)−2] =2−(π/2)$
	$a r c \sin x + \arccos x = \frac{π}{2}$ $\arcsin x . \arccos x = \frac{π}{2} \arcsin x - {(\arcsin x)}^{2}$ $\arcsin x = t$ $d x = \cos t . d t$ $x = \sin t$ $\int_{0}^{\frac{π}{2}} (\frac{π}{2} t - t^{2}) \cos t . d t$ $= \frac{π}{2} \int_{0}^{\frac{π}{2}} t \cos t . d t - \int_{0}^{\frac{π}{2}} t^{2} \cos t . d t$ $= \frac{π}{2} [t \int_{0}^{\frac{π}{2}} \cos t . d t - \int_{0}^{\frac{π}{2}} {\int_{0}^{\frac{π}{2}} \cos t . d t} . d t] - [\int_{0}^{\frac{π}{2}} \cos t . d t - \int_{0}^{\frac{π}{2}} 2 t \sin t . d t]$ $= \frac{π}{2} {[t \sin t + \cos t]}_{0}^{\frac{π}{2}} - {[t^{2} \sin t + 2 t \cos t - 2 \sin t]}_{0}^{\frac{π}{2}}$ $= \frac{π}{2} [\frac{π}{2} - 1] - [\frac{π^{2}}{4} - 2]$ $= 2 - \frac{π}{2}$
	Commented by Ar Brandon last updated on 05/Apr/20

	$n i c e m e t h o d!$
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